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1 2 16 ✔ 4 3 log 3 log log 16 12 4 2x 2x 2x 1 1 1 2 2 2 2x 104 Given equation Subtract 4 Divide by 3 Exponential form (or raise 10 to each side) x 5000 ✎ Practice what you’ve learned: Do Exercise 43. Divide by 2 ▲ SE CTI ON 5. 4 \| Exponential and Logarithmic Equations 405 E X AM P L E 8 \| Solving a Logarithmic Equation Algebraically and Graphically Solve the equation log x 2 log x 1 1 2 ▼ SO LUTI O N 1: Algebraic We ﬁrst combine the logarithmic terms, using the Laws of Logarithms. 1 2 1 algebraically and graphically. Check Your Answers x 4: log 10 Law 1 Exponential form (or raise 10 to each side) log 4 2 log 4 1 1 log 2 2 1 log 1 2 2 5 1 undeﬁned 2 x 2 x 2 10 x 2 x 12 or x 3 2 Expand left side Subtract 10 Factor x 3: log 2 3 2 3 1 log 2 1 log 5 log 2 log log 10 1 1 3 0 _3 FIGURE 2 5 # 2 1 2 We check these potential solutions in the original equation and ﬁnd that x 4 is not a solution (because logarithms of negative numbers are undeﬁned), but x 3 is a solution. (See Check Your Answers.) ✔ ▼ SO LUTI O N 2: Graphical We ﬁrst move all terms to one side of the equation: x 1 log x 2 log 1 2 1 1 0 2 Then we graph 6 y log x 2 log x 1 1 2 2 1 1 as in Figure 2. The solutions are the x-intercepts of the graph. Thus, the only solution is x 3. ✎ Practice what you’ve learned: Do Exercise 49. ▲ In Example 9 it’s not possible to isolate x algebraically, so we must solve the equation graphically. E X AM P L E 9 \| Solving a Logarithmic Equation Graphically x 2 Solve the equation x 2 2 ln. 1 2 ▼ SO LUTI O N We ﬁrst move all terms to one side of the equation Then we graph x 2 2 ln x 2 1 2 0 y x 2
2 ln x 2 1 2 as in Figure 3. The solutions are the x-intercepts of the graph. Zooming in on the x-intercepts, we see that there are two solutions: x 0.71 and x 1.60 2 _2 3 FIGURE 3 _2 ✎ Practice what you’ve learned: Do Exercise 59. ▲ 406 CHAPTER 5 \| Exponential and Logarithmic Functions Logarithmic equations are used in determining the amount of light that reaches various depths in a lake. (This information helps biologists to determine the types of life a lake can support.) As light passes through water (or other transparent materials such as glass or plastic), some of the light is absorbed. It’s easy to see that the murkier the water, the more light is absorbed. The exact relationship between light absorption and the distance light travels in a material is described in the next example. The intensity of light in a lake diminishes with depth. E X AM P L E 10 \| Transparency of a Lake If I0 and I denote the intensity of light before and after going through a material and x is the distance (in feet) the light travels in the material, then according to the Beer-Lambert Law, 1 k ln a I I0 b x where k is a constant depending on the type of material. (a) Solve the equation for I. (b) For a certain lake k 0.025, and the light intensity is I0 light intensity at a depth of 20 ft. ▼ SO LUTI O N (a) We ﬁrst isolate the logarithmic term. 14 lumens (lm). Find the 1 k ln a ln a I I0 b I I0 b x Given equation kx Multiply by –k ekx I I0 I I0ekx Exponential form Multiply by I0 (b) We ﬁnd I using the formula from part (a). I I0ekx 14e1 0.025 20 2 21 From part (a) I0 = 14, k = 0.025, x = 20 8.49 Calculator The light intensity at a depth of 20 ft is about 8.5 lm. ✎ Practice what you’ve learned: Do Exercise 83. ▲ SE CTI ON 5. 4 \| Exponential and Logarithmic Equations 407 ■ Compound Interest
Recall the formulas for interest that we found in Section 5.1. If a principal P is invested at an interest rate r for a period of t years, then the amount A of the investment is given by Simple interest (for one year) nt Interest compounded n times per year Pert Interest compounded continuously A A t t 1 1 2 2 We can use logarithms to determine the time it takes for the principal to increase to a given amount. E X AM P L E 11 \| Finding the Term for an Investment to Double A sum of $5000 is invested at an interest rate of 5% per year. Find the time required for the money to double if the interest is compounded according to the following method. (a) Semiannual (b) Continuous ▼ SO LUTI O N (a) We use the formula for compound interest with P $5000, A 1 r 0.05, and n 2 and solve the resulting exponential equation for t. t 2 $10,000, 5000 2t 1 0.05 2 b a 10,000 1.025 2t 2 1 2 log 1.0252t log 2 nt 1 r n b P a A Divide by 5000 Take log of each side 2t log 1.025 log 2 Law 3 (bring down the exponent) t log 2 2 log 1.025 Divide by 2 log 1.025 t 14.04 Calculator The money will double in 14.04 years. (b) We use the formula for continuously compounded interest with P $5000, $10,000, and r 0.05 and solve the resulting exponential equation for t. A t 1 2 5000e0.05t 10,000 Pert = A e0.05t 2 ln e0.05t ln 2 0.05t ln 2 t ln 2 0.05 t 13.86 Divide by 5000 Take ln of each side Property of ln Divide by 0.05 Calculator The money will double in 13.86 years. ✎ Practice what you’ve learned: Do Exercise 71. ▲ 408 CHAPTER 5 \| Exponential and Logarithmic Functions E X AM P L E 12 \| Time Required to Grow an Investment A sum of $1000 is invested at an interest rate of 4% per year. Find the time required for the amount to grow to $4000 if interest is compounded continuously. ▼ SO LUTI O N We use the formula for continuously compounded interest with P $1000,
, and r 0.04 and solve the resulting exponential equation for t. $4000 A t 1 2 1000e0.04t 4000 e0.04t 4 0.04t ln 4 t ln 4 0.04 t 34.66 Pert = A Divide by 1000 Take ln of each side Divide by 0.04 Calculator The amount will be $4000 in about 34 years and 8 months. ✎ Practice what you’ve learned: Do Exercise 73. ▲ If an investment earns compound interest, then the annual percentage yield (APY) is the simple interest rate that yields the same amount at the end of one year. E X AM P L E 13 \| Calculating the Annual Percentage Yield Find the annual percentage yield for an investment that earns interest at a rate of 6% per year, compounded daily. ▼ SO LUTI O N After one year, a principal P will grow to the amount A P 365 1 0.06 365 b a P 1 1.06183 2 The formula for simple interest is A P 1 r 1 Comparing, we see that 1 r 1.06183, so r 0.06183. Thus, the annual percentage yield is 6.183%. ✎ Practice what you’ve learned: Do Exercise 77. ▲ 2 5. ▼ CONCE PTS 1. Let’s solve the exponential equation 2ex 50. (b) Next, we write each side in exponential form to get the equivalent equation. (a) First, we isolate ex to get the equivalent equation. (c) Now we ﬁnd x. (b) Next, we take ln of each side to get the equivalent equation. (c) Now we use a calculator to ﬁnd x. 2. Let’s solve the logarithmic equation x 2 log 3 log log x. 1 2 (a) First, we combine the logarithms to get the equivalent equation. ▼ SKI LLS 3–28 ■ Find the solution of the exponential equation, correct to four decimal places. 10x 25 e2x 7 10x 4 4. 6. e3x 12 5. 3. 9. 11. 13. 15. 17. 19. 21. 23. 25. 21x 3 3ex 10 e14x 2 4 35x 8 80.4x 5 5x/100 2 e2x1 200 5x 4x1 23x1 3x2 50
1 ex 4 27. 100 1.04 1 2 2t 300 8. 10. 12. 14. 16. 18. 20. 22. 24. 26. 28. 9 2 32x1 5 2e12x 17 1 105x 4 1 23x 34 3x/14 0.1 e35x 16 x 75 1 4B A 101x 6x 7x/2 51x 10 1 ex 2 1.00625 1 12t 2 2 29. 31. 29–36 ■ Solve the equation. e2x 3ex 2 0 e4x 4e2x 21 0 x22x 2x 0 4x3e3x 3x4e3x 0 33. 35. 30. 32. 34. 36. e2x ex 6 0 ex 12ex 1 0 x210x x10x 2 10x x2ex xex ex 0 1 2 37. 39. 37–54 ■ Solve the logarithmic equation for x x2 x 2 ln x 10 log x 2 3x 5 2 3 x 2 43. 40. 41. 42. 38. ln 2 2 1 log 1 log31 log21 x 2 2 1 x 1 3x 4 log 1 4 log 3 44. 2 log2 3 log2 x log2 5 log21 2 log x log 2 log 1 2 log log x log 4x 2 1 2 1 log5 20 x 1 log5 x log51 2 x 1 log51 x 1 2 log51 2 2 x 1 x 15 log31 2 log31 2 log2 x log21 2 x 3 2 log x log x 3 1 1 2 x 5 log9 1 x 3 log9 1 2 x 2 1 ln 1 1 2 ln x 1 1 2 2 2 45. 46. 47. 48. 49. 50. 51. 52. 53. 54. 55. For what value of x is the following true? log 1 x 3 2 log x log 3 56. For what value of x is it true that 57. Solve for x: 58. Solve for x: 22/log5 x 1 16 log2 1 log3 x 2 4 log x 1 2 3 3 log x? 59–66 ■ Use a graphing device to ﬁnd all solutions of the equation, correct to two decimal places. ✎ 59. 61. ln x 3 x x 3 x log x 1 1 2 60. 62. x ln log x x2 2 4
x 2 1 2 SE CTI ON 5. 4 \| Exponential and Logarithmic Equations 409 63. 65. ex x 4x 1x 64. 66. 2x x 1 ex 2 2 x 3 x 67–70 ■ Solve the inequality. 9 x log 67. 1 2 1 2 x 2 log 3 log2 x 4 2 10x 5 68. 69. 1 70. x 2e x 2e x 0 ▼ APPLICATIONS 71. Compound Interest A man invests $5000 in an account that ✎ pays 8.5% interest per year, compounded quarterly. (a) Find the amount after 3 years. (b) How long will it take for the investment to double? 72. Compound Interest A woman invests $6500 in an account that pays 6% interest per year, compounded continuously. (a) What is the amount after 2 years? (b) How long will it take for the amount to be $8000? ✎ 73. Compound Interest Find the time required for an invest- ment of $5000 to grow to $8000 at an interest rate of 7.5% per year, compounded quarterly. 74. Compound Interest Nancy wants to invest $4000 in saving certiﬁcates that bear an interest rate of 9.75% per year, compounded semiannually. How long a time period should she choose to save an amount of $5000? 75. Doubling an Investment How long will it take for an in- vestment of $1000 to double in value if the interest rate is 8.5% per year, compounded continuously? 76. Interest Rate A sum of $1000 was invested for 4 years, and the interest was compounded semiannually. If this sum amounted to $1435.77 in the given time, what was the interest rate? ✎ 77. Annual Percentage Yield Find the annual percentage yield for an investment that earns 8% per year, compounded monthly. 78. Annual Percentage Yield Find the annual percentage yield for an investment that earns 5 % per year, compounded continuously. 1 2 79. Radioactive Decay A 15-g sample of radioactive iodine decays in such a way that the mass remaining after t days is given by, where After how many days is there only 5 g remaining? 15e0.087t is measured in grams. m m t t 2 2 1 1 80. Sky Diving The velocity of a sky diver t seconds after 80 jumping is given by 1 seconds
is the velocity 70 ft/s? √ t 2 1 1 e0.2t. After how many 2 81. Fish Population A small lake is stocked with a certain species of ﬁsh. The ﬁsh population is modeled by the function P 10 1 4e0.8t where P is the number of ﬁsh in thousands and t is measured in years since the lake was stocked. (a) Find the ﬁsh population after 3 years. 410 CHAPTER 5 \| Exponential and Logarithmic Functions (b) After how many years will the ﬁsh population reach 5000 85. Electric Circuits An electric circuit contains a battery that ﬁsh? 82. Transparency of a Lake Environmental scientists measure the intensity of light at various depths in a lake to ﬁnd the “transparency” of the water. Certain levels of transparency are required for the biodiversity of the submerged macrophyte population. In a certain lake the intensity of light at depth x is given by I 10e0.008x where I is measured in lumens and x in feet. (a) Find the intensity I at a depth of 30 ft. (b) At what depth has the light intensity dropped to I 5? ✎ 83. Atmospheric Pressure Atmospheric pressure P (in kilopascals, kPa) at altitude h (in kilometers, km) is governed by the formula ln a P P0 b h k where k 7 and P0 (a) Solve the equation for P. (b) Use part (a) to ﬁnd the pressure P at an altitude of 4 km. 100 kPa are constants. 84. Cooling an Engine Suppose you’re driving your car on a cold winter day (20F outside) and the engine overheats (at about 220F). When you park, the engine begins to cool down. The temperature T of the engine t minutes after you park satisﬁes the equation ln T 20 a 200 b 0.11t (a) Solve the equation for T. (b) Use part (a) to ﬁnd the temperature of the engine after 20 min t 20 1. 2 produces a voltage of 60 volts (V), a resistor with a resistance of 13 ohms (), and an inductor with an inductance of 5 henrys (H), as shown in the ﬁgure. Using calculus, it can be
shown that the current switch is closed is (in amperes, A) t seconds after the 1 e13t/5 (a) Use this equation to express the time t as a function of the t 2 1 I 60 131 I I. 2 current I. (b) After how many seconds is the current 2 A? 13 60 V 5 H Switch 86. Learning Curve A learning curve is a graph of a function 1 2 t P that measures the performance of someone learning a skill as a function of the training time t. At ﬁrst, the rate of learning is rapid. Then, as performance increases and approaches a maximal value M, the rate of learning decreases. It has been found that the function M Cekt P t 1 2 where k and C are positive constants and C M is a reasonable model for learning. (a) Express the learning time t as a function of the per- formance level P. (b) For a pole-vaulter in training, the learning curve is given by 20 14e0.024t P t 1 2 1 t P is the height he is able to pole-vault after where t months. After how many months of training is he able to vault 12 ft? 2 (c) Draw a graph of the learning curve in part (b). ▼ DISCOVE RY • DISCUSSION • WRITI NG 87. Estimating a Solution Without actually solving the equation, ﬁnd two whole numbers between which the solution of 9x 20 reached your conclusions. must lie. Do the same for 9x 100. Explain how you SE CTI ON 5. 5 \| Modeling with Exponential and Logarithmic Functions 411 88. A Surprising Equation Take logarithms to show that the equation x1/log x 5 has no solution. For what values of k does the equation x1/log x k have a solution? What does this tell us about the graph of the function device.? Conﬁrm your answer using a graphing x1/log x x f 2 1 89. Disguised Equations Each of these equations can be transformed into an equation of linear or quadratic type by applying the hint. Solve each equation. (a) (b) x 1 x 1 2 1 log2 x log4 x log8 x 11 2 100 x1 log 1 2 1 [Take log of each side.] [Change all logs to base 2
.] [Write as a quadratic in 2x.] (c) 4x 2x1 3 Modeling with Exponential and Logarithmic Functions 5.5 LEARNING OBJECTIVES After completing this section, you will be able to: ■ Find exponential models of population growth ■ Find exponential models of radioactive decay ■ Find models using Newton’s Law of Cooling ■ Use logarithmic scales (pH, Richter, and decibel scales) Many processes that occur in nature, such as population growth, radioactive decay, heat diffusion, and numerous others, can be modeled by using exponential functions. Logarithmic functions are used in models for the loudness of sounds, the intensity of earthquakes, and many other phenomena. In this section we study exponential and logarithmic models. ■ Exponential Models of Population Growth Biologists have observed that the population of a species doubles its size in a ﬁxed period of time. For example, under ideal conditions a certain population of bacteria doubles in size every 3 hours. If the culture is started with 1000 bacteria, then after 3 hours there will be 2000 bacteria, after another 3 hours there will be 4000, and so on. If we let be the number of bacteria after t hours, then 1000 1000 # 2 1000 # 2 2 1000 # 22 1000 # 23 # 2 1000 # 22 # 2 1000 # 23 # 2 1000 # 24 n n 1 12 From this pattern it appears that the number of bacteria after t hours is modeled by the function 1000 # 2t/3 In general, suppose that the initial size of a population is n0 and the doubling period is n t 2 1 a. Then the size of the population at time t is modeled by n n03ct where c 1/a. If we knew the tripling time b, then the formula would be, where c 1/b. These formulas indicate that the growth of the bacteria is modeled by an n t t 1 2 1 2 n02ct 412 CHAPTER 5 \| Exponential and Logarithmic Functions exponential function. But what base should we use? The answer is e, because then it can be shown (using calculus) that the population is modeled by n0ert n t 1 2 where r is the relative rate of growth of population, expressed as a proportion of the population at any time. For instance, if r 0.02, then at any time t the growth rate is 2% of the population at time t. Notice that the
formula for population growth is the same as that for continuously compounded interest. In fact, the same principle is at work in both cases: The growth of a population (or an investment) per time period is proportional to the size of the population (or the amount of the investment). A population of 1,000,000 will increase more in one year than a population of 1000; in exactly the same way, an investment of $1,000,000 will increase more in one year than an investment of $1000. EXPONENTIAL GROWTH MODEL A population that experiences exponential growth increases according to the model where n n0ert t n 1 population at time t initial size of the population t 2 1 n0 r relative rate of growth (expressed as a proportion of the 2 population) t time In the following examples we assume that the populations grow exponentially. E X AM P L E 1 \| Predicting the Size of a Population The initial bacterium count in a culture is 500. A biologist later makes a sample count of bacteria in the culture and ﬁnds that the relative rate of growth is 40% per hour. (a) Find a function that models the number of bacteria after t hours. (b) What is the estimated count after 10 hours? (c) Sketch the graph of the function n t. 1 2 ▼ SO LUTI O N (a) We use the exponential growth model with n0 500 and r 0.4 to get 5000 n(t)=500eº—¢‰ 500 0 FIGURE 1 where t is measured in hours. 500e0.4t n t 1 2 (b) Using the function in part (a), we ﬁnd that the bacterium count after 10 hours is 10 n 1 2 500e0.4 1 10 2 500e4 27,300 6 (c) The graph is shown in Figure 1. ✎ Practice what you’ve learned: Do Exercise 1. ▲ E X AM P L E 2 \| Comparing Different Rates of Population Growth The relative growth of world population has been declining over the past few decades—from 2% in 1995 to 1.3% in 2006. In 2000 the population of the world was 6.1 billion, and the relative rate of growth was 1.4% per year. It is claimed that a rate of 1.0% per year would make a signiﬁcant difference in the total population in just a few decades.
Test this claim by estimating the population of the world in the year 2050 using a relative rate of growth of (a) 1.4% per year and (b) 1.0% per year. SE CTI O N 5. 5 \| Modeling with Exponential and Logarithmic Functions 413 Graph the population functions for the next 100 years for the two relative growth rates in the same viewing rectangle. ▼ SO LUTI O N (a) By the exponential growth model we have 6.1e0.014t n t 1 2 is measured in billions and t is measured in years since 2000. Because the 2 1 t n where year 2050 is 50 years after 2000, we ﬁnd 6.1e0.014 1 50 n 50 1 2 2 6.1e0.7 12.3 n(t)=6.1e0.014t (b) We use the function The estimated population in the year 2050 is about 12.3 billion. n(t)=6.1e0.01t and ﬁnd 100 6.1e0.010t n t 1 2 50 n 1 2 6.1e0.010 1 50 2 6.1e0.50 10.1 The estimated population in the year 2050 is about 10.1 billion. The graphs in Figure 2 show that a small change in the relative rate of growth will, over time, make a large difference in population size. ✎ Practice what you’ve learned: Do Exercise 3. ▲ E X AM P L E 3 \| World Population Projections 30 0 FIGURE 2 Standing Room Only The population of the world was about 6.1 billion in 2000 and was increasing at 1.4% per year. Assuming that each person occupies an average of 4 ft2 of the surface of the earth, the exponential model for population growth projects that by the year 2801 there will be standing room only! (The total land surface area of the world is about 1.8 1015 ft2.) The population of the world in 2000 was 6.1 billion, and the estimated relative growth rate was 1.4% per year. If the population continues to grow at this rate, when will it reach 122 billion? ▼ SO LUTI O N We use the population growth function with n0 and billion. This leads to an exponential equation, which we solve for t. 6.1 billion, r 0.014, 122 n t 1 2 6.
1e0.014t 122 e0.014t 20 ln e0.014t ln 20 0.014t ln 20 t ln 20 0.014 n0ert n(t) Divide by 6.1 Take ln of each side Property of ln Divide by 0.014 t 213.98 Calculator Thus, the population will reach 122 billion in approximately 214 years, that is, in the year 2000 214 2214. ✎ Practice what you’ve learned: Do Exercise 11. ▲ E X AM P L E 4 \| Finding the Initial Population A certain breed of rabbit was introduced onto a small island about 8 years ago. The current rabbit population on the island is estimated to be 4100, with a relative growth rate of 55% per year. (a) What was the initial size of the rabbit population? (b) Estimate the population 12 years from now. 414 CHAPTER 5 \| Exponential and Logarithmic Functions ▼ SO LUTI O N (a) From the exponential growth model we have n0e0.55t n t 1 2 and we know that the population at time t 8 is we know into the equation and solve for n0: 4100. We substitute what n 8 1 2 4100 n0e0.55 4100 e0.55 n0 8 2 1 8 1 2 4100 81.45 50 Thus, we estimate that 50 rabbits were introduced onto the island. (b) Now that we know n0, we can write a formula for population growth: Another way to solve part (b) is to let t be the number of years from now. In 4100 (the current poputhis case n0 lation), and the population 12 years from now will be 12 n 1 2 4100e0.55 1 12 2 3 million Twelve years from now, t 8 12 20 and 50e0.55t n t 1 2 20 n 1 2 50e0.55 1 20 2 2,993,707 We estimate that the rabbit population on the island 12 years from now will be about 3 million. ✎ Practice what you’ve learned: Do Exercise 5. ▲ Can the rabbit population in Example 4(b) actually reach such a high number? In reality, as the island becomes overpopulated with rabbits, the rabbit population growth will be slowed because of food shortage and other factors. One model that takes into account such factors is the logistic growth
model described in the Focus on Modeling, page 436. E X AM P L E 5 \| The Number of Bacteria in a Culture A culture starts with 10,000 bacteria, and the number doubles every 40 minutes. (a) Find a function that models the number of bacteria at time t. (b) Find the number of bacteria after one hour. (c) After how many minutes will there be 50,000 bacteria? (d) Sketch a graph of the number of bacteria at time t. ▼ SO LUTI O N (a) To ﬁnd the function that models this population growth, we need to ﬁnd the rate r. 10,000, t 40, and To do this, we use the formula for population growth with n0 n and then solve for r. 20,000 t 1 2 10,000er 1 40 2 20,000 n0ert = n(t) e40r 2 ln e40r ln 2 40r ln 2 r ln 2 40 r 0.01733 Divide by 10,000 Take ln of each side Property of ln Divide by 40 Calculator Now that we know that r 0.01733, we can write the function for the population growth: 10,000e0.01733t n t 1 2 SE CTI O N 5. 5 \| Modeling with Exponential and Logarithmic Functions 415 (b) Using the function that we found in part (a) with t 60 min (one hour), we get 10,000e0.01733 2 28,287 60 n 60 1 1 2 Thus, the number of bacteria after one hour is approximately 28,000. (c) We use the function we found in part (a) with n exponential equation for t. t 1 2 50,000 and solve the resulting 50,000 (t)=10,000 eº.º¡¶££ t 80 Time (min) FIGURE 3 The half-lives of radioactive elements vary from very long to very short. Here are some examples. Element Half-life 14.5 billion years Thorium-232 4.5 billion years Uranium-235 Thorium-230 80,000 years Plutonium-239 24,360 years Carbon-14 Radium-226 Cesium-137 Strontium-90 Polonium-210 Thorium-234 Iodine-135 Radon-222 Lead-211 Krypton-91 5,730 years
1,600 years 30 years 28 years 140 days 25 days 8 days 3.8 days 3.6 minutes 10 seconds 10,000e0.01733t 50,000 e0.01733t 5 n0ert = n(t) Divide by 10,000 ln e0.01733t ln 5 0.01733t ln 5 t ln 5 0.01733 t 92.9 Take ln of each side Property of ln Divide by 0.01733 Calculator The bacterium count will reach 50,000 in approximately 93 min. (d) The graph of the function ✎ Practice what you’ve learned: Do Exercise 9. 10,000e0.01733t n t 2 1 is shown in Figure 3. ▲ ■ Radioactive Decay Radioactive substances decay by spontaneously emitting radiation. The rate of decay is directly proportional to the mass of the substance. This is analogous to population growth, t except that the mass of radioactive material decreases. It can be shown that the mass remaining at time t is modeled by the function m 1 2 m0ert m t 1 2 where r is the rate of decay expressed as a proportion of the mass and m0 is the initial mass. Physicists express the rate of decay in terms of half-life, the time required for half the mass to decay. We can obtain the rate r from this as follows. If h is the half-life, then a mass of 1 unit becomes unit when t h. Substituting this into the model, we get 1 2 ln A 1 2 1 # erh rh 1 2B r 1 h ln 1 m(t) = moe–rt Take ln of each side 21 2 Solve for r r ln 2 h ln 2–1 = In 2 by Law 3 This last equation allows us to ﬁnd the rate r from the half-life h. RADIOACTIVE DECAY MODEL If m0 is the initial mass of a radioactive substance with half-life h, then the mass remaining at time t is modeled by the function where r ln 2 h. m0ert m t 1 2 416 CHAPTER 5 \| Exponential and Logarithmic Functions E X AM P L E 6 \| Radioactive Decay has a half-life of 140 days. Suppose a sample of this substance has 210Po 1 2 Polonium-210 a mass of 300 mg. (a) Find a function that models
the amount of the sample remaining at time t. (b) Find the mass remaining after one year. (c) How long will it take for the sample to decay to a mass of 200 mg? (d) Draw a graph of the sample mass as a function of time. ▼ SO LUTI O N (a) Using the model for radioactive decay with m0 we have m (b) We use the function we found in part (a) with t 365 (one year). 300e0.00495t t 2 1 300 and r ln 2/140 1 2 0.00495, 2 49.256 2 Thus, approximately 49 mg of 210Po remains after one year. 300e0.00495 365 m 365 1 1 (c) We use the function that we found in part (a) with m ing exponential equation for t. 200 and solve the result- t 1 2 300e0.00495t 200 e0.00495t 2 3 ln e0.00495t ln 2 3 0.00495t ln 2 3 t ln 2 3 0.00495 t 81.9 m(t) = moe rt Divided by 300 Take ln of each side Property of ln Divide by – 0.00495 Calculator The time required for the sample to decay to 200 mg is about 82 days. m (d) A graph of the function 1 ✎ Practice what you’ve learned: Do Exercise 15. 300e0.00495t t 2 is shown in Figure 4. ▲ m(t) 300 200 100 ) (t)=300 e_º.ºº¢ª∞ t 0 50 150 Time (days) t FIGURE 4 ■ Newton’s Law of Cooling Newton’s Law of Cooling states that the rate of cooling of an object is proportional to the temperature difference between the object and its surroundings, provided that the temperature difference is not too large. Using calculus, the following model can be deduced from this law. Radioactive Waste Harmful radioactive isotopes are produced whenever a nuclear reaction occurs, whether as the result of an atomic bomb test, a nuclear accident such as the one at Chernobyl in 1986, or the uneventful production of electricity at a nuclear power plant. One radioactive material that is produced in atomic bombs is the iso90Sr tope strontium-90, with a half2 life of 28 years. This is deposited
1 like calcium in human bone tissue, where it can cause leukemia and other cancers. However, in the decades since atmospheric testing of nuclear weapons was halted, 90Sr levels in the environment have fallen to a level that no longer poses a threat to health. Nuclear power plants produce radioactive plutonium-239 239Pu, which has a half-life of 24,360 years. Because of its long 1 2 half-life, 239Pu could pose a threat to the environment for thousands of years. So great care must be taken to dispose of it properly. The difﬁculty of ensuring the safety of the disposed radioactive waste is one reason that nuclear power plants remain controversial © SE CTI O N 5. 5 \| Modeling with Exponential and Logarithmic Functions 417 NEWTON’S LAW OF COOLING If D0 is the initial temperature difference between an object and its surroundings, and if its surroundings have temperature Ts, then the temperature of the object at time t is modeled by the function where k is a positive constant that depends on the type of object. Ts D0ekt T t 1 2 E X AM P L E 7 \| Newton’s Law of Cooling A cup of coffee has a temperature of 200F and is placed in a room that has a temperature of 70F. After 10 min the temperature of the coffee is 150F. (a) Find a function that models the temperature of the coffee at time t. (b) Find the temperature of the coffee after 15 min. (c) When will the coffee have cooled to 100F? (d) Illustrate by drawing a graph of the temperature function. ▼ SO LUTI O N (a) The temperature of the room is Ts 70F, and the initial temperature difference is D0 200 70 130°F So by Newton’s Law of Cooling, the temperature after t minutes is modeled by the function 70 130ekt T t 1 2 We need to ﬁnd the constant k associated with this cup of coffee. To do this, we use the fact that when t 10, the temperature is T 150. So we have 10 1 2 70 130e10k 150 Ts + Doe–kt = T(t) 130e10k 80 e10k 8 13 10k ln 8 13 k 1 10 ln 8 13 Subtract 70 Divide by 130 Take ln of each side Divide by –10 k 0.04855 Calculator Substituting this
value of k into the expression for T t 1 70 130e0.04855t T t 1 2, we get 2 (b) We use the function that we found in part (a) with t 15. 70 130e0.04855 1 15 2 133°F 15 T 1 2 (c) We use the function that we found in part (a) with T exponential equation for t. 100 and solve the resulting t 1 2 418 CHAPTER 5 \| Exponential and Logarithmic Functions T (°F) 200 70 T=70+130e_º.º¢•∞∞ t T=70 70 130e0.04855t 100 130e0.04855t 30 e0.04855t 3 13 0.04855t ln 3 13 ln 3 13 0.04855 t Ts + Doe– kt = T(t) Subtract 70 Divide by 130 Take ln of each side Divide by – 0.04855 t 30.2 Calculator 0 10 20 30 40 t (min) The coffee will have cooled to 100F after about half an hour. (d) The graph of the temperature function is sketched in Figure 5. Notice that the line t 70 is a horizontal asymptote. (Why?) FIGURE 5 Temperature of coffee after t minutes ✎ Practice what you’ve learned: Do Exercise 23. ▲ ■ Logarithmic Scales When a physical quantity varies over a very large range, it is often convenient to take its logarithm in order to have a more manageable set of numbers. We discuss three such situations: the pH scale, which measures acidity; the Richter scale, which measures the intensity of earthquakes; and the decibel scale, which measures the loudness of sounds. Other quantities that are measured on logarithmic scales are light intensity, information capacity, and radiation. The pH Scale Chemists measured the acidity of a solution by giving its hydrogen ion concentration until Søren Peter Lauritz Sørensen, in 1909, proposed a more convenient measure. He deﬁned pH log H 4 3 where this to avoid very small numbers and negative exponents. For instance, is the concentration of hydrogen ions measured in moles per liter (M). He did 3 H 4 if H 104 M, then 3 4 pH log101 104 4 4 2 1 2 Solutions with a pH of 7 are deﬁned as neutral,
those with pH 7 are acidic, and those decreases by with pH 7 are basic. Notice that when the pH increases by one unit, a factor of 10. H 3 4 E X AM P L E 8 \| pH Scale and Hydrogen Ion Concentration pH for Some Common Substances Substance Milk of magnesia Seawater Human blood Crackers Hominy Cow’s milk Spinach Tomatoes Oranges Apples Limes Battery acid pH 10.5 8.0–8.4 7.3–7.5 7.0–8.5 6.9–7.9 6.4–6.8 5.1–5.7 4.1–4.4 3.0–4.0 2.9–3.3 1.3–2.0 1.0 (a) The hydrogen ion concentration of a sample of human blood was measured to be Find the pH and classify the blood as acidic or basic. H 3.16 108 M. 3 4 (b) The most acidic rainfall ever measured occurred in Scotland in 1974; its pH was 2.4. Find the hydrogen ion concentration. ▼ SO LUTI O N (a) A calculator gives pH log H 3 4 log 1 3.16 108 7.5 2 Since this is greater than 7, the blood is basic. (b) To ﬁnd the hydrogen ion concentration, we need to solve for equation H 4 3 in the logarithmic log 3 H 4 pH SE CTI O N 5. 5 \| Modeling with Exponential and Logarithmic Functions 419 So we write it in exponential form. In this case pH 2.4, so H 3 4 10pH H 3 ✎ Practice what you’ve learned: Do Exercise 27. 102.4 4.0 103 M 4 ▲ Largest Earthquakes Location Date Magnitude 1960 Chile 1964 Alaska 2004 Sumatra 1957 Alaska 1952 Kamchatka 1906 Ecuador 1965 Alaska 2005 Sumatra 1950 Tibet 1923 Kamchatka Indonesia 1938 Kuril Islands 1963 9.5 9.2 9.1 9.1 9.0 8.8 8.7 8.7 8.6 8.5 8.5 8.5 The Richter Scale the magnitude M of an earthquake to be In 1935 the American geologist Charles Richter (1900–1984) deﬁned M log I S where I is the intensity of the earthquake (measured by the amplitude of a seismograph reading taken 100 km
from the epicenter of the earthquake) and S is the intensity of a “standard” earthquake (whose amplitude is 1 micron 104 cm). The magnitude of a standard earthquake is M log S S log 1 0 Richter studied many earthquakes that occurred between 1900 and 1950. The largest had magnitude 8.9 on the Richter scale, and the smallest had magnitude 0. This corresponds to a ratio of intensities of 800,000,000, so the Richter scale provides more manageable numbers to work with. For instance, an earthquake of magnitude 6 is ten times stronger than an earthquake of magnitude 5. E X AM P L E 9 \| Magnitude of Earthquakes The 1906 earthquake in San Francisco had an estimated magnitude of 8.3 on the Richter scale. In the same year a powerful earthquake occurred on the Colombia-Ecuador border that was four times as intense. What was the magnitude of the Colombia-Ecuador earthquake on the Richter scale? ▼ SO LUTI O N If I is the intensity of the San Francisco earthquake, then from the deﬁnition of magnitude we have M log I S 8.3 The intensity of the Colombia-Ecuador earthquake was 4I, so its magnitude was 4I S ✎ Practice what you’ve learned: Do Exercise 33. log 4 log M log I S log 4 8.3 8.9 ▲ E X AM P L E 10 \| Intensity of Earthquakes The 1989 Loma Prieta earthquake that shook San Francisco had a magnitude of 7.1 on the Richter scale. How many times more intense was the 1906 earthquake (see Example 9) than the 1989 event? ▼ SO LUTI O N If I1 and I2 are the intensities of the 1906 and 1989 earthquakes, then we are required to ﬁnd I1/I2. To relate this to the deﬁnition of magnitude, we divide the numerator and denominator by S. 420 CHAPTER 5 \| Exponential and Logarithmic Functions © The intensity levels of sounds that we can hear vary from very loud to very soft. Here are some examples of the decibel levels of commonly heard sounds. Source of sound Jet takeoff Jackhammer Rock concert Subway Heavy trafﬁc Ordinary trafﬁc Normal conversation Whisper Rustling leaves Threshold of hearing B dB 1 140 130 120 100 80 70 50 30 10–20 0 log I1 I2 log I1
/S I2/S Divide numerator and denominator by S log I1 S log I2 S 8.3 7.1 1.2 Law 2 of logarithms Deﬁnition of earthquake magnitude Therefore, I1 I2 10log I1/I22 101.2 16 1 The 1906 earthquake was about 16 times as intense as the 1989 earthquake. ✎ Practice what you’ve learned: Do Exercise 35. ▲ The Decibel Scale The ear is sensitive to an extremely wide range of sound intensities. 1012 W/m2 (watts per square meter) at a frequency We take as a reference intensity I0 of 1000 hertz, which measures a sound that is just barely audible (the threshold of hearing). The psychological sensation of loudness varies with the logarithm of the intensity (the Weber-Fechner Law), so the intensity level B, measured in decibels (dB), is deﬁned as 2 B 10 log I I0 The intensity level of the barely audible reference sound is B 10 log I0 I0 10 log 1 0 dB E X AM P L E 11 \| Sound Intensity of a Jet Takeoff Find the decibel intensity level of a jet engine during takeoff if the intensity was measured at 100 W/m2. ▼ SO LUTI O N From the deﬁnition of intensity level we see that B 10 log I I0 10 log 102 1012 10 log 1014 140 dB Thus, the intensity level is 140 dB. ✎ Practice what you’ve learned: Do Exercise 39. ▲ The table in the margin lists decibel intensity levels for some common sounds ranging from the threshold of human hearing to the jet takeoff of Example 11. The threshold of pain is about 120 dB. 5. ▼ APPLICATIONS 1–13 ■ These exercises use the population growth model. ✎ 1. Bacteria Culture The number of bacteria in a culture is modeled by the function 500e0.45t n t 1 2 where t is measured in hours. (a) What is the initial number of bacteria? (b) What is the relative rate of growth of this bacterium popu- lation? Express your answer as a percentage. (c) How many bacteria are in the culture after 3 hours? 05-W4525.qxd 1/8/08 11:32 AM Page 421 SE CTI O N 5. 5 \| Modeling with
Exponential and Logarithmic Functions 421 (d) After how many hours will the number of bacteria reach 10,000? (c) What is the projected deer population in 2011? (d) In what year will the deer population reach 100,000? 2. Fish Population The number of a certain species of ﬁsh is modeled by the function 12e0.012t n t 1 2 where t is measured in years and is measured in millions. (a) What is the relative rate of growth of the ﬁsh population? n t 2 1 Express your answer as a percentage. (b) What will the ﬁsh population be after 5 years? (c) After how many years will the number of ﬁsh reach 30 million? ✎ (d) Sketch a graph of the ﬁsh population function n. t 1 2 3. Population of a Country The population of a country has a relative growth rate of 3% per year. The government is trying to reduce the growth rate to 2%. The population in 1995 was approximately 110 million. Find the projected population for the year 2020 for the following conditions. (a) The relative growth rate remains at 3% per year. (b) The relative growth rate is reduced to 2% per year. 4. Fox Population The fox population in a certain region has a relative growth rate of 8% per year. It is estimated that the population in 2005 was 18,000. (a) Find a function that models the population t years after ✎ 2005. n(t) 30,000 20,000 10,000 Deer population (4, 31,000) 0 1 2 3 4 Years since 2003 t 8. Bacteria Culture A culture contains 1500 bacteria initially and doubles every 30 min. (a) Find a function that models the number of bacteria after t minutes. n t 1 2 (b) Find the number of bacteria after 2 hours. (c) After how many minutes will the culture contain 4000 bacteria? 9. Bacteria Culture A culture starts with 8600 bacteria. After one hour the count is 10,000. (a) Find a function that models the number of bacteria n t 1 2 (b) Use the function from part (a) to estimate the fox popula- after t hours. tion in the year 2013. (c) Sketch a graph of the fox population function for the years (b) Find the number of bacteria after 2 hours. (c) After how many hours
will the number of bacteria double? 2005–2013. ✎ 5. Frog Population Some frogs were introduced into a small pond 6 years ago. The current frog population in the pond is estimated to be 100, with a relative growth rate of 42% per year. (a) What was the initial size of the frog population? (b) Estimate the frog population 5 years from now. 6. Population of a City The population of a certain city was 112,000 in 2006, and the observed relative growth rate is 4% per year. (a) Find a function that models the population after t years. (b) Find the projected population in the year 2012. (c) In what year will the population reach 200,000? 7. Deer Population The graph shows the deer population in a Pennsylvania county between 2003 and 2007. Assume that the population grows exponentially. (a) What was the deer population in 2003? (b) Find a function that models the deer population t years after 2003. 10. Bacteria Culture The count in a culture of bacteria was 400 after 2 hours and 25,600 after 6 hours. (a) What is the relative rate of growth of the bacteria popula- tion? Express your answer as a percentage. (b) What was the initial size of the culture? (c) Find a function that models the number of bacteria after t hours. (d) Find the number of bacteria after 4.5 hours. (e) When will the number of bacteria be 50,000? n t 1 2 ✎ 11. World Population The population of the world was 5.7 billion in 1995, and the observed relative growth rate was 2% per year. (a) By what year will the population have doubled? (b) By what year will the population have tripled? 12. Population of California The population of California was 10,586,223 in 1950 and 23,668,562 in 1980. Assume that the population grows exponentially. (a) Find a function that models the population t years after 1950. (b) Find the time required for the population to double. (c) Use the function from part (a) to predict the population of California in the year 2006. Look up California’s actual population in 2006, and compare. 13. Infectious Bacteria An infectious strain of bacteria in- creases in number at a relative growth rate of 200% per hour. When a certain critical number of bacteria are present in the bloodstream, a person becomes
ill. If a single bacterium infects a person, the critical level is reached in 24 hours. How long will it take for the critical level to be reached if the same person is infected with 10 bacteria? 422 CHAPTER 5 \| Exponential and Logarithmic Functions 14–22 ■ These exercises use the radioactive decay model. 14. Radioactive Radium The half-life of radium-226 is 1600 years. Suppose we have a 22-mg sample. (a) Find a function that models the mass remaining after t years. (b) How much of the sample will remain after 4000 years? (c) After how long will only 18 mg of the sample remain? ✎ 15. Radioactive Cesium The half-life of cesium-137 is 30 years. Suppose we have a 10-g sample. (a) Find a function that models the mass remaining after t years. (b) How much of the sample will remain after 80 years? (c) After how long will only 2 g of the sample remain? 16. Radioactive Thorium The mass m t remaining after t days from a 40-g sample of thorium-234 is given by 1 2 40e0.0277t m t 1 2 (a) How much of the sample will remain after 60 days? (b) After how long will only 10 g of the sample remain? (c) Find the half-life of thorium-234. 17. Radioactive Strontium The half-life of strontium-90 is 28 years. How long will it take a 50-mg sample to decay to a mass of 32 mg? 18. Radioactive Radium Radium-221 has a half-life of 30 s. How long will it take for 95% of a sample to decay? 19. Finding Half-life If 250 mg of a radioactive element decays to 200 mg in 48 hours, ﬁnd the half-life of the element. ✎ 20. Radioactive Radon After 3 days a sample of radon-222 has decayed to 58% of its original amount. (a) What is the half-life of radon-222? (b) How long will it take the sample to decay to 20% of its original amount? 21. Carbon-14 Dating A wooden artifact from an ancient tomb contains 65% of the carbon-14 that is present in living trees. How long ago was the artifact made? (The half-
life of carbon14 is 5730 years.) 22. Carbon-14 Dating The burial cloth of an Egyptian mummy is estimated to contain 59% of the carbon-14 it contained originally. How long ago was the mummy buried? (The half-life of carbon-14 is 5730 years.) 23–26 ■ These exercises use Newton’s Law of Cooling. ✎ 23. Cooling Soup A hot bowl of soup is served at a dinner party. It starts to cool according to Newton’s Law of Cooling, so its temperature at time t is given by 65 145e0.05t T t 1 2 where t is measured in minutes and T is measured in F. (a) What is the initial temperature of the soup? (b) What is the temperature after 10 min? (c) After how long will the temperature be 100F? 24. Time of Death Newton’s Law of Cooling is used in homi- cide investigations to determine the time of death. The normal body temperature is 98.6 F. Immediately following death, the body begins to cool. It has been determined experimentally that the constant in Newton’s Law of Cooling is approximately k 0.1947, assuming that time is measured in hours. Suppose that the temperature of the surroundings is 60F. (a) Find a function t after death. that models the temperature t hours T 2 1 (b) If the temperature of the body is now 72F, how long ago was the time of death? 25. Cooling Turkey A roasted turkey is taken from an oven when its temperature has reached 185F and is placed on a table in a room where the temperature is 75F. (a) If the temperature of the turkey is 150F after half an hour, what is its temperature after 45 min? (b) When will the turkey cool to 100F? 26. Boiling Water A kettle full of water is brought to a boil in a room with temperature 20C. After 15 min the temperature of the water has decreased from 100C to 75C. Find the temperature after another 10 min. Illustrate by graphing the temperature function. 27–41 ■ These exercises deal with logarithmic scales. 27. Finding pH The hydrogen ion concentration of a sample of each substance is given. Calculate the pH of the substance. (a) Lemon juice: 3 (b) Tomato juice: 3 H (c) Seawater: 4 H 4 H 4 5
.0 109 M 5.0 103 M 3.2 104 M 3 28. Finding pH An unknown substance has a hydrogen ion con- H centration of 3 the substance as acidic or basic. 4 3.1 108 M. Find the pH and classify 29. Ion Concentration The pH reading of a sample of each substance is given. Calculate the hydrogen ion concentration of the substance. (a) Vinegar: pH 3.0 (b) Milk: pH 6.5 30. Ion Concentration The pH reading of a glass of liquid is given. Find the hydrogen ion concentration of the liquid. (a) Beer: pH 4.6 (b) Water: pH 7.3 31. Finding pH The hydrogen ion concentrations in cheeses range from 4.0 107 M to 1.6 105 M. Find the corresponding range of pH readings. ✎ ✎ 32. Ion Concentration in Wine The pH readings for wines vary from 2.8 to 3.8. Find the corresponding range of hydrogen ion concentrations. ✎ 33. Earthquake Magnitudes If one earthquake is 20 times as intense as another, how much larger is its magnitude on the Richter scale? 34. Earthquake Magnitudes The 1906 earthquake in San Francisco had a magnitude of 8.3 on the Richter scale. At the same time in Japan an earthquake with magnitude 4.9 caused only minor damage. How many times more intense was the San Francisco earthquake than the Japanese earthquake? 35. Earthquake Magnitudes The Alaska earthquake of 1964 had a magnitude of 8.6 on the Richter scale. How many times more intense was this than the 1906 San Francisco earthquake? (See Exercise 34.) 36. Earthquake Magnitudes The Northridge, California, earthquake of 1994 had a magnitude of 6.8 on the Richter scale. A year later, a 7.2-magnitude earthquake struck Kobe, Japan. How many times more intense was the Kobe earthquake than the Northridge earthquake? 37. Earthquake Magnitudes The 1985 Mexico City earthquake had a magnitude of 8.1 on the Richter scale. The 1976 earthquake in Tangshan, China, was 1.26 times as intense. What was the magnitude of the Tangshan earthquake? CHAPTER 5 \| Review 423 38. Subway Noise The intensity of the sound of a subway train was measured at 98 dB. Find the intensity in W/m2. 39. Trafﬁc Noise The intensity of the sound of trafﬁc at a busy intersection was measured
at 2.0 105 W/m2. Find the intensity level in decibels. 40. Comparing Decibel Levels The noise from a power mower was measured at 106 dB. The noise level at a rock concert was measured at 120 dB. Find the ratio of the intensity of the rock music to that of the power mower. 41. Inverse Square Law for Sound A law of physics states that the intensity of sound is inversely proportional to the square of the distance d from the source: I k/d 2. (a) Use this model and the equation B 10 log I I0 (described in this section) to show that the decibel levels B1 and B2 at distances d1 and d2 from a sound source are related by the equation B2 B1 20 log d1 d2 (b) The intensity level at a rock concert is 120 dB at a distance 2 m from the speakers. Find the intensity level at a distance of 10 m. CHAPTER 5 \| REVIEW ▼ P R O P E RTI LAS Exponential Functions (pp. 371–373) The exponential function f with base a (where deﬁned for all real numbers x by a 0, a 1 ) is f x ax The domain of f is has one of the following shapes, depending on the value of a: The graph of f 1 2 2, and the range of f is 1 0, q y 0 (0, 1) x y 0 (0, 1) x Ï=a˛ for a>1 Ï=a˛ for 0<a<1 The Natural Exponential Function (p. 375) The natural exponential function is the exponential function with base e: ex f x 1 2 The number e is deﬁned to be the number that the expression 1 1/n n q. 2 1 irrational number e is An approximate value for the approaches as n e 2.7182818284590 p Compound Interest (pp. 377–378) If a principal P is invested in an account paying an annual interest rate r, compounded n times a year, then after t years the amount A in the account is t 1 2 P A t 1 2 nt 1 r n If the interest is compounded continuously, then the amount is ¢ Pert A ≤ t 1 2 Logarithmic Functions (pp. 384–386) The logarithmic function is deﬁned for x 0 log
a with base a (where a 0, a 1 ) by loga x y 3 ay x is the exponent to which the base a must be raised to loga x So give y. 424 CHAPTER 5 \| Exponential and Logarithmic Functions The domain of is graph of the function loga 0, q 1 loga, and the range is has the following shape: 2. For y a 1, the Guidelines for Solving Logarithmic Equations (p. 404) 1. Isolate the logarithmic term(s) on one side of the equation, and use the Laws of Logarithms to combine logarithmic terms if necessary. 0 1 x y loga x, a 1 Common and Natural Logarithms (pp. 388–390) The logarithm function with base 10 is called the common logarithm and is denoted log. So log x log10 x The logarithm function with base e is called the natural logarithm and is denoted ln. So ln x loge x Properties of Logarithms (p. 390) 1. 2. loga a 1 x x aloga 4. loga 1 0 loga ax x 3. Laws of Logarithms (p. 394) a 0, a 1 Let a be a logarithm base any real numbers or algebraic expressions that represent real numbers, with and let A, B, and C be 1 B 0 A 0. Then: and, 2 2. Rewrite the equation in exponential form. 3. Solve for the variable. Exponential Growth Model (p. 412) A population experiences exponential growth if it can be modeled by the exponential function n t 1 2 n0ert n0 1 t is the population at time t, n where is the initial population (at time t = 0), and r is the relative growth rate (expressed as a proportion of the population). 2 Radioactive Decay Model (p. 416) If a radioactive substance with half-life h has initial mass m0, then at time t the mass of the substance that remains is modeled by the exponential function m t 2 1 m0ert m t 1 2 where r ln 2 h. Newton’s Law of Cooling (p. 417) If an object has an initial temperature that is than the surrounding temperature T of the object is modeled by the function D0 ekt Ts Ts, t t
2 1 where the constant T k 0 2 1 depends on the size and type of the object. D0 degrees warmer then at time t the temperature 1. 2. 3. loga1 loga1 loga1 AB 2 A/B AC 2 loga A loga B loga A loga B 2 C loga A Change of Base Formula (p. 397) logb x loga x loga b Guidelines for Solving Exponential Equations (p. 401) 1. Isolate the exponential term on one side of the equation. 2. Take the logarithm of each side, and use the Laws of Logarithms to “bring down the exponent.” 3. Solve for the variable. Logarithmic Scales (pp. 418–420) The pH scale measures the acidity of a solution: The Richter scale measures the intensity of earthquakes: pH log H 3 4 M log I S The decibel scale measures the intensity of sound: B 10 log I I0 ▼ CO N C E P T S U M MARY Section 5.1 ■ Evaluate exponential functions ■ Graph exponential functions ■ Evaluate and graph the natural exponential function ■ Find compound interest ■ Find continuously compounded interest Review Exercises 1–4 5–8, 73–74, 80 4, 13–14 89–90 89(d), 90(c), 92 Section 5.2 ■ Evaluate logarithmic functions ■ Change between logarithmic and exponential forms of an expression ■ Use basic properties of logarithms ■ Graph logarithmic functions ■ Use common and natural logarithms Section 5.3 ■ Use the Laws of Logarithms to evaluate logarithmic expressions ■ Use the Laws of Logarithms to expand logarithmic expressions ■ Use the Laws of Logarithms to combine logarithmic expressions ■ Use the Change of Base Formula Section 5.4 ■ Solve exponential equations ■ Solve logarithmic equations ■ Solve problems involving compound interest ■ Calculate annual percentage yield Section 5.5 ■ Find exponential models of population growth ■ Find exponential models of radioactive decay ■ Find models using Newton’s Law of Cooling ■ Use logarithmic scales (pH, Richter, and decibel scales) ▼ E X E RC I S E S 1–4 ■ Use a calculator to ﬁnd the indicated values of the exponential function, correct to three
decimal places. CHAPTER 5 \| Review 425 Review Exercises 29–44 21–28 30, 31, 33, 36 9–12, 15–16, 75–76 17–20, 23, 24, 27, 28 Review Exercises 29, 32, 34, 35, 37–44 45–50 51–56 83–86 Review Exercises 57–64, 69–72, 78 65–68, 77 91–92 93–94 Review Exercises 95–96, 101 97–100 102 103–106 25–28 ■ Write the equation in logarithmic form. 491/2 1 7 26 64 26. 25. 27. 10x 74 28. ek m 2 29–44 ■ Evaluate the expression without using a calculator. 1. 2. 3. 4 22, f 2 2.2 1.5 5x; f 1 3 # 2x; f 4 # 7 4e 2 3B A x1; g 1 x2.7, g 1 2 1, g 2 23 1 2, f 2.5 2 27 1, f 2 5..6 1 2 2 5–16 ■ Sketch the graph of the function. State the domain, range, and asymptote. 6. 8. 10. 7. 9 2x1 3 2x log31 x 1 2 log2 x ex 1 2 ln x f x 12. 11. 13 17–20 ■ Find the domain of the function. 1 x 15. 16. 14. g 2 1 2 2 x 3x2 5x 5 log x 2 1 3 log51 2 ex1 1 x2 ln 1 2 2 17. 19. f h x x 1 1 2 2 10x 2 log ln x2 4 1 2 1 2x 1 2 18. g 20. k log ln 0 1 x 21–24 ■ Write the equation in exponential form. log6 37 x 21. log2 1024 10 log x y 23. 22. 24. ln c 17 2 2 2 x 1 x 1 29. log2 128 31. 10log 45 33. 35. 37. ln e6 2 1 1 log3A 27 B log5 15 30. log8 1 32. log 0.000001 34. log4 8 36. 2log 213 38. e2ln7 x 4 2 39. log 25 log 4 41. 43. log2 1623 log8 6 log8 3 log8 2 40
. 42. log3 1243 log5 250 log5 2 44. log log10100 45–50 ■ Expand the logarithmic expression. 2 x x2 2 45. log 1 0 47. ln B AB2C3 2 x 2 1 x 2 1 46. log2 1 x 2x 2 1 2 48. log a y 2 4x 3 x 1 5 b 2 1 23 x 4 12 x2 49. log5 a 3/2 1 5x 1 2x3 x b 2 50. ln a 1 x 16 2 1x 3 b 51. 53. 54. 55. 56. 57. 59. 61. 63. 65. 66. 67. 68. 426 CHAPTER 5 \| Exponential and Logarithmic Functions 51–56 ■ Combine into a single logarithm. 52. log x log 3 log y x2y 1 2 log 6 4 log 2 x y 3 2 log2 1 2 log5 2 log5 1 x 2 log 2 x 4 1 ln 1 2 3 1 2 x2 y2 2 log2 1 x 1 1 3 log5 1 2 x 2 1 2 x2 4x log 1 5 ln 1 2 4 2 3x 7 2 x2 4 2 2 log 1 57–68 ■ Solve the equation. Find the exact solution if possible; otherwise, use a calculator to approximate to two decimals. 58. 60. 62. 64. 54x 1 125 1063x 18 e3x/4 10 32x 3x 6 0 32x7 27 23x5 7 41x 32x5 x2e2x 2xe2x 8e2x 1 x log21 2 log x log x 5 log81 2x 3 ln 1 4 x 1 2 log81 1 0 1 2 2 log 12 x 2 1 2 69–72 ■ Use a calculator to ﬁnd the solution of the equation, correct to six decimal places. 69. 71. 52x/3 0.63 52x1 34x1 70. 72. 23x5 7 e15k 10,000 73–76 ■ Draw a graph of the function and use it to determine the asymptotes and the local maximum and minimum values. x2 2 73. 75. 1 y ex/ y log x 3 x 1 2 74. 76. y 10x 5x y 2x2 ln x 77–78 ■ Find the solutions of the equation, correct to two
decimal places. 77. 3 log x 6 2x 78. 4 x2 e2x 79–80 ■ Solve the inequality graphically. 79. ln x x 2 81. Use a graph of f 80. ex 4x2 e x 3ex 4x to ﬁnd, approximately, x 1 2 the intervals on which f is increasing and on which f is decreasing. 82. Find an equation of the line shown in the ﬁgure. y 0 y=ln x e a x 83–86 ■ Use the Change of Base Formula to evaluate the logarithm, correct to six decimal places. 83. log4 15 84. log71 3 4 2 85. log9 0.28 86. log100 250 87. Which is larger, log4 258 or log5 620? 88. Find the inverse of the function f domain and range. 23x, and state its x 1 2 89. If $12,000 is invested at an interest rate of 10% per year, ﬁnd the amount of the investment at the end of 3 years for each compounding method. (a) Semiannual (c) Daily (b) Monthly (d) Continuous 90. A sum of $5000 is invested at an interest rate of % per year, 8 1 2 compounded semiannually. (a) Find the amount of the investment after (b) After what period of time will the investment amount years. 11 2 to $7000? (c) If interest were compounded continously instead of semiannually, how long would it take for the amount to grow to $7000? 91. A money market account pays 5.2% annual interest, compounded daily. If $100,000 is invested in this account, how long will it take for the account to accumulate $10,000 in interest? 92. A retirement savings plan pays 4.5% interest, compounded continuously. How long will it take for an investment in this plan to double? 93–94 ■ Determine the annual percentage yield (APY) for the given nominal annual interest rate and compounding frequency. 93. 4.25%; daily 94. 3.2%; monthly 95. The stray-cat population in a small town grows exponentially. In 1999 the town had 30 stray cats, and the relative growth rate was 15% per year. (a) Find a function that models the stray-cat population n t 1 2 after t years. (b) Find the
projected population after 4 years. (c) Find the number of years required for the stray-cat popula- tion to reach 500. 96. A culture contains 10,000 bacteria initially. After an hour the bacteria count is 25,000. (a) Find the doubling period. (b) Find the number of bacteria after 3 hours. 97. Uranium-234 has a half-life of 2.7 105 years. (a) Find the amount remaining from a 10-mg sample after a thousand years. (b) How long will it take this sample to decompose until its mass is 7 mg? 98. A sample of bismuth-210 decayed to 33% of its original mass after 8 days. (a) Find the half-life of this element. (b) Find the mass remaining after 12 days. 99. The half-life of radium-226 is 1590 years. (a) If a sample has a mass of 150 mg, ﬁnd a function that models the mass that remains after t years. (b) Find the mass that will remain after 1000 years. (c) After how many years will only 50 mg remain? CHAPTER 5 \| Review 427 102. A car engine runs at a temperature of 190F. When the engine is turned off, it cools according to Newton’s Law of Cooling with constant k 0.0341, where the time is measured in minutes. Find the time needed for the engine to cool to 90F if the surrounding temperature is 60F. 103. The hydrogen ion concentration of fresh egg whites was measured as H 3 4 1.3 108 M Find the pH, and classify the substance as acidic or basic. 104. The pH of lime juice is 1.9. Find the hydrogen ion concentration. 105. If one earthquake has magnitude 6.5 on the Richter scale, what is the magnitude of another quake that is 35 times as intense? 106. The drilling of a jackhammer was measured at 132 dB. The sound of whispering was measured at 28 dB. Find the ratio of the intensity of the drilling to that of the whispering. 100. The half-life of palladium-100 is 4 days. After 20 days a sample has been reduced to a mass of 0.375 g. (a) What was the initial mass of the sample? (b) Find a function that models the mass remaining after t days. (c) What is the mass after 3 days? (d
) After how many days will only 0.15 g remain? 101. The graph shows the population of a rare species of bird, where t represents years since 1999 and thousands. (a) Find a function that models the bird population at time t is measured in 2 1 n t n in the form. (b) What is the bird population expected to be in the year t 1 2 n0e rt 2010? n(t) 4000 3000 2000 1000 Bird population (5, 3200) 0 5432 1 Years since 1999 t ■ CHAPTER 5 \| TEST 1. Sketch the graph of each function, and state its domain, range, and asymptote. Show the x- and y-intercepts on the graph. (a) f x 1 2 2x 4 (b) g x 1 2 log31 x 3 2 2. (a) Write the equation 62x 25 in logarithmic form. (b) Write the equation ln A 3 in exponential form. 3. Find the exact value of each expression. (a) (c) (e) 10log 36 log3 127 log8 4 (b) (d) (f) ln e3 log2 80 log2 10 log6 4 log6 9 4. Use the Laws of Logarithms to expand the expression. 5. Combine into a single logarithm: 3 log B x4 ln x 2 ln x 2 x2 ln 1 3 x 4 2 (a) 6. Find the solution of the equation, correct to two decimal places. 4 log2 1 2x1 10 x3 6 10 3 x 2 x 2 5 ln 1 log2 1 (b) (d) (c) 2x 2 x 1 2 2 7. The initial size of a culture of bacteria is 1000. After one hour the bacteria count is 8000. (a) Find a function that models the population after t hours. (b) Find the population after 1.5 hours. (c) When will the population reach 15,000? (d) Sketch the graph of the population function. 8. Suppose that $12,000 is invested in a savings account paying 5.6% interest per year. (a) Write the formula for the amount in the account after t years if interest is compounded monthly. (b) Find the amount in the account after 3 years if interest is compounded daily. (c) How long will it take for the amount in the
account to grow to $20,000 if interest is com- pounded semiannually? 9. The half-life of krypton-91 (91Kr) is 10 seconds. At time t = 0 a heavy canister contains 3 g of this radioactive gas. (a) Find a function that models the amount A(t) of 91Kr remaining in the canister after t seconds. (b) How much 91Kr remains after one minute? (c) When will the amount of 91Kr remaining be reduced to 1 mg (1 microgram, or 106 g)? 10. An earthquake measuring 6.4 on the Richter scale struck Japan in July 2007, causing extensive damage. Earlier that year, a minor earthquake measuring 3.1 on the Richter scale was felt in parts of Pennsylvania. How many times more intense was the Japanese earthquake than the Pennsylvania earthquake? 428 ● CUMUL ATIVE REVIE W TEST \| CHAPTERS 3, 4, and 5 1. Let f x x2 4x (a) The domain of f 1 2 and (b) The domain of g 1x 4. Find each of the followingcd) (e) The average rate of change of g between x 5 and x 21 (f) 12 1, f 6 12,, g f, f 1 (g) The inverse of g 1 2 2 1 1 2 2 2. Let f x 1 2 (a) Evaluate 4 x 3 f, f 0 1 2 if x 2 if x 2, f 1 2 1 2 (b) Sketch the graph of f. b, f, and. Let f be the quadratic function f (a) Express f in standard form. 2x2 8x 5. x 1 2 (b) Find the maximum or minimum value of f. (c) Sketch the graph of f. (d) Find the interval on which f is increasing and the interval on which f is decreasing. (e) How is the graph of g h (f) How is the graph of of f? x x 1 1 2 2 2x2 8x 10 2 x 3 2 8 1 2 obtained from the graph of f? x 3 obtained from the graph 5 2 1 4. Without using a graphing calculator, match each of the following functions to the graphs below. Give reasons for your choices. f x 1 2 x s 1 2 x3 8x 2x 3 x2 x4
8x2 2x 5 2x 3 x2 9 2x. Let P x 2x3 11x2 10x 8. (a) List all possible rational zeros of P. (b) Determine which of the numbers you listed in part (a) actually are zeros of P. (c) Factor P completely. (d) Sketch a graph of P. 6. Let Q x x5 3x4 3x3 x2 4x 2. (a) Find all zeros of Q, real and complex, and state their multiplicities. (b) Factor Q completely. (c) Factor Q into linear and irreducible quadratic factors. r 7. Let 3x2 6x x2 x 2 x 1 2. Find the x- and y-intercepts and the horizontal and vertical asymptotes. Then sketch the graph of r. 429 430 CUMUL ATIVE REVIE W TEST \| Chapters 3, 4, and 5 8. Sketch graphs of the following functions on the same coordinate plane. (a) f x 1 2 2 ex 9. (a) Find the exact value of (b) g log3 16 2 log3 36. x 1 2 ln 1 x 1 2 (b) Use the Laws of Logarithms to expand the expression log x5 1x 1 a 2x 3 b 10. Solve the equations. log2 x log21 2e3x 11e2x 10ex 8 0 x 2 3 (b) (a) 2 [Hint: Compare to the polynomial in problem 5.] 11. A sum of $25,000 is deposited into an account paying 5.4% interest per year, compounded daily. (a) What will the amount in the account be after 3 years? (b) When will the account have grown to $35,000? (c) How long will it take for the initial deposit to double? 12. After a shipwreck, 120 rats manage to swim from the wreckage to a deserted island. The rat population on the island grows exponentially, and after 15 months there are 280 rats on the island. (a) Find a function that models the population t months after the arrival of the rats. (b) What will the population be 3 years after the shipwreck? (c) When will the population reach 2000 FITTING EXPONENTIAL AND POWER CURVES TO DATA In a previous Focus on Model
ing (page 364) we learned that the shape of a scatter plot helps us to choose the type of curve to use in modeling data. The ﬁrst plot in Figure 1 fairly screams for a line to be ﬁtted through it, and the second one points to a cubic polynomial. For the third plot it is tempting to ﬁt a second-degree polynomial. But what if an exponential curve ﬁts better? How do we decide this? In this section we learn how to ﬁt exponential and power curves to data and how to decide which type of curve ﬁts the data better. We also learn that for scatter plots like those in the last two plots in Figure 1, the data can be modeled by logarithmic or logistic functions. FIGURE 1 TABLE 1 World population Year (t) 1900 1910 1920 1930 1940 1950 1960 1970 1980 1990 2000 World population (P in millions) 1650 1750 1860 2070 2300 2520 3020 3700 4450 5300 6060 Modeling with Exponential Functions If a scatter plot shows that the data increase rapidly, we might want to model the data using an exponential model, that is, a function of the form Cekx f x 1 2 where C and k are constants. In the ﬁrst example we model world population by an exponential model. Recall from Section 5.5 that population tends to increase exponentially. E X AM P L E 1 \| An Exponential Model for World Population Table 1 gives the population of the world in the 20th century. (a) Draw a scatter plot, and note that a linear model is not appropriate. (b) Find an exponential function that models population growth. (c) Draw a graph of the function that you found together with the scatter plot. How well does the model ﬁt the data? (d) Use the model that you found to predict world population in the year 2020. ▼ SO LUTI O N (a) The scatter plot is shown in Figure 2. The plotted points do not appear to lie along a straight line, so a linear model is not appropriate. 6500 1900 1900 0 0 2000 2000 FIGURE 2 Scatter plot of world population 431 © 432 Focus On Modeling (b) Using a graphing calculator and the ExpReg command (see Figure 3(a)), we get the exponential model P t 0.0082543 1.0137186 t #
2 This is a model of the form y Cbt. To convert this to the form y Cekt, we use the properties of exponentials and logarithms as follows: 2 1 1 2 1 1.0137186t eln 1.0137186 t et ln 1.0137186 e0.013625t A = elnA ln AB = B ln A ln 1.0137186 0.013625 The population of the world increases exponentially 0.0082543e0.013625t P t 1 2 Thus, we can write the model as (c) From the graph in Figure 3(b) we see that the model appears to ﬁt the data fairly well. The period of relatively slow population growth is explained by the depression of the 1930s and the two world wars. 6500 1900 0 2000 (a) (b) FIGURE 3 Exponential model for world population (d) The model predicts that the world population in 2020 will be 2020 P 1 2 0.013625 0.0082543e1 7,405,400,000 2020 2 2 1 ▲ Modeling with Power Functions If the scatter plot of the data we are studying resembles the graph of y ax 2, y ax1.32, or some other power function, then we seek a power model, that is, a function of the form ax n f x 1 2 Mercury Sun Mars Earth Venus Saturn Jupiter where a is a positive constant and n is any real number. In the next example we seek a power model for some astronomical data. In astronomy, distance in the solar system is often measured in astronomical units. An astronomical unit (AU) is the mean distance from the earth to the sun. The period of a planet is the time it takes the planet to make a complete revolution around the sun (measured in earth years). In this example we derive the remarkable relationship, ﬁrst discovered by Johannes Kepler (see page 585), between the mean distance of a planet from the sun and its period. E X AM P L E 2 \| A Power Model for Planetary Periods Table 2 gives the mean distance d of each planet from the sun in astronomical units and its period T in years. TABLE 2 Distances and periods of the planets Planet Mercury Venus Earth Mars Jupiter Saturn Uranus Neptune Pluto d 0.387 0.723 1.000 1.523 5.203 9.541 19.190
30.086 39.507 FIGURE 4 Scatter plot of planetary data Fitting Exponential and Power Curves to Data 433 (a) Sketch a scatter plot. Is a linear model appropriate? (b) Find a power function that models the data. (c) Draw a graph of the function you found and the scatter plot on the same graph. How T well does the model ﬁt the data? 0.241 0.615 1.000 1.881 11.861 29.457 84.008 164.784 248.350 (d) Use the model that you found to calculate the period of an asteroid whose mean dis- tance from the sun is 5 AU. ▼ SO LUTI O N (a) The scatter plot shown in Figure 4 indicates that the plotted points do not lie along a straight line, so a linear model is not appropriate. 260 0 0 45 (b) Using a graphing calculator and the PwrReg command (see Figure 5(a)), we get the power model T 1.000396d1.49966 If we round both the coefﬁcient and the exponent to three signiﬁcant ﬁgures, we can write the model as T d1.5 This is the relationship discovered by Kepler (see page 585). Sir Isaac Newton later used his Law of Gravity to derive this relationship theoretically, thereby providing strong scientiﬁc evidence that the Law of Gravity must be true. (c) The graph is shown in Figure 5(b). The model appears to ﬁt the data very well. 260 0 0 45 (a) (b) FIGURE 5 Power model for planetary data (d) In this case d 5 AU, so our model gives T 1.00039 # 51.49966 11.22 The period of the asteroid is about 11.2 years. ▲ Linearizing Data We have used the shape of a scatter plot to decide which type of model to use: linear, exponential, or power. This works well if the data points lie on a straight line. But it’s difﬁcult to distinguish a scatter plot that is exponential from one that requires a power model. 434 Focus On Modeling So to help decide which model to use, we can linearize the data, that is, apply a function that “straightens” the scatter plot. The inverse of the linearizing function is then an
appropriate model. We now describe how to linearize data that can be modeled by exponential or power functions. ■ LINEARIZING EXPONENTIAL DATA If we suspect that the data points 1 x, y 2 lie on an exponential curve y Ce kx, then the points x, ln y 1 2 should lie on a straight line. We can see this from the following calculations: ln y ln Cekx Assume that y = Cekx and take ln ln ekx ln C kx ln C Property of ln Property of ln TABLE 3 World population data To see that ln y is a linear function of x, let Y ln y and A ln C; then t 1900 1910 1920 1930 1940 1950 1960 1970 1980 1990 2000 Population P (in millions) 1650 1750 1860 2070 2300 2520 3020 3700 4450 5300 6060 ln P 21.224 21.283 21.344 21.451 21.556 21.648 21.829 22.032 22.216 22.391 22.525 TABLE 4 Log-log table ln d 0.94933 0.32435 0 0.42068 1.6492 2.2556 2.9544 3.4041 3.6765 ln T 1.4230 0.48613 0 0.6318 2.4733 3.3829 4.4309 5.1046 5.5148 Y kx A We apply this technique to the world population data in Table 3. The scatter plot in Figure 6 shows that the linearized data lie approximately on a straight line, so an exponential model should be appropriate. to obtain the points t, ln P t, P 2 1 2 1 23 FIGURE 6 1900 21 2010 ■ LINEARIZING POWER DATA If we suspect that the data points x, y 1 2 lie on a power curve y ax n, then the points ln x, ln y 1 2 should be on a straight line. We can see this from the following calculations: ln y ln ax n Assume that y = axn and take ln ln a ln x n ln a n ln x Property of ln Property of ln To see that ln y is a linear function of ln x, let Y ln y, X ln x, and A ln a; then Y nX A We apply this technique to the planetary
data ln d, ln T 1 so a power model seems appropriate. in Table 2 to obtain the points 1 in Table 4. The scatter plot in Figure 7 shows that the data lie on a straight line, d, T 2 2 6 _2 4 _2 FIGURE 7 Log-log plot of data in Table 4 Fitting Exponential and Power Curves to Data 435 An Exponential or Power Model? x, y Suppose that a scatter plot of the data points shows a rapid increase. Should we use 1 an exponential function or a power function to model the data? To help us decide, we draw. If the two scatter plots: one for the points ﬁrst scatter plot appears to lie along a line, then an exponential model is appropriate. If the second plot appears to lie along a line, then a power model is appropriate. and the other for the points ln x, ln y x, ln y 2 2 1 1 2 E X AM P L E 3 \| An Exponential or Power Model? 1 x, y are shown in Table 5. Data points 2 (a) Draw a scatter plot of the data. ln x, ln y (b) Draw scatter plots of (c) Is an exponential function or a power function appropriate for modeling this data? (d) Find an appropriate function to model the data. x, ln y and. 2 1 2 1 ▼ SO LUTI O N (a) The scatter plot of the data is shown in Figure 8. 140 TABLE 10 y 2 6 14 22 34 46 64 80 102 130 TABLE 6 (b) We use the values from Table 6 to graph the scatter plots in Figures 9 and 10. FIGURE 8 0 0 11 10 140 0 0 FIGURE 11 ln x ln y 6 5 0 0.7 1.1 1.4 1.6 1.8 1.9 2.1 2.2 2.3 0.7 1.8 2.6 3.1 3.5 3.8 4.2 4.4 4.6 4.9 11 0 0 FIGURE 9 11 0 0 FIGURE 10 2.5 (c) The scatter plot of x, ln y in Figure 9 does not appear to be linear, so an exponen- tial model is not appropriate. On the other hand, the scatter plot of ure 10 is very nearly linear, so a power model is appropriate. ln x, ln y 1
2 in Fig- 1 2 (d) Using the PwrReg command on a graphing calculator, we ﬁnd that the power func- tion that best ﬁts the data point is y 1.85x1.82 The graph of this function and the original data points are shown in Figure 11. ▲ Before graphing calculators and statistical software became common, exponential and power models for data were often constructed by ﬁrst ﬁnding a linear model for the linearized data. Then the model for the actual data was found by taking exponentials. For instance, if we ﬁnd that ln y A ln x B, then by taking exponentials we get the model y eB e A ln x, or y Cx A (where C eB). Special graphing paper called “log paper” or “loglog paper” was used to facilitate this process. 436 Focus On Modeling TABLE 7 Week Catﬁsh 0 15 30 45 60 75 90 105 120 1000 1500 3300 4400 6100 6900 7100 7800 7900 Modeling with Logistic Functions A logistic growth model is a function of the form f t 1 2 c 1 aebt where a, b, and c are positive constants. Logistic functions are used to model populations where the growth is constrained by available resources. (See Exercises 63–66 of Section 5.1.) E X AM P L E 4 \| Stocking a Pond with Catﬁsh Much of the ﬁsh that is sold in supermarkets today is raised on commercial ﬁsh farms, not caught in the wild. A pond on one such farm is initially stocked with 1000 catﬁsh, and the ﬁsh population is then sampled at 15-week intervals to estimate its size. The population data are given in Table 7. (a) Find an appropriate model for the data. (b) Make a scatter plot of the data and graph the model that you found in part (a) on the scatter plot. (c) How does the model predict that the ﬁsh population will change with time? ▼ SO LUTI O N (a) Since the catﬁsh population is restricted by its habitat (the pond), a logistic model is appropriate. Using the Logistic command on a calculator (see Figure 12(a)), we ﬁnd the following
model for the catﬁsh population P t : 1 2 P t 1 2 7925 1 7.7e0.052t 9000 FIGURE 12 (a) 0 180 0 (b) Catfish population y = P(t) (b) The scatter plot and the logistic curve are shown in Figure 12(b). (c) From the graph of P in Figure 12(b) we see that the catﬁsh population increases rapidly until about t 80 weeks. Then growth slows down, and at about t 120 weeks the population levels off and remains more or less constant at slightly over 7900. ▲ The behavior that is exhibited by the catﬁsh population in Example 4 is typical of logistic growth. After a rapid growth phase, the population approaches a constant level called the carrying capacity of the environment. This occurs because as t q, we have ebt 0 (see Section 5.1), and so P t 1 2 c 1 aebt c 1 0 c Thus, the carrying capacity is c. Fitting Exponential and Power Curves to Data 437 Problems 1. U.S. Population The U.S. Constitution requires a census every 10 years. The census data for 1790–2000 are given in the table. (a) Make a scatter plot of the data. (b) Use a calculator to ﬁnd an exponential model for the data. (c) Use your model to predict the population at the 2010 census. (d) Use your model to estimate the population in 1965. (e) Compare your answers from parts (c) and (d) to the values in the table. Do you think an exponential model is appropriate for these data? Year 1790 1800 1810 1820 1830 1840 1850 1860 Population (in millions) 3.9 5.3 7.2 9.6 12.9 17.1 23.2 31.4 Year 1870 1880 1890 1900 1910 1920 1930 1940 Population (in millions) 38.6 50.2 63.0 76.2 92.2 106.0 123.2 132.2 Year 1950 1960 1970 1980 1990 2000 Population (in millions) 151.3 179.3 203.3 226.5 248.7 281.4 2. A Falling Ball In a physics experiment a lead ball is dropped from a height of 5 m. The students record the distance the ball has fallen every one-tenth of a second. (This can be done by using a camera
and a strobe light.) (a) Make a scatter plot of the data. (b) Use a calculator to ﬁnd a power model. (c) Use your model to predict how far a dropped ball would fall in 3 s. 3. Health-Care Expenditures The U.S. health-care expenditures for 1970–2001 are given in the table below, and a scatter plot of the data is shown in the ﬁgure. (a) Does the scatter plot shown suggest an exponential model? (b) Make a table of the values linear? t, ln E 1 2 and a scatter plot. Does the scatter plot appear to be (c) Find the regression line for the data in part (b). (d) Use the results of part (c) to ﬁnd an exponential model for the growth of health-care expenditures. (e) Use your model to predict the total health-care expenditures in 2009. U.S. health-care expenditures (in billions of dollars) E 1400 1200 1000 800 600 400 200 1970 1980 1990 Year 2000 t Time Distance (s) 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 1.0 (m) 0.048 0.197 0.441 0.882 1.227 1.765 2.401 3.136 3.969 4.902 Health-Care expenditures (in billions of dollars) 74.3 251.1 434.5 506.2 696.6 820.3 937.2 1039.4 1150.0 1310.0 1424.5 Year 1970 1980 1985 1987 1990 1992 1994 1996 1998 2000 2001 438 Focus On Modeling Time (h) Amount of 131I g 1 2 0 8 16 24 32 40 48 4.80 4.66 4.51 4.39 4.29 4.14 4.04 4. Half-Life of Radioactive Iodine A student is trying to determine the half-life of radioactive iodine-131. He measures the amount of iodine-131 in a sample solution every 8 hours. His data are shown in the table in the margin. (a) Make a scatter plot of the data. (b) Use a calculator to ﬁnd an exponential model. (c) Use your model to ﬁnd the half-life of iodine-131. 5. The Beer-Lambert
Law As sunlight passes through the waters of lakes and oceans, the light is absorbed, and the deeper it penetrates, the more its intensity diminishes. The light intensity I at depth x is given by the Beer-Lambert Law: I I0ekx where I0 is the light intensity at the surface and k is a constant that depends on the murkiness of the water (see page 406). A biologist uses a photometer to investigate light penetration in a northern lake, obtaining the data in the table. (a) Use a graphing calculator to ﬁnd an exponential function of the form given by the BeerLambert Law to model these data. What is the light intensity I0 at the surface on this day, [Hint: If your calculator gives you and what is the “murkiness” constant k for this lake? a function of the form I abx, convert this to the form you want using the identities bx eln 1. See Example 1(b).] 2 e x ln b bx Light intensity decreases exponentially with depth. (b) Make a scatter plot of the data, and graph the function that you found in part (a) on your scatter plot. (c) If the light intensity drops below 0.15 lumens (lm), a certain species of algae can’t survive because photosynthesis is impossible. Use your model from part (a) to determine the depth below which there is insufﬁcient light to support this algae. Depth (ft) Light intensity (lm) Depth (ft) Light intensity (lm) 5 10 15 20 13.0 7.6 4.5 2.7 25 30 35 40 1.8 1.1 0.5 0.3 6. Experimenting with “Forgetting” Curves Every one of us is all too familiar with the phenomenon of forgetting. Facts that we clearly understood at the time we ﬁrst learned them sometimes fade from our memory by the time the ﬁnal exam rolls around. Psychologists have proposed several ways to model this process. One such model is Ebbinghaus’ Law of Forgetting, described on pages 396–397. Other models use exponential or logarithmic functions. To develop her own model, a psychologist performs an experiment on a group of volunteers by asking them to memorize a list of 100 related words. She then tests how many of these words they can recall
after various periods of time. The average results for the group are shown in the table. (a) Use a graphing calculator to ﬁnd a power function of the form y at b that models the average number of words y that the volunteers remember after t hours. Then ﬁnd an exponential function of the form y abt to model the data. (b) Make a scatter plot of the data, and graph both the functions that you found in part (a) on your scatter plot. (c) Which of the two functions seems to provide the better model? Time Words recalled 15 min 1 h 8 h 1 day 2 days 3 days 5 days 64.3 45.1 37.3 32.8 26.9 25.6 22.9 Fitting Exponential and Power Curves to Data 439 7. Lead Emissions The table below gives U.S. lead emissions into the environment in millions of metric tons for 1970–1992. (a) Find an exponential model for these data. (Use t 0 for the year 1970.) (b) Find a fourth-degree polynomial model for these data. (c) Which of these curves gives a better model for the data? Use graphs of the two models to decide. (d) Use each model to estimate the lead emissions in 1972 and 1982. Year 1970 1975 1980 1985 1988 1989 1990 1991 1992 Lead emissions 199.1 143.8 68.0 18.3 5.9 5.5 5.1 4.5 4.7 8. Auto Exhaust Emissions A study by the U.S. Ofﬁce of Science and Technology in 1972 estimated the cost of reducing automobile emissions by certain percentages. Find an exponential model that captures the “diminishing returns” trend of these data shown in the table below. Reduction in emissions (%) Cost per car ($) 50 55 60 65 70 75 80 85 90 95 45 55 62 70 80 90 100 200 375 600 9. Exponential or Power Model? Data points (a) Draw a scatter plot of the data. x, y 1 2 are shown in the table. (b) Draw scatter plots of x, ln y and ln x, ln y. 1 2 1 2 (c) Which is more appropriate for modeling this data: an exponential function or a power function? (d) Find an appropriate function to model the data. x 2 4 6 8 10 12 14 16 y 0.08 0.12 0.18 0
.25 0.36 0.52 0.73 1.06 440 Focus On Modeling x 10 20 30 40 50 60 70 80 90 y 29 82 151 235 330 430 546 669 797 Time (days) Number of ﬂies 0 2 4 6 8 10 12 16 18 10 25 66 144 262 374 446 492 498 10. Exponential or Power Model? Data points (a) Draw a scatter plot of the data. x, y 1 2 are shown in the table in the margin. (b) Draw scatter plots of x, ln y and ln x, ln y. 1 2 1 2 (c) Which is more appropriate for modeling this data: an exponential function or a power function? (d) Find an appropriate function to model the data. 11. Logistic Population Growth The table and scatter plot give the population of black ﬂies in a closed laboratory container over an 18-day period. (a) Use the Logistic command on your calculator to ﬁnd a logistic model for these data. (b) Use the model to estimate the time when there were 400 ﬂies in the container. Number of flies N 500 400 300 200 100 0 2 4 6 8 12 14 16 18 t 10 Days 12. Logarithmic Models A logarithmic model is a function of the form y a b ln x Many relationships between variables in the real world can be modeled by this type of function. The table and the scatter plot show the coal production (in metric tons) from a small mine in northern British Columbia. (a) Use the LnReg command on your calculator to ﬁnd a logarithmic model for these pro- duction ﬁgures. (b) Use the model to predict coal production from this mine in 2010. Year 1950 1960 1970 1980 1990 2000 Metric tons of coal 882 889 894 899 905 909 C 905 900 895 890 885 Metric tons of coal 1940 1960 1980 2000 t Year 6.1 Systems of Equations 6.2 Systems of Linear Equations in Two Variables 6.3 Systems of Linear Equations in Several Variables 6.4 Partial Fractions 6.5 Systems of Inequalities © CHAPTER 6 Systems of Equations and Inequalities Do you know where you are? Most of the time, we pretty much know where we are—in town, at the library, in
the classroom, and so on. But if we get lost during a hike in the woods, it’s not enough just to know that we’re in the woods. To get back home, we would need to know exactly where we are. Fortunately, with modern technology we can carry a small handheld GPS (Global Positioning System) device that can pinpoint our exact location. At the heart of its operation, the GPS device solves a system of equations. The device measures its distance from three GPS satellites, thus determining three equations. These equations work together to pinpoint the latitude, longitude, and elevation of the device (see Exercise 61 of Section 6.1). We will see in this chapter how systems of equations are also useful in solving many other problems in which several equations are needed to model a real-world situation. 441441 441 442 CHAPTER 6 \| Systems of Equations and Inequalities 6.1 Systems of Equations LEARNING OBJECTIVES After completing this section, you will be able to: ■ Solve a system of equations using the substitution method ■ Solve a system of equations using the elimination method ■ Solve a system of equations using the graphical method We have already seen how a real-world situation can be modeled by an equation (Section 1.2). But many such situations involve too many variables to be modeled by a single equation. For example, weather depends on the relationship among many variables, including temperature, wind speed, air pressure, and humidity. So to model (and forecast) the weather, scientists use many equations, each having many variables. Such collections of equations, called systems of equations, work together to describe the weather. Systems of equations with hundreds or even thousands of variables are used extensively by airlines to establish consistent ﬂight schedules and by telecommunications companies to ﬁnd efﬁcient routings for telephone calls. In this chapter we learn how to solve systems of equations that consist of several equations in several variables. In this section we learn how to solve systems of two equations in two variables. We learn three different methods of solving such systems: by substitution, by elimination, and graphically. ■ Systems of Equations and Their Solutions A system of equations is a set of equations that involve the same variables. A solution of a system is an assignment of values for the variables that makes each equation in the system true. To solve a system means to ﬁnd all solutions of the system. Here is an example of a system
of two equations in two variables: 2x y 5 x 4y 7 Equation 1 Equation 2 We can check that x 3 and y 1 is a solution of this system. b Equation 1 2x y 5 1 5 3 2 ✔ Equation 2 x 4y The solution can also be written as the ordered pair 1 1 3, 1 Note that the graphs of Equations 1 and 2 are lines (see Figure 1). Since the solution lies on each line. So it is the point of inter- satisﬁes each equation, the point 3, 1 3, 1 1 section of the two lines. 2 1 2 y x+4y=7 1 (3, 1) 0 1 3 x 2x-y=5 FIGURE 1 SE CTI O N 6.1 \| Systems of Equations 443 ■ Substitution Method In the substitution method we start with one equation in the system and solve for one variable in terms of the other variable. The following box describes the procedure. SUBSTITUTION METHOD 1. Solve for One Variable. Choose one equation, and solve for one variable in terms of the other variable. 2. Substitute. Substitute the expression you found in Step 1 into the other equa- tion to get an equation in one variable, then solve for that variable. 3. Back-Substitute. Substitute the value you found in Step 2 back into the expression found in Step 1 to solve for the remaining variable. E X AM P L E 1 \| Substitution Method Find all solutions of the system. 2x y 1 3x 4y 14 Equation 1 Equation 2 Solve for one variable b y 1 2x Solve for y in Equation 1 ▼ SO LUTI O N We solve for y in the ﬁrst equation. Substitute Back-substitute Substitute y 1 2x into Equation 2 2 1 3x 4 1 2x Now we substitute for y in the second equation and solve for x: 14 3x 4 8x 14 5x 4 14 5x 10 x 2 Next we back-substitute x 2 into the equation y 1 2x: Subtract 4 Solve for x Simplify Expand y 1 2 Thus, x 2 and y 5, so the solution is the ordered pair 2, 5 the graphs of the two equations intersect at the point 5 2 1 2 Back-substitute 2, 5 2 1.. Figure 2 shows that 1 2
Check Your Answer x 2, y 5: y 2 1 2 3 2 5 1 14 5 2 4 1 2 1 2 b ✔ (-2, 5) 3x+4y=14 2x+y=1 1 0 1 x FIGURE 2 ✎ Practice what you’ve learned: Do Exercise 5. ▲ 444 CHAPTER 6 \| Systems of Equations and Inequalities E X AM P L E 2 \| Substitution Method Find all solutions of the system. x2 y2 100 3x y 10 Equation 1 Equation 2 ▼ SO LUTI O N We start by solving for y in the second equation. Solve for one variable y 3x 10 b Solve for y in Equation 2 Next we substitute for y in the ﬁrst equation and solve for x: Substitute 2 x x 2 3x 10 2 1 9x2 60x 100 2 100 100 1 10x 2 2 60x 0 x 6 0 1 x 0 or x 6 10x 2 Substitute y 3x 10 into Equation 1 Expand Simplify Factor Solve for x Now we back-substitute these values of x into the equation y 3x 10. Back-substitute For x 0: y 3 For x 6: y 3 10 10 10 8 Back-substitute Back-substitute 0 1 2 6 2 1 and So we have two solutions:. 2 The graph of the ﬁrst equation is a circle, and the graph of the second equation is a line; 6, 8 1 2 1 0, 10 Figure 3 shows that the graphs intersect at the two points 1 0, 10 and 6, 8. 2 1 2 Check Your Answers x 0, y 10 : 2 10 2 10 2 0 2 x 6, y 8: 0 1 3 2 1 1 1 2 100 10 2 36 64 100 18 8 10 ≈+¥=100 6 0 (6, 8) 6 x FIGURE 3 3x-y=10 (0, _10) ✎ Practice what you’ve learned: Do Exercise 11. ▲ ■ Elimination Method To solve a system using the elimination method, we try to combine the equations using sums or differences so as to eliminate one of the variables. ELIMINATION METHOD 1. Adjust the Coefﬁcients. Multiply one or more of the equations by appropriate numbers so that the coefﬁcient of one variable in one equation
is the negative of its coefﬁcient in the other equation. 2. Add the Equations. Add the two equations to eliminate one variable, then solve for the remaining variable. 3. Back-Substitute. Substitute the value that you found in Step 2 back into one of the original equations, and solve for the remaining variable. SE CTI O N 6.1 \| Systems of Equations 445 E X AM P L E 3 \| Elimination Method Find all solutions of the system. 3x 2y 14 0x 2y 2 Equation 1 Equation 2 ▼ SO LUTI O N Since the coefﬁcients of the y-terms are negatives of each other, we can add the equations to eliminate y. b 3x 2y 14 0x 2y 2 4x 16 x 4 b System Add Solve for x 3x+2y=14 Now we back-substitute x 4 into one of the original equations and solve for y. Let’s choose the second equation because it looks simpler. y 7 1 0 (4, 1) 1 x-2y=2 x FIGURE 4 y 3≈+2y=26 5 0 2 x 5≈+7y=3 (_4, _11) (4, _11) x 2y 2 4 2y 2 Equation 2 Back-substitute x = 4 into Equation 2 2y 2 y 1 Subtract 4 Solve for y 4, 1 2 4, 1 The solution is 1 sect at the point ✎ Practice what you’ve learned: Do Exercise 15.. Figure 4 shows that the graphs of the equations in the system inter. ▲ 2 1 E X AM P L E 4 \| Elimination Method Find all solutions of the system. 3x 2 2y 26 5x 2 7y 3 Equation 1 Equation 2 ▼ SO LUTI O N We choose to eliminate the x-term, so we multiply the ﬁrst equation by 5 and the second equation by 3. Then we add the two equations and solve for y. b 15x 15x 2 10y 130 2 21y 9 5 Equation 1 (3) Equation 2 11y 121 y 11 Add Solve for y b Now we back-substitute y 11 into one of the original equations, say and solve for x: 3x 2 2y 26, 1 3x 11 2 2 26 2 2
48 3x 2 16 x 4 or x 4 x Back-substitute y 11 into Equation 1 Add 22 Divide by 3 Solve for x So we have two solutions: 4, 11 The graphs of both equations are parabolas (see Section 4.1). Figure 5 shows that the 4, 11 and. 1 2 1 4, 11 2 4, 11. 2 and 1 2 FIGURE 5 graphs intersect at the two points 1 446 CHAPTER 6 \| Systems of Equations and Inequalities Check Your Answers x 4, y 11: x 4, y 11 11 11 1 1 2 2 26 11 11 1 1 2 2 26 3 ✔ ✎ Practice what you’ve learned: Do Exercise 21. b b ▲ ■ Graphical Method In the graphical method we use a graphing device to solve the system of equations. Note that with many graphing devices, any equation must ﬁrst be expressed in terms of one or more functions of the form before we can use the calculator to graph it. Not all equations can be readily expressed in this way, so not all systems can be solved by this method. y f x 1 2 GRAPHICAL METHOD 1. Graph Each Equation. Express each equation in a form suitable for the graphing calculator by solving for y as a function of x. Graph the equations on the same screen. 2. Find the Intersection Points. The solutions are the x- and y-coordinates of the points of intersection. It may be more convenient to solve for x in terms of y in the equations. In that case, in Step 1 graph x as a function of y instead. E X AM P L E 5 \| Graphical Method Find all solutions of the system. 2 y 2 x 2x y 1 Graph each equation ▼ SO LUTI O N Solving for y in terms of x, we get the equivalent system b y x 2 2 y 2x 1 Find intersection points Figure 6 shows that the graphs of these equations intersect at two points. Zooming in, we see that the solutions are 8 1 2x-y=_1 (3, 7) Check Your Answers b 1, 1 and 2 3, 7 1 2 _3 4 ≈-y=2 (_1, _1) _3 FIGURE 6 x 1, y 1, y 7: 2 1 2 2 ✔ 32 7 2 7 1 3 2 2 1 ✔ ✎ Practice what you’ve learned: Do Exercise 47
. b b ▲ SE CTI O N 6.1 \| Systems of Equations 447 E X AM P L E 6 \| Solving a System of Equations Graphically Find all solutions of the system, correct to one decimal place. Graph each equation x 2 y y 2x 2 12 2 5x Equation 1 Equation 2 ▼ SO LUTI O N The graph of the ﬁrst equation is a circle, and the graph of the second is a parabola. To graph the circle on a graphing calculator, we must ﬁrst solve for y in terms of x (see Section 2.3). b 2 y x 2 12 2 12 x y 2 Isolate y2 on LHS y 212 x 2 Take square roots To graph the circle, we must graph both functions: Find intersection points y 212 x 2 and y 212 x 2 In Figure 7 the graph of the circle is shown in red and the parabola in blue. The graphs intersect in Quadrants I and II. Zooming in, or using the Intersect command, we see that. There also appears to be an and the intersection points are 2 intersection point in quadrant IV. However, when we zoom in, we see that the curves come close to each other but don’t intersect (see Figure 8). Thus, the system has two solutions; correct to the nearest tenth, they are 0.559, 3.419 1 2.847, 1.974 1 2 5 5 0.6, 3.4 and 2 2.8, 2.0 1 2 1 _7 7 _7 7 Intersection X=-.5588296 Y=3.4187292 Intersection X=2.8467004 Y=1.973904 _5 (a) _5 (b) _2 0.5 _4 FIGURE 7 x2 y2 12, y 2x2 5x FIGURE 8 Zooming in ✎ Practice what you’ve learned: Do Exercise 51. 2.0 ▲ Text not available due to copyright restrictions 448 CHAPTER 6 \| Systems of Equations and Inequalities 6. ▼ CONCE PTS 1. A set of equations involving the same variables is called a of equations. 2. The system of equations 2x 3y 7 5x y 9 is a system of two equations in the two variables and this system, we check whether. To determine whether x 5 b 5, 1 1
and is a solution of satisfy each 2 y 1 in the system. Which of the following are solutions of this system? 5, 1 1, 3, 2 1, 2 2, 1 1 2 1 3. A system of equations in two variables can be solved by the or the method, the method. 4. The system of equations method, ✎ 11 13. x y 2 0 b 2x 5y 2 75 12. x 2 y 9 x y 3 0 14. x 2 y 1 b 2x 2 3y 17 b 15–24 ■ Use the elimination method to ﬁnd all solutions of the system of equations. b ✎ 15. 3x 4y 10 4x 4y 2 17. x 2y 5 b 2x 3y 8 19. x 2 2y 01 b x 2 5y 29 ✎ 21. 3x2 y2 11 b x2 4y2 8 16. 2x 5y 15 4x 4y 21 18. 4x 3y 11 b 8x 4y 12 20. 3x2 4y 17 b 2x2 5y 2 22. 2x2 4y 13 b x2 y2 7 2 2y x 2 0 y x 4 23. x y2 3 0 b 2x2 y2 4 0 24. x2 y2 1 b 2x2 y2 x 3 is graphed below. (a) Use the graph to ﬁnd the solution(s) of the system. (b) Check that the solutions you found in part (a) satisfy the b system. y 1 0 1 x ▼ SKI LLS 5–14 ■ Use the substitution method to ﬁnd all solutions of the system of equations. 0x 0y 81 4x 3y 18 3x 0y 1 5x 2y 1 6. 5. ✎ 7. 9. x y 2 2x 3y 9 b y x 2 y x 12 b 8. 2x y 7 x 2y 2 b 10. x2 y2 25 y 2x b b b b 25–30 ■ Two equations and their graphs are given. Find the intersection point(s) of the graphs by solving the system. b 25. 2x y 1 x 2y 8 26. x y 2 2x 27. x 2 y 8 x 2y 6 28 SE CTI O N 6.1 \| Systems of Equations 449 29. x 2 y 0 x 3 2x y 0
30. x 2 y 2 4x x y 2 ✎ 51. • 1 y 2 18 x 2 9 y x 2 6x2 b y 0 11 x b y 1 0 1 x 52. x2 y2 3 y x2 2x 8 53. x4 16y4 32 x2 2x y 0 b 54. y ex ex y 5 x2 b 31–44 ■ Find all solutions of the system of equations. 31. y x 2 4x y 4x 16 33. x 2y 2 y 2 x 2 2x 4 b 35. x y 4 b xy 12 37. x 2y 16 b x 2 4y 16 0 39 41. 2x 2 8y 3 19 b 4x 2 16y 3 34 43 32. x y 2 0 y x 2 0 34 36. xy 24 b 2x 2 y 2 4 0 38. x 1y 0 y 2 4x 2 12 b 40. x 2 2y 2 2 2x 2 3y 15 b 42. x4 y3 17 b 3x4 5y3 53 44. b µ 4 x2 1 x2 6 y4 2 y4 7 2 0 45–54 ■ Use the graphical method to ﬁnd all solutions of the system of equations, correct to two decimal places. 45. 46. ✎ 47. y 2x 6 y x 5 y 2x 12 y x 3 y x2 8x y 2x 16 b b 48. y x2 4x 2x y 2 b 49. x2 y2 25 x 3y 2 b 50. x 2 y 2 17 x 2 2x y 2 13 b b b ▼ APPLICATIONS 55. Dimensions of a Rectangle A rectangle has an area of 180 cm2 and a perimeter of 54 cm. What are its dimensions? 56. Legs of a Right Triangle A right triangle has an area of 84 ft2 and a hypotenuse 25 ft long. What are the lengths of its other two sides? 57. Dimensions of a Rectangle The perimeter of a rectangle is 70, and its diagonal is 25. Find its length and width. 58. Dimensions of a Rectangle A circular piece of sheet metal has a diameter of 20 in. The edges are to be cut off to form a rectangle of area 160 in2 (see the ﬁgure). What are the dimensions of the rectangle? 59. Flight of a Rocket A hill is inclined so that its “slope”
1 2 is, as shown in the ﬁgure. We introduce a coordinate system with the origin at the base of the hill and with the scales on the axes measured in meters. A rocket is ﬁred from the base of the hill in such a way that its trajectory is the parabola y x2 401x. At what point does the rocket strike the hillside? How far is this point from the base of the hill (to the nearest centimeter)? y 0 x rise run rise run = 1 2 60. Making a Stovepipe A rectangular piece of sheet metal with an area of 1200 in2 is to be bent into a cylindrical length 450 CHAPTER 6 \| Systems of Equations and Inequalities of stovepipe having a volume of 600 in3. What are the dimensions of the sheet metal? x y 61. Global Positioning System (GPS) The Global Positioning System determines the location of an object from its distances to satellites in orbit around the earth. In the simpliﬁed, twodimensional situation shown in the ﬁgure, determine the coordinates of P from the fact that P is 26 units from satellite A and 20 units from satellite B. y A(22, 32) 26 B(28, 20) 20 P(x, y) Planet x ▼ DISCOVE RY • DISCUSSION • WRITI NG 62. Intersection of a Parabola and a Line On a sheet of graph paper or using a graphing calculator, draw the parabola y x 2. Then draw the graphs of the linear equation y x k on the same coordinate plane for various values of k. Try to choose values of k so that the line and the parabola intersect at two points for some of your k’s, and not for others. For what value of k is there exactly one intersection point? Use the results of your experiment to make a conjecture about the values of k for which the following system has two solutions, one solution, and no solution. Prove your conjecture. y x2 y x k 63. Some Trickier Systems Follow the hints and solve the (a) (b) (c) (d) systems. b log x log y 3 2 2 log x log y 0 2x 2y 10 4x 4y 68 b x y 3 x 3 y 3 387 b x 2 xy 1 xy y 2 3 b b [Hint: Add the
equations.] ”Hint: Note that 4x 22x Ó2xÔ2.’ [Hint: Factor the left-hand side of the second equation.] [Hint: Add the equations, and factor the result.] 6.2 Systems of Linear Equations in Two Variables LEARNING OBJECTIVES After completing this section, you will be able to: ■ Solve a system of two linear equations in two variables ■ Determine whether a system of two linear equations in two variables has one solution, inﬁnitely many solutions, or no solution ■ Model with linear systems Recall that an equation of the form Ax By C is called linear because its graph is a line (see Section 2.4). In this section we study systems of two linear equations in two variables. ■ Systems of Linear Equations in Two Variables A system of two linear equations in two variables has the form a1x b1y c1 a2x b2y c2 We can use either the substitution method or the elimination method to solve such systems algebraically. But since the elimination method is usually easier for linear systems, we use elimination rather than substitution in our examples. b SE CTI ON 6. 2 \| Systems of Linear Equations in Two Variables 451 The graph of a linear system in two variables is a pair of lines, so to solve the system graphically, we must ﬁnd the intersection point(s) of the lines. Two lines may intersect in a single point, they may be parallel, or they may coincide, as shown in Figure 1. So there are three possible outcomes in solving such a system. NUMBER OF SOLUTIONS OF A LINEAR SYSTEM IN TWO VARIABLES For a system of linear equations in two variables, exactly one of the following is true. (See Figure 1.) 1. The system has exactly one solution. 2. The system has no solution. 3. The system has inﬁnitely many solutions. A system that has no solution is said to be inconsistent. A system with inﬁnitely many solutions is called dependent. y 0 x (a) Linear system with one solution. Lines intersect at a single point. FIGURE 1 y 0 (b) x Linear system with no solution. Lines are parallel—they do not intersect. y 0 (c) x Linear system with inﬁnitely many solutions. Lines coincide— equations are for the same line. y 3
x-y=0 Solve the system and graph the lines. E X AM P L E 1 \| A Linear System with One Solution 3x y 0 5x 2y 22 Equation 1 Equation 2 ▼ SO LUTI O N We eliminate y from the equations and solve for x. 6 (2, 6) 2 x 5x+2y=22 FIGURE 2 Check Your Answer x 2, y 6 : b 6x 2y 0 5x 2y 22 11x 2y 22 x 2 b 2 Equation 1 Add Solve for x Now we back-substitute into the ﬁrst equation and solve for y: 2y 0 2y 12 y 6 Subtract 6 2 = 12 Back-substitute x = 2 Solve for y 2 6 2 1 The solution of the system is the ordered pair, that is, 2, 6 1 2 x 2 22 ✔ The graph in Figure 2 shows that the lines in the system intersect at the point ✎ Practice what you’ve learned: Do Exercise 11. 1 2, 6. 2 ▲ b 452 CHAPTER 6 \| Systems of Equations and Inequalities y _12x+3y=7 1 8x-2y=5 0 1 x FIGURE t, 1 2 t-2) t = 1 FIGURE 4 E X AM P L E 2 \| A Linear System with No Solution Solve the system. 8x 2y 5 12x 3y 7 Equation 1 Equation 2 ▼ SO LUTI O N This time we try to ﬁnd a suitable combination of the two equations to eliminate the variable y. Multiplying the ﬁrst equation by 3 and the second equation by 2 gives b 24x 6y 15 24x 6y 14 3 Equation 1 2 Equation 2 124x 60 29 b Add Adding the two equations eliminates both x and y in this case, and we end up with 0 29, which is obviously false. No matter what values we assign to x and y, we cannot make this statement true, so the system has no solution. Figure 3 shows that the lines in the system are parallel and do not intersect. The system is inconsistent. ✎ Practice what you’ve learned: Do Exercise 23. ▲ E X AM P L E 3 \| A Linear System with Inﬁnitely Many Solutions Solve the system. 3x 6y 12 4x 8y 16
Equation 1 Equation 2 ▼ SO LUTI O N We multiply the ﬁrst equation by 4 and the second by 3 to prepare for subtracting the equations to eliminate x. The new equations are b 12x 24y 48 12x 24y 48 4 Equation 1 3 Equation 2 We see that the two equations in the original system are simply different ways of expressing the equation of one single line. The coordinates of any point on this line give a solution of the system. Writing the equation in slope-intercept form, we have So if we let t represent any real number, we can write the solution as 2 x 2 We can also write the solution in ordered-pair form as t, 1 2 t 2 A B where t is any real number. The system has inﬁnitely many solutions (see Figure 4). ✎ Practice what you’ve learned: Do Exercise 25. ▲ In Example 3, to get speciﬁc solutions, we have to assign values to t. For instance, if. For every value of we get the solution 4, 0, we get the solution t 1 t 4, t we get a different solution. (See Figure 4.) 1, 3 2B. If A 1 2 ■ Modeling with Linear Systems Frequently, when we use equations to solve problems in the sciences or in other areas, we obtain systems like the ones we’ve been considering. When modeling with systems of equations, we use the following guidelines, which are similar to those in Section 1.2. SE CTI ON 6. 2 \| Systems of Linear Equations in Two Variables 453 GUIDELINES FOR MODELING WITH SYSTEMS OF EQUATIONS 1. Identify the Variables. Identify the quantities that the problem asks you to ﬁnd. These are usually determined by a careful reading of the question posed at the end of the problem. Introduce notation for the variables (call them x and y or some other letters). 2. Express All Unknown Quantities in Terms of the Variables. Read the problem again, and express all the quantities mentioned in the problem in terms of the variables you deﬁned in Step 1. 3. Set Up a System of Equations. Find the crucial facts in the problem that give the relationships between the expressions you found in Step 2. Set up a system of equations (or a model) that expresses these relationships. 4. Solve the System
and Interpret the Results. Solve the system you found in Step 3, check your solutions, and state your ﬁnal answer as a sentence that answers the question posed in the problem. The next two examples illustrate how to model with systems of equations. E X AM P L E 4 \| A Distance-Speed-Time Problem current 4 mi Identify the variables Express unknown quantities in terms of the variable Set up a system of equations A woman rows a boat upstream from one point on a river to another point 4 mi away in 11 hours. The return trip, traveling with the current, takes only 45 min. How fast does she 2 row relative to the water, and at what speed is the current ﬂowing? ▼ SO LUTI O N We are asked to ﬁnd the rowing speed and the speed of the current, so we let x rowing speed (mi/h) y current speed (mi/h) The woman’s speed when she rows upstream is her rowing speed minus the speed of the current; her speed downstream is her rowing speed plus the speed of the current. Now we translate this information into the language of algebra. In Words In Algebra Rowing speed Current speed Speed upstream Speed downstream x y x y x y The distance upstream and downstream is 4 mi, so using the fact that speed time distance for both legs of the trip, we get speed upstream time upstream distance traveled speed downstream time downstream distance traveled In algebraic notation this translates into the following equations Equation 1 Equation 2 454 CHAPTER 6 \| Systems of Equations and Inequalities (The times have been converted to hours, since we are expressing the speeds in miles per hour.) We multiply the equations by 2 and 4, respectively, to clear the denominators. Solve the system 3x 3y 8 3x 3y 16 16x 3y 24 b 1x 3y 4 2 Equation 1 4 Equation 2 Add Solve for x Back-substituting this value of x into the ﬁrst equation (the second works just as well) and solving for y gives 3y 8 3 4 1 2 Back-substitute x = 4 3y 8 12 Subtract 12 y 4 3 Solve for y The woman rows at 4 mi/h, and the current ﬂows at mi/h. 11 3 Check Your Answer Speed upstream is distance time 4 mi 11 h 2 22 3 mi/h Speed downstream is
distance time 4 mi h 3 4 51 mi/h 3 and this should equal and this should equal rowing speed current ﬂow rowing speed current ﬂow 4 mi/h mi/h mi/h 22 3 4 3 4 mi/h mi/h mi/h 51 3 4 3 ✎ Practice what you’ve learned: Do Exercise 51. ✔ ▲ E X AM P L E 5 \| A Mixture Problem A vintner fortiﬁes wine that contains 10% alcohol by adding a 70% alcohol solution to it. The resulting mixture has an alcoholic strength of 16% and ﬁlls 1000 one-liter bottles. How many liters (L) of the wine and of the alcohol solution does the vintner use? ▼ SO LUTI O N Since we are asked for the amounts of wine and alcohol, we let Identify the variables x amount of wine used (L) y amount of alcohol solution used (L) From the fact that the wine contains 10% alcohol and the solution contains 70% alcohol, we get the following. Express all unknown quantities in terms of the variable In Words In Algebra Amount of wine used (L) Amount of alcohol solution used (L) Amount of alcohol in wine (L) Amount of alcohol in solution (L) x y 0.10x 0.70y SE CTI ON 6. 2 \| Systems of Linear Equations in Two Variables 455 The volume of the mixture must be the total of the two volumes the vintner is adding together, so x y 1000 Also, the amount of alcohol in the mixture must be the total of the alcohol contributed by the wine and by the alcohol solution, that is, 0.10x 0.70y 0.16 0.10x 0.70y 160 x 7y 1600 Multiply by 10 to clear decimals Simplify 1000 2 1 Set up a system of equations Thus, we get the system x y 1000 x 7y 1600 Equation 1 Equation 2 Solve the system Subtracting the ﬁrst equation from the second eliminates the variable x, and we get 6y 600 b y 100 Subtract Equation 1 from Equation 2 Solve for y We now back-substitute y 100 into the ﬁrst equation and solve for x. x 100 1000 x 900 Back-substitute y = 100 Solve for x The vintner uses 900
L of wine and 100 L of the alcohol solution. ✎ Practice what you’ve learned: Do Exercise 53. ▲ 6. ▼ CONCE PTS 1. A system of two linear equations in two variables can have one solution, solutions. solution, or 2. The following is a system of two linear equations in two variables. x y 1 2x 2y 2 The graph of the ﬁrst equation is the same as the graph of the b second equation, so the system has solutions. We express these solutions by writing x t y where t is any real number. Some of the solutions of this 1, system are 1, 2 3, 1, and 2 5, 1. 2 ▼ SKI LLS 3–8 ■ Graph each linear system, either by hand or using a graphing device. Use the graph to determine whether the system has one solution, no solution, or inﬁnitely many solutions. If there is exactly one solution, use the graph to ﬁnd it. 3. 5. 7. x y 4 2x y 2 2x 3y 12 2x 1 2 y 10 b 4. 6. 8. 2x y 4 3x y 6 2x 6y 0 3x 9y 18 b 12x 15y 18 2x 5 2 y 3 b b 9–36 ■ Solve the system, or show that it has no solution. If the system has inﬁnitely many solutions, express them in the orderedpair form given in Example 3. b 9. ✎ 11. 13. 15. x y 4 x y 0 2x 3y 9 4x 3y 9 b x 3y 5 2x y 3 b x y 2 4x 3y 3 b 10. 12. 14. 16. x y 3 x 3y 7 3x 2y 0 x 2y 8 b x y 7 2x 3y 1 b 4x 3y 28 9x y 6 b b b 456 CHAPTER 6 \| Systems of Equations and Inequalities 17. 19. 21. ✎ 23. ✎ 25. 27. 29. 31. 33. 35. 1 b x 2y 7 5x 3x 2y 8 x 2y 0 x 4y 8 3x 12y 2 b b 2x 6y 10 3x 9y 15 b 6x 4y 12 9x 6y 18 b 8s 3t 3 5s 2t 2y 10 b
0.4x 1.2y 14 12x 5y 10 4 y 2 3 x 1 8x 6y 10 b b 5 1 18. 20. 22. 24. 26. 28. 30. 32. 34. 36. b 4x 12y 0 12x 4y 160 0.2x 0.2y 1.8 0.3x 0.5y 3.3 4x 2y 16 x 5y 70 b 3x 5y 2 9x 15y 6 b 2x 3y 8 14x 21y 3 b 25x 75y 100 10x 30y 40 b u 30√ 5 3u 80√ 5 b 2 x 1 3 y 3 2x 1 2 y 1 b 26x 10y 4 0.6x 1.2y 3 b 1 2 y 4 2x 10y 80 b 10 x 1 1 2 2 b 37–40 ■ Use a graphing device to graph both lines in the same viewing rectangle. (Note that you must solve for y in terms of x before graphing if you are using a graphing calculator.) Solve the system correct to two decimal places, either by zooming in and using or by using Intersect. TRACE b 37. 38. 39. 40. 0.21x 3.17y 9.51 2.35x 1.17y 5.89 18.72x 14.91y 12.33 6.21x 12.92y 17.82 b 2371x 6552y 13,591 9815x 992y 618,555 b 435x 912y 0 132x 455y 994 b 41–44 ■ Find x and y in terms of a and b. 41. 42. 43. 44. 1 x y 0 b x ay 1 ax by 0 x y 1 b ax by 1 bx ay 1 b ax by 0 a2x b2y 1 b a 1 2 a b 2 a2 b2 0 2 1 1 a 0, b 0, a b 1 2 47. Value of Coins A man has 14 coins in his pocket, all of which are dimes and quarters. If the total value of his change is $2.75, how many dimes and how many quarters does he have? 48. Admission Fees The admission fee at an amusement park is $1.50 for children and $4.00 for adults. On a certain day, 2200 people entered the park, and the admission fees that were
collected totaled $5050. How many children and how many adults were admitted? 49. Gas Station A gas station sells regular gas for $2.20 per gallon and premium gas for $3.00 a gallon. At the end of a business day 280 gallons of gas were sold, and receipts totaled $680. How many gallons of each type of gas were sold? 50. Fruit Stand A fruit stand sells two varieties of strawberries: standard and deluxe. A box of standard strawberries sells for $7, and a box of deluxe strawberries sells for $10. In one day the stand sells 135 boxes of strawberries for a total of $1110. How many boxes of each type were sold? ✎ 51. Airplane Speed A man ﬂies a small airplane from Fargo to Bismarck, North Dakota—a distance of 180 mi. Because he is ﬂying into a head wind, the trip takes him 2 hours. On the way back, the wind is still blowing at the same speed, so the return trip takes only 1 h 12 min. What is his speed in still air, and how fast is the wind blowing? wind Bismarck Fargo 180 mi 52. Boat Speed A boat on a river travels downstream between two points, 20 mi apart, in one hour. The return trip against the current takes 21 hours. What is the boat’s speed, and how fast 2 does the current in the river ﬂow? current 20 mi ▼ APPLICATIONS 45. Number Problem Find two numbers whose sum is 34 and b ✎ whose difference is 10. 46. Number Problem The sum of two numbers is twice their difference. The larger number is 6 more than twice the smaller. Find the numbers. 53. Nutrition A researcher performs an experiment to test a hypothesis that involves the nutrients niacin and retinol. She feeds one group of laboratory rats a daily diet of precisely 32 units of niacin and 22,000 units of retinol. She uses two types of commercial pellet foods. Food A contains 0.12 unit of niacin and 100 units of retinol per gram. Food B contains SE CTI ON 6. 3 \| Systems of Linear Equations in Several Variables 457 0.20 unit of niacin and 50 units of retinol per gram. How many grams of each food does she feed this group of rats each day? 54. Coffee Blends A customer in a coffee shop purchases a blend of two
coffees: Kenyan, costing $3.50 a pound, and Sri Lankan, costing $5.60 a pound. He buys 3 lb of the blend, which costs him $11.55. How many pounds of each kind went into the mixture? 55. Mixture Problem A chemist has two large containers of sulfuric acid solution, with different concentrations of acid in each container. Blending 300 mL of the ﬁrst solution and 600 mL of the second gives a mixture that is 15% acid, whereas blending 100 mL of the ﬁrst with 500 mL of the second gives a % acid mixture. What are the concentrations of sulfuric acid in the original containers? 121 2 56. Mixture Problem A biologist has two brine solutions, one containing 5% salt and another containing 20% salt. How many milliliters of each solution should she mix to obtain 1 L of a solution that contains 14% salt? 57. Investments A woman invests a total of $20,000 in two accounts, one paying 5% and the other paying 8% simple interest per year. Her annual interest is $1180. How much did she invest at each rate? 58. Investments A man invests his savings in two accounts, one paying 6% and the other paying 10% simple interest per year. He puts twice as much in the lower-yielding account because it is less risky. His annual interest is $3520. How much did he invest at each rate? 59. Distance, Speed, and Time John and Mary leave their house at the same time and drive in opposite directions. John drives at 60 mi/h and travels 35 mi farther than Mary, who drives at 40 mi/h. Mary’s trip takes 15 min longer than John’s. For what length of time does each of them drive? 60. Aerobic Exercise A woman keeps ﬁt by bicycling and running every day. On Monday she spends hour at each 121 activity, covering a total of 2 12 min and cycles for 45 min, covering a total of 16 mi. Assuming that her running and cycling speeds don’t change from day to day, ﬁnd these speeds. mi. On Tuesday she runs for 1 2 62. Area of a Triangle Find the area of the triangle that lies in the ﬁrst quadrant (with its base on the x-axis) and that is bounded by the lines y 2x 4 and y 4 x 20. y
0 y=2x-4 x y=_4x+20 ▼ DISCOVE RY • DISCUSSION • WRITI NG 63. The Least Squares Line The least squares line or regression line is the line that best ﬁts a set of points in the plane. We studied this line in the Focus on Modeling that follows Chapter 2 (see page 192). By using calculus, it can be shown that the line that best ﬁts the n data points xn, yn2 is the line y ax b, where the coefﬁcients a and b satisfy k1 xk the following pair of linear equations. (The notation stands for the sum of all the x’s. See Section 9.1 for a complete description of sigma x2, y2 2 x1, y1 2 notation.) 1 © k1 xk n a nb a k1 yk n a k1 2 xk ¢ a ≤ n a k1 xk n b a k1 xk yk Use these equations to ﬁnd the least squares line for the following data points. ¢ ≤ ¢ 1, 3 1, 2 2, 5 1, 2 3, 6 1 2 5, 6 1, 2 7, 9 1 2 Sketch the points and your line to conﬁrm that the line ﬁts these points well. If your calculator computes regression lines, see whether it gives you the same line as the formulas. ≤, 61. Number Problem The sum of the digits of a two-digit number is 7. When the digits are reversed, the number is increased by 27. Find the number. 6.3 Systems of Linear Equations in Several Variables LEARNING OBJECTIVES After completing this section, you will be able to: ■ Use Gaussian elimination to solve a system of three (or more) linear equations ■ Determine whether a system of three (or more) linear equations has one solution, inﬁnitely many solutions, or no solution ■ Model with linear systems in three (or more) variables A linear equation in n variables is an equation that can be put in the form a2x2 p anxn c a1x1 458 CHAPTER 6 \| Systems of Equations and Inequalities where a1, a2,..., an and c are real numbers, and x 1, x 2,...,
x n are the variables. If we have only three or four variables, we generally use x, y, z, and „ instead of x 1, x 2, x 3, and x 4. Such equations are called linear because if we have just two variables, the equation is a1x a2y c, which is the equation of a line. Here are some examples of equations in three variables that illustrate the difference between linear and nonlinear equations. Linear equations 15x3 3x2 6x1 Nonlinear equations x2 3y 1z 5 10 Not linear because it contains the square and the square root of a variable x y z 2„ 1 2 x1x2 6x3 6 Not linear because it contains a product of variables In this section we study systems of linear equations in three or more variables. ■ Solving a Linear System The following are two examples of systems of linear equations in three variables. The second system is in triangular form; that is, the variable x doesn’t appear in the second equation, and the variables x and y do not appear in the third equation. A system of linear equations x 2y z 1 x 3y 3z 4 2x 3y z 10 A system in triangular form x 2y z 1 y 2z 5 z 3 It’s easy to solve a system that is in triangular form by using back-substitution. So our goal in this section is to start with a system of linear equations and change it to a system in triangular form that has the same solutions as the original system. We begin by showing how to use back-substitution to solve a system that is already in triangular form. c c E X AM P L E 1 \| Solving a Triangular System Using Back-Substitution Solve the system using back-substitution: x 2y z 1 y 2z 5 z 3 Equation 1 Equation 2 Equation 3 ▼ SO LUTI O N From the last equation we know that z 3. We back-substitute this into the second equation and solve for y Back-substitute z = 3 into Equation 2 Solve for y Then we back-substitute y 1 and z 3 into the ﬁrst equation and solve for x. Back-substitute y = –1 and z = 3 into Equation Solve for x x 2, y 1, z 3 The solution of the system is ordered triple ✎ Practice
what you’ve learned: Do Exercise 7. 2, 1, 3. 1 2. We can also write the solution as the ▲ SE CTI ON 6. 3 \| Systems of Linear Equations in Several Variables 459 To change a system of linear equations to an equivalent system (that is, a system with the same solutions as the original system), we use the elimination method. This means that we can use the following operations. OPERATIONS THAT YIELD AN EQUIVALENT SYSTEM 1. Add a nonzero multiple of one equation to another. 2. Multiply an equation by a nonzero constant. 3. Interchange the positions of two equations. To solve a linear system, we use these operations to change the system to an equivalent triangular system. Then we use back-substitution as in Example 1. This process is called Gaussian elimination. E X AM P L E 2 \| Solving a System of Three Equations in Three Variables Solve the system using Gaussian elimination. 3x 2y 3z 1 3x 2y z 13 3x 2y 5z 3 Equation 1 Equation 2 Equation 3 ▼ SO LUTI O N We need to change this to a triangular system, so we begin by eliminating the x-term from the second equation. c x 2y 0z 13 x 2y 3z 1 4y 4z 12 Equation 2 Equation 1 Equation 2 + (–1) Equation 1 = new Equation 2 This gives us a new, equivalent system that is one step closer to triangular form: x 2y 3z 1 4y 4z 12 3x 2y 5z 3 Equation 1 Equation 2 Equation 3 Now we eliminate the x-term from the third equation. c x 2y 03z 01 4y 14z 12 8y 14z 00 Equation 3 + (3) Equation 1 = new Equation 3 Then we eliminate the y-term from the third equation. c x 2y 3z 1 x 4y 4z 12 x 2y 6z 24 Equation 3 + (2) Equation 2 = new Equation 3 The system is now in triangular form, but it will be easier to work with if we divide the second and third equations by the common factors of each term. c 3x 2y 05z 3 3x 6y 09z 3 8y 14z 0 8y 14z 0 8y 08z 24 6z 24 x 2y
3z 1 y 3z 3 z 4 c 1 4 Equation 2 = new Equation 2 – Equation 3 = new Equation 3 1 6 460 CHAPTER 6 \| Systems of Equations and Inequalities Intersection of Three Planes When you study calculus or linear algebra, you will learn that the graph of a linear equation in three variables is a plane in a three-dimensional coordinate system. For a system of three equations in three variables the following situations arise: 1. The three planes intersect in a single point. The system has a unique solution. 2. The three planes intersect in more than one point. The system has inﬁnitely many solutions. Now we use back-substitution to solve the system. From the third equation we get z 4. We back-substitute this into the second equation and solve for y Back-substitute z = 4 into Equation 2 Solve for y and z 4 into the ﬁrst equation and solve for x. Back-substitute y = 7 and z = 4 into Equation 1 Solve for x Now we back-substitute x 2 3 7 1 2 The solution of the system is x 3, y 7, z 4, which we can write as the ordered triple 3, 7, 4 1. 2 Check Your Answer x 3, y 7, z 4 01 13 3 3 ✔ 2 1 ✎ Practice what you’ve learned: Do Exercise 17. 2 1 1 2 ▲ ■ The Number of Solutions of a Linear System Just as in the case of two variables, a system of equations in several variables may have one solution, no solution, or inﬁnitely many solutions. The graphical interpretation of the solutions of a linear system is analogous to that for systems of equations in two variables (see the margin note). NUMBER OF SOLUTIONS OF A LINEAR SYSTEM 3. The three planes have no point in For a system of linear equations, exactly one of the following is true. common. The system has no solution. 1. The system has exactly one solution. 2. The system has no solution. 3. The system has inﬁnitely many solutions. A system with no solutions is said to be inconsistent, and a system with inﬁnitely many solutions is said to be dependent. As we see in the next example, a linear system has no solution if we end up with a false equation after applying Gaussian elimination to the system
. E X AM P L E 3 \| A System with No Solution Solve the following system. x 2y 2z 1 2x 2y z 6 3x 4y 3z 5 Equation 1 Equation 2 Equation 3 ▼ SO LUTI O N To put this in triangular form, we begin by eliminating the x-terms from the second equation and the third equation. c SE CTI ON 6. 3 \| Systems of Linear Equations in Several Variables 461 x 2y 2z 1 2y 3z 4 3x 4y 3z 5 c x 2y 2z 1 2y 3z 4 2y 3z 2 Equation 2 + (–2) Equation 1 = new Equation 2 Equation 3 + (–3) Equation 1 = new Equation 3 Now we eliminate the y-term from the third equation. c x 2y 2z 1 2y 3z 4 0 2 Equation 3 + (–1) Equation 2 = new Equation 3 c The system is now in triangular form, but the third equation says 0 2, which is false. No matter what values we assign to x, y, and z, the third equation will never be true. This means that the system has no solution. ✎ Practice what you’ve learned: Do Exercise 27. ▲ E X AM P L E 4 \| A System with Inﬁnitely Many Solutions Solve the following system. x y 5z 2 2x y 4z 2 2x 4y 2z 8 Equation 1 Equation 2 Equation 3 ▼ SO LUTI O N To put this in triangular form, we begin by eliminating the x-terms from the second equation and the third equation. c x 3y 5z 2 3y 6z 6 2x 4y 2z 8 x y 5z 2 c 3y 6z 6 6y 12z 12 Equation 2 + (–2) Equation 1 = new Equation 2 Equation 3 + (–2) Equation 1 = new Equation 3 Now we eliminate the y-term from the third equation. c x y 5z 2 3y 6z 6 0 0 Equation 3 + (–2) Equation 2 = new Equation 3 The new third equation is true, but it gives us no new information, so we can drop it from the system. Only two equations are left. We can use them to solve for x and y in terms
of z, but z can take on any value, so there are inﬁnitely many solutions. c To ﬁnd the complete solution of the system, we begin by solving for y in terms of z, us- ing the new second equation. 3y 6z 6 y 2z 2 y 2z 2 Equation 2 1 Multiply by 3 Solve for y Then we solve for x in terms of z, using the ﬁrst equation. 462 CHAPTER 6 \| Systems of Equations and Inequalities x 1 2z 2 2 5z 2 x 3z 2 2 x 3z Substitute y = 2z + 2 into Equation 1 Simplify Solve for x To describe the complete solution, we let t represent any real number. The solution is x 3t y 2t 2 z t 3t, 2t 2, t We can also write this as the ordered triple. 2 ✎ Practice what you’ve learned: Do Exercise 25. 1 ▲ In the solution of Example 4 the variable t is called a parameter. To get a speciﬁc solution, we give a speciﬁc value to the parameter t. For instance, if we set t 2, we get Thus, obtained by substituting other values for the parameter t. 6, 6, 2 1 2 is a solution of the system. Here are some other solutions of the system Parameter t 1 0 3 10 Solution 3t, 2t 2, t 2 2 1 3, 0, 1 1 0, 2, 0 2 1 9, 8, 3 1 2 30, 22, 10 1 2 You should check that these points satisfy the original equations. There are inﬁnitely many choices for the parameter t, so the system has inﬁnitely many solutions. MATHEMATICS IN THE MODERN WORLD Global Positioning System (GPS) On a cold, foggy day in 1707 a British naval ﬂeet was sailing home at a fast clip. The ﬂeet’s navigators didn’t know it, but the ﬂeet was only a few yards from the rocky shores of England. In the ensuing disaster the ﬂeet was totally destroyed. This tragedy could have been avoided had the navigators known their po- sitions. In those days latitude was determined by the position of the North Star (and this could
only be done at night in good A S A N weather), and longitude by the position of the sun relative to where it would be in England at that same time. So navigation required an accurate method of telling time on ships. (The invention of the spring-loaded clock brought about the eventual solution.) Since then, several different methods have been developed to determine position, and all rely heavily on mathematics (see LORAN, page 578). The latest method, called the Global Positioning System (GPS), uses triangulation. In this system, 24 satellites are strategically located above the surface of the earth. A handheld GPS device measures distance from a satellite, using the travel time of radio signals emitted from the satellite. Knowing the distances to three different satellites tells us that we are at the point of intersection of three different spheres. This uniquely determines our position (see Exercise 61, page 450). SE CTI ON 6. 3 \| Systems of Linear Equations in Several Variables 463 ■ Modeling Using Linear Systems Linear systems are used to model situations that involve several varying quantities. In the next example we consider an application of linear systems to ﬁnance. E X AM P L E 5 \| Modeling a Financial Problem Using a Linear System Jason receives an inheritance of $50,000. His ﬁnancial advisor suggests that he invest this in three mutual funds: a money-market fund, a blue-chip stock fund, and a high-tech stock fund. The advisor estimates that the money-market fund will return 5% over the next year, the blue-chip fund 9%, and the high-tech fund 16%. Jason wants a total ﬁrst-year return of $4000. To avoid excessive risk, he decides to invest three times as much in the moneymarket fund as in the high-tech stock fund. How much should he invest in each fund? ▼ SO LUTI O N Let x amount invested in the money-market fund y amount invested in the blue-chip stock fund z amount invested in the high-tech stock fund We convert each fact given in the problem into an equation. x y z 50,000 Total amount invested is $50,000 0.05x 0.09y 0.16z 4000 x 3z Total investment return is $4000 Money-market amount is 3 high-tech amount Multiplying the second equation by 100 and rewriting the third gives the following system, which we solve using Gaussian elimination. x y
z 0 50,000 5x 9y 16z 400,000 x 3z 0 100 Equation 2 Subtract 3z x y z 50,000 c 4y 11z 150,000 y 4z 50,000 x y z 50,000 c 5z 50,000 y 4z 50,000 c x y z 50,000 z 10,000 y 4z 50,000 x y z 50,000 c y 4z 50,000 z 10,000 Equation 2 + (–5) Equation 1 = new Equation 2 Equation 3 + (–1) Equation 1 = new Equation 3 Equation 2 + 4 Equation 3 = new Equation 3 1) Equation 2 (– 5 (–1) Equation 3 Interchange Equations 2 and 3 Now that the system is in triangular form, we use back-substitution to ﬁnd that x 30,000, y 10,000, and z 10,000. This means that Jason should invest c $30,000 in the money-market fund $10,000 in the blue-chip stock fund $10,000 in the high-tech stock fund ✎ Practice what you’ve learned: Do Exercise 37. ▲ 464 CHAPTER 6 \| Systems of Equations and Inequalities 6. ▼ CONCE PTS 1–2 ■ These exercises refer to the following system. x y z 2 x 2y z 3 3x y 2z 2 1. If we add 2 times the ﬁrst equation to the second equation, the second equation becomes c. 2. To eliminate x from the third equation, we add times the ﬁrst equation to the third equation. The third equation becomes. ▼ SKI LLS 3–6 ■ State whether the equation or system of equations is linear. 3. 5. 6x 13y 1 2 z 0 xy 3y z 5 x y2 5z 0 yz 3 2x 4. x2 y2 z2 4 6. x 2y 3z 10 2x 5y 2 y 2z 4 7–12 ■ Use back-substitution to solve the triangular system. c c ✎ 7. x 2y 4z 3 y 2z 7 z 2 c 9. x 2y z 7 y 3z 9 2z 6 11. 2x y 6z 5 c y 4z 0 2z 1 8. x y 3z 8
y 3z 5 z 1 c 10. x 2y 3z 10 2y z 2 3z 12 c 12. 4x 3z 10 2y 3z 6 2z 4 1 c 13–16 ■ Perform an operation on the given system that eliminates the indicated variable. Write the new equivalent system. c 13. x 2y z 4 x y 3z 0 2x y z 0 Eliminate the x-term from the second equation. c 15. 2x y 3z 2 x 2y z 4 4x 5y z 10 Eliminate the x-term from the third equation. c 14. 16. x y 3z 3 2x 3y z 2 x y 2z 0 Eliminate the x-term from the second equation. c x 4y z 3 y 3z 10 3y 8z 24 Eliminate the y-term from the third equation. c 17–36 ■ Find the complete solution of the linear system, or show that it is inconsistent. 18. 20. 22. x y 2z 0 x y 2z 2y 5z 3 6 3x y c x y 2z 2 3x y 5z 8 2x y 2z 7 c 24. 2x y 4z 8 x y 4z 3 2x y 4z 18 26. 28. 30. 2y 3z 3 c 5x 4y 3z 1 x 3y 3z 2 c x 2y 5z 4 x 2z 0 4x 2y 11z 2 x 2y 3z 5 c 2x y z 5 4x 3y 7z 5 c ✎ 17. x y z 4 2y z 1 x y 2z 5 c 19. x y z 4 x 3y 3z 10 2x y z 3 c 21. 4z 1 x 2x y 6z 4 2x 3y 2z 8 c 23. 2x 4y z 2 x 2y 3z 4 3x y z 1 ✎ 25. c y 2z 0 2x 3y 2 x 2y z 1 ✎ 27. x 2y z 1 c 2x 3y 4z 3 3x 6y 3z 4 c 29. 2x 3y z 1 3 x 2y x 3y z 4 c 31. x y z 0 x 2y 3z 3 2x 3y 4z 3 c 32. x 2y z 3 2x 5y 6z 7 2x 3y 2z 5 c 33. 2x 3
y 2z 0 2x 3y 4z 4 4x 6y 2z 4 c 34. 2x 4y z 3 x 2y 4z 6 x 2y 2z 0 2x 2y 2z 2„ 6 2x 2y 2z 2„ 3 2x 2y 2z 2„ 2 2x 2y 3z 2„ 0 x y z „ 0 x y 2z 2„ 0 2x 2y 3z 4„ 1 2x 3y 4z 5„ 2 35. 36. c d d SE CTI ON 6. 3 \| Systems of Linear Equations in Several Variables 465 ▼ APPLICATIONS 37–38 ■ Finance An investor has $100,000 to invest in three types of bonds: short-term, intermediate-term, and long-term. How much should she invest in each type to satisfy the given conditions? ✎ 37. Short-term bonds pay 4% annually, intermediate-term bonds pay 5%, and long-term bonds pay 6%. The investor wishes to realize a total annual income of 5.1%, with equal amounts invested in short- and intermediate-term bonds. 38. Short-term bonds pay 4% annually, intermediate-term bonds pay 6%, and long-term bonds pay 8%. The investor wishes to have a total annual return of $6700 on her investment, with equal amounts invested in intermediate- and long-term bonds. 39. Agriculture A farmer has 1200 acres of land on which he grows corn, wheat, and soybeans. It costs $45 per acre to grow corn, $60 to grow wheat, and $50 to grow soybeans. Because of market demand the farmer will grow twice as many acres of wheat as of corn. He has allocated $63,750 for the cost of growing his crops. How many acres of each crop should he plant? 40. Gas Station A gas station sells three types of gas: Regular for $3.00 a gallon, Performance Plus for $3.20 a gallon, and Premium for $3.30 a gallon. On a particular day 6500 gallons of gas were sold for a total of $20,050. Three times as many gallons of Regular as Premium gas were sold. How many gallons of each type of gas were sold that day? 41. Nutrition A biologist is performing an experiment on the effects of various combinations of vitamins. She wishes to feed each of her laboratory rabbits a diet that contains exactly 9
mg of niacin, 14 mg of thiamin, and 32 mg of riboﬂavin. She has available three different types of commercial rabbit pellets; their vitamin content (per ounce) is given in the table. How many ounces of each type of food should each rabbit be given daily to satisfy the experiment requirements? Type A Type B Type C Niacin (mg) Thiamin (mg) Riboﬂavin (mg 42. Diet Program Nicole started a new diet that requires each meal to have 460 calories, 6 grams of ﬁber, and 11 grams of fat. The table shows the ﬁber, fat, and calorie content of one serving of each of three breakfast foods. How many servings of each food should Nicole eat to follow her diet? Food Fiber Fat Calories Toast Cottage cheese Fruit 2 0 2 1 5 0 100 120 60 43. Juice Blends The Juice Company offers three kinds of smoothies: Midnight Mango, Tropical Torrent, and Pineapple Power. Each smoothie contains the amounts of juices shown in the table. Smoothie Midnight Mango Tropical Torrent Pineapple Power Mango juice (oz) Pineapple Orange juice (oz) juice (oz On a particular day the Juice Company used 820 oz of mango juice, 690 oz of pineapple juice, and 450 oz of orange juice. How many smoothies of each kind were sold that day? 44. Appliance Manufacturing Kitchen Korner produces refrigerators, dishwashers, and stoves at three different factories. The table gives the number of each product produced at each factory per day. Kitchen Korner receives an order for 110 refrigerators, 150 dishwashers, and 114 ovens. How many days should each plant be scheduled to ﬁll this order? Appliance Factory A Factory B Factory C Refrigerators Dishwashers Stoves 8 16 10 10 12 18 14 10 6 45. Stock Portfolio An investor owns three stocks: A, B, and C. The closing prices of the stocks on three successive trading days are given in the table. Stock A Stock B Stock C Monday Tuesday Wednesday $10 $12 $16 $25 $20 $15 $29 $32 $32 Despite the volatility in the stock prices, the total value of the investor’s stocks remained unchanged at $74,000 at the end of each of these three days. How many shares of each stock does 466 CHAPTER 6 \| Systems of Equations and Inequalities the investor own?
46. Electricity By using Kirchhoff’s Laws, it can be shown that the currents I1, I2, and I3 that pass through the three branches of the circuit in the ﬁgure satisfy the given linear system. Solve the system to ﬁnd I1, I2, and I3. I2 I1 8I2 16I1 8I2 I3 4I3 4I3 0 4 5 c I⁄ 16 I¤ 8 I‹ 4 4 V 5 V ▼ DISCOVE RY • DISCUSSION • WRITI NG 47. Can a Linear System Have Exactly Two Solutions? x0, y0, z02 1 x1, y1, z12 and 1 are solutions of the (a) Suppose that system DISCOVERY PR OJECT BEST FIT VERSUS EXACT FIT Given several points in the plane, we can ﬁnd the line that best ﬁts them (see the Focus on Modeling, page 192). Of course, not all the points will necessarily lie on the line. We can also ﬁnd the quadratic polynomial that best ﬁts the points. Again, not every point will necessarily lie on the graph of the polynomial. However, if we are given just two points, we can ﬁnd a line of exact ﬁt, that is, a line that actually passes through both points. Similarly, given three points (not all on the same line), we can ﬁnd the quadratic polynomial of exact ﬁt. For example, suppose we are given the following three points: 1, 6 1, 2 2 _1 (_1, 6) _2 FIGURE 1 From Figure 1 we see that the points do not lie on a line. Let’s ﬁnd the quadratic polynomial that ﬁts these points exactly. The polynomial must have the form y ax2 bx c (2, 3) (1, 2) We need to ﬁnd values for a, b, and c so that the graph of the resulting polynomial contains the given points. Substituting the given points into the equation, we get the following. 4 Point 2 1, 6 1 1, 2 1 2, 3 1 2 2 Substitute Equation x 1, x 1, x 2 These three
equations simplify into the following system. a b c 6 a b c 2 4a 2b c 3 Using Gaussian elimination, we obtain the solution a 1, b 2, and c 3. So the required quadratic polynomial is c y x 2 2x 3 From Figure 2 we see that the graph of the polynomial passes through the given points. 9 (2, 3) (1, 2) 4 (_1, 6) _2 _1 FIGURE 2 1. Find the quadratic polynomial y ax 2 bx c whose graph passes through (a) the given points. 2, 3, 2 1 1, 3, 2 1 (b) 1, 1, 2 1 2, 0 1, 2 1, 9 2 1 3, 3 2 1 (CONTINUES) 467 467 BEST FIT VERSUS EXACT FIT (CONTINUED) 2. Find the cubic polynomial through the given points. y ax 3 bx 2 cx d whose graph passes (a) (b) 1, 4, 2 1 2, 10, 2 1, 1, 2 2 1 1, 1 1, 2, 2, 11 2 1 1, 1 1, 2 3, 32 1 2 3, 45 1 2 3. A stone is thrown upward with velocity √ from a height h. Its elevation d above the ground at time t is given by d at2 √t h The elevation is measured at three different times as shown in the table. Time (s) Elevation (ft) 1.0 144 2.0 192 6.0 64 (a) Find the constants a, √, and h. (b) Find the elevation of the stone when t 4 s. 4. (a) Find the quadratic function y ax2 bx c whose graph passes through the given points. (This is the quadratic curve of exact ﬁt.) Graph the points and the quadratic curve that you found. 2, 10 2, 6 4, 2 1,5,,, 2 1 2 (b) Now use the QuadReg command on your calculator to ﬁnd the quadratic curve that best ﬁts the points in part (a). How does this compare to the function that you found in part (a)? 1 1 1 2 2 (c) Show that no quadratic function passes through the points 1, 6 2, 11 2, 5
4d) Use the QuadReg command on your calculator to ﬁnd the quadratic curve that best ﬁts the points in part (b). Graph the points and the quadratic curve that you found. (e) Explain how the curve of exact ﬁt differs from the curve of best ﬁt. 468 6.4 Partial Fractions SE CTION 6.4 \| Partial Fractions 469 LEARNING OBJECTIVES After completing this section, you will be able to: ■ Find the form of the partial fraction decomposition of a rational expression in the following cases. —Denominator contains distinct linear factors —Denominator contains repeated linear factors —Denominator contains distinct quadratic factors —Denominator contains repeated quadratic factors ■ Find the partial fraction decomposition of a rational expression in the above cases To write a sum or difference of fractional expressions as a single fraction, we bring them to a common denominator. For example, 1 x 1 1 2x 1 1 2x 2x 1 2 2 3x 2x2 x 1 But for some applications of algebra to calculus we must reverse this process—that is, we 2x2 x 1 must express a fraction such as as the sum of the simpler fractions These simpler fractions are called partial fractions; we learn 1/ how to ﬁnd them in this section. Let r be the rational function 2x 1 x 1 and 1/ 3x Common denominator 1 x 1 1 2x 1 3x 2x2 x 1 Partial fractions where the degree of P is less than the degree of Q. By the Linear and Quadratic Factors Theorem in Section 4.5, every polynomial with real coefﬁcients can be factored completely into linear and irreducible quadratic factors, that is, factors of the form ax b and ax2 bx c, where a, b, and c are real numbers. For instance, x 1 x2 1 x4 1 x2 x2 1 2 2 1 After we have completely factored the denominator Q of r, we can express of partial fractions of the form x r 1 2 as a sum A ax b 1 i 2 and Ax B ax2 bx c 1 j 2 This sum is called the partial fraction decomposition of r. Let’s examine the details of the four possible cases. CASE 1: THE DENOMINATOR IS A PRODUCT OF DIST
INCT LINEAR FACTORS x Suppose that we can factor Q as 1 2 a1x b12 1 Q x 1 1 2 a2x b2 2 # # # anx bn2 1 with no factor repeated. In this case the partial fraction decomposition of P takes the form / a1x b1 A2 a2x b2 p An anx bn The constants A1, A2,..., An are determined as in the following example. 470 CHAPTER 6 \| Systems of Equations and Inequalities E X AM P L E 1 \| Distinct Linear Factors Find the partial fraction decomposition of 5x 7 x3 2x2 x 2. ▼ SO LUTI O N The denominator factors as follows: The Rhind papyrus is the oldest known mathematical document. It is an Egyptian scroll written in 1650 B.C. by the scribe Ahmes, who explains that it is an exact copy of a scroll written 200 years earlier. Ahmes claims that his papyrus contains “a thorough study of all things, insight into all that exists, knowledge of all obscure secrets.” Actually, the document contains rules for doing arithmetic, including multiplication and division of fractions and several exercises with solutions. The exercise shown below reads: “A heap and its seventh make nineteen; how large is the heap?” In solving problems of this sort, the Egyptians used partial fractions because their number system required all fractions to be written as sums of reciprocals of whole numbers. For example, would be 1. written as 3 1 4 7 12 The papyrus gives a correct formula for the volume of a truncated pyramid (page 63). It also gives A the formula of a circle with diameter d. How close is this to the actual area? for the area 8 9 d B A 2 x3 2x2 x 2 x2 x2 1 x 2 2 2 1 1 This gives us the partial fraction decomposition 5x 7 x3 2x2 Multiplying each side by the common denominator, Óx 1ÔÓx 1ÔÓx 2Ô, we get 5x 7 A 1 x 1 x 2 2 1 x2 3x 2 B x 1 2 B 2 1 1 x2 x 2 1 A B C 2 x2 1 3A x2 1 1 2 2A 2B C x 1 2 2 1 Expand Combine like terms 2 If two polynomials are equal, then
their coefﬁcients are equal. Thus, since 5x 7 has no x 2-term, we have A B C 0. Similarly, by comparing the coefﬁcients of x, we see that 3A B 5, and by comparing constant terms, we get 2A 2B C 7. This leads to the following system of linear equations for A, B, and C. A B C 0 3A B 5 2A 2B C 7 Equation 1: Coefﬁcients of x 2 Equation 2: Coefﬁcients of x Equation 3: Constant coefﬁcients We use Gaussian elimination to solve this system. c A 2B C 0 A 2B 3C 5 A 4B 3C 7 A 2B 3C 0 c A 2B 3C 5 A 2B 3C 3 Equation 2 + (–3) Equation 1 Equation 3 + (–2) Equation 1 Equation 3 + (–2) Equation 2 From the third equation we get C 1. Back-substituting, we ﬁnd that B 1 and A 2. So the partial fraction decomposition is c 1 x 2 ✎ Practice what you’ve learned: Do Exercises 3 and 13. 5x 7 x3 2x2 x 2 1 x 1 2 x 1 ▲ The same approach works in the remaining cases. We set up the partial fraction decomposition with the unknown constants A, B, C,.... Then we multiply each side of the resulting equation by the common denominator, simplify the right-hand side of the equation, and equate coefﬁcients. This gives a set of linear equations that will always have a unique solution (provided that the partial fraction decomposition has been set up correctly). SE CTION 6.4 \| Partial Fractions 471 CASE 2: THE DENOMINATOR IS A PRODUCT OF LINEAR FACTORS, SOME OF WHICH ARE REPEATED Suppose the complete factorization of repeated k times; that is, Óax bÔk is a factor of x such factor, the partial fraction decomposition for contains the linear factor ax b Q. Then, corresponding to each 2 P contains Q /Q x x x 1 2 1 A1 ax b A2 ax b 1 2 2 p 1 E X AM P L E 2 \| Repeated Linear Factors 1 2 2 1 Ak ax b
k 2 Find the partial fraction decomposition of x2 1 x 1 x. 3 1 ▼ SO LUTI O N Because the factor x 1 is repeated three times in the denominator, the partial fraction decomposition has the form 2 D x 1 Multiplying each side by the common denominator, xÓx 1Ô3, gives C x 1 A x B x 1 x2 x2 1 A A x 1 3 Bx x 1 x3 3x2 3x 1 x3 1 A B 2 1 2 Cx 2 B 1 2 x 1 x3 2x2 x x2 2 3A 2B C 1 Dx C x2 x 2 1 3A B C D 2 1 2 1 2 1 Dx x A 2 Expand Combine like terms Equating coefﬁcients, we get the following equations. A B 3A 2B C 0 1 3A B C D 0 1 A Coefﬁcients of x3 Coefﬁcients of x2 Coefﬁcients of x Constant coefﬁcients If we rearrange these equations by putting the last one in the ﬁrst position, we can easily see (using substitution) that the solution to the system is A 1, B 1, C 0, D 2, so the partial fraction decomposition is d x2 ✎ Practice what you’ve learned: Do Exercises 5 and 29. ▲ CASE 3: THE DENOMINATOR HAS IRREDUCIBLE QUADRATIC FACTORS, NONE OF WHICH IS REPEATED Suppose the complete factorization of contains the quadratic factor ax 2 bx c (which can’t be factored further). Then, corresponding to this, the x /Q partial fraction decomposition of 1 Ax B ax2 bx c will have a term of the form P x Q x 1 1 2 2 2 472 CHAPTER 6 \| Systems of Equations and Inequalities E X AM P L E 3 \| Distinct Quadratic Factors Find the partial fraction decomposition of 2x2 x 4 x3 4x. ▼ SO LUTI O N Since x3 4x x x2 4, which can’t be factored further, we write 1 2x2 x 4 x3 4x 2 A x Bx C x2 4 Multiplying by x 1 x2 4 we get, 2 2x
2 x 4 A x2 4 1 A B Bx C 2 2 x2 Cx 4A 1 x 1 2 Equating coefﬁcients gives us the equations A B 2 A C 1 A 4A 4 Coefﬁcients of x2 Coefﬁcients of x Constant coefﬁcient so A 1, B 1, and C 1. The required partial fraction decomposition is c 2x2 x 4 x3 4x 1 x x 1 x2 4 ✎ Practice what you’ve learned: Do Exercises 7 and 37. ▲ CASE 4: THE DENOMINATOR HAS A REPEATED IRREDUCIBLE QUADRATIC FACTOR Suppose the complete factorization of QÓxÔ contains the factor Óax 2 bx cÔk, where ax 2 bx c can’t be factored further. Then the partial fraction decomposition of will have the terms P 1 x x / cÔk 2 2 2 b b x bx ax 2 B 1 B k B 2 Óax Óax cÔ2 x c E X AM P L E 4 \| Repeated Quadratic Factors Write the form of the partial fraction decomposition of x 5 3x 2 12x 1 x 3Óx 2 x 1ÔÓx 2 2Ô3 ▼ SO LUTI O N x 5 3x 2 12x 1 x 3Óx 2 x 1ÔÓx 2 2Ô3 A x B x 2 C x 3 Dx E x 2 x 1 Fx G x 2 2 Hx I Óx 2 2Ô2 Jx K Óx 2 2Ô3 ✎ Practice what you’ve learned: Do Exercises 11 and 41. ▲ SE CTION 6.4 \| Partial Fractions 473 To ﬁnd the values of A, B, C, D, E, F, G, H, I, J, and K in Example 4, we would have to solve a system of 11 linear equations. Although possible, this would certainly involve a great deal of work! The techniques that we have described in this section apply only to rational functions PÓxÔQÓxÔ in which the degree of P is less than the degree of Q. If this isn’t the case, we must �
�rst use long division to divide Q into P. E X AM P L E 5 \| Using Long Division to Prepare for Partial Fractions Find the partial fraction decomposition of 2x4 4x3 2x2 x 7 x3 2x2 x 2 2x x 3 2x 2 x 2 2 x 4 4x3 2x2 x 7 2x 4 4x 3 2x 2 4x 5x 7 ▼ SO LUTI O N Since the degree of the numerator is larger than the degree of the denominator, we use long division to obtain 2x4 4x3 2x2 x 7 x3 2x2 x 2 2x 5x 7 x3 2x2 x 2 The remainder term now satisﬁes the requirement that the degree of the numerator is less than the degree of the denominator. At this point we proceed as in Example 1 to obtain the decomposition 2x4 4x3 2x2 x 7 x3 2x2 x 2 ✎ Practice what you’ve learned: Do Exercise 43. 2x. ▼ CONCE PTS 1–2 ■ For each rational function r, choose from (i)–(iv) the appropriate form for its partial fraction decomposition. ▼ SKI LLS 3–12 ■ Write the form of the partial fraction decomposition of the function (as in Example 4). Do not determine the numerical values of the coefﬁcients 2x 8 (ii) (iv. r x 1 2 ii(i) (iii) 2. r x 1 2 ii(i) i(ii) (iii) x 1 x2 4 2 2 1 B x2 4 Bx C x2 x2 4 x 2 Cx D x2 4 (iv) Ax B x 1 2 2 Cx D x 2 2 1 2 ✎ 3. ✎ 5. ✎ 7. 9. ✎ 11 x2 3x 5 x 4 x 2 2 2 1 2 x2 x2 4 x 3 2 2 1 x3 4x2 2 x2 2 x2 1 2 2 1 x3 x 1 2x 5 x 1 3 2 1 x2 2x 5 2 2 4. 6. 6. 8. 8. 10. 10. 12. 12. x x2 3x 4 1 x4 x3 1 x4 1 x4 x2 1 x2 4 2 x2 1 2 1 2 1 x2 1
2 x3 1 1 13–44 ■ Find the partial fraction decomposition of the rational function. ✎ 13 14. 14. 2x x 1 1 2 1 x 1 2 474 CHAPTER 6 \| Systems of Equations and Inequalities 15. 17. 19. 21. 23. 25. 27. ✎ 29. 31. 33. 35. ✎ 37. 39. ✎ 41 12 x2 9 4 x2 4 x 14 x2 2 x 8 x 8x2 10x 3 9x2 9x 6 2x3 x2 8x 4 x2 1 x3 x2 2x 4x2 12x 9 4x2 x 2 x4 2x3 10x2 27x 14 3 x 2 x 1 2 2 1 1 3x3 22x2 53x 41 x3 3x 2x3 7x 5 2 1 x2 1 x2 x 2 2 1 x4 x3 x2 x 1 x2 1 x 2 1 2 16. 18. 20. 22. 24. 26. 28. 30. 32. 34. 36. 38. 40. 42. x 6 x 3 x 2 1 x 12 x2 4x 2x 1 x2 x 2 8x 3 2x2 x 7x 3 x3 2x2 3x 3x2 3x 27 2x2 3x 9 x 2 2 1 1 3x2 5x 13 x2 4x 4 2 1 2 1 3x 2 x 4 2x 5 2 1 x3 2x2 4x 3 x4 2x2 5x 1 x4 2x3 2x 1 3x2 12x 20 x4 8x2 16 3x2 2x 8 x3 x2 2x 2 x2 x 1 2x4 3x2 1 2x2 x 8 x2 4 2 1 2 ✎ 43. x5 2x4 x3 x 5 x3 2x2 x 2 44. x5 3x4 3x3 4x2 4x 12 2 x 2 x2 2 2 1 1 45. Determine A and B in terms of a and b. B A 2 ax b x2 1 x 1 x 1 46. Determine A, B, C, and D in terms of a and b. Cx D x2 1 2 Ax B x2 1 ax3 bx2 x2 1 2 1 2 1 2 2 2 ▼ DISCOVE RY • DISCUSS
ION • WRITI NG 47. Recognizing Partial Fraction Decompositions For each expression, determine whether it is already a partial fraction decomposition or whether it can be decomposed further. (a) (c) x x2 b) (d) x x 1 2 x 2 x2 1 1 1 2 2 2 48. Assembling and Disassembling Partial Fractions The following expression is a partial fraction decomposition Use a common denominator to combine the terms into one fraction. Then use the techniques of this section to ﬁnd its partial fraction decomposition. Did you get back the original expression? 6.5 Systems of Inequalities LEARNING OBJECTIVES After completing this section, you will be able to: ■ Graph the solution of an inequality ■ Graph the solution of a system of inequalities ■ Graph the solution of a system of linear inequalities y 1 0 y=≈ 1 x In this section we study systems of inequalities in two variables from a graphical point of view. ■ Graphing an Inequality We begin by considering the graph of a single inequality. We already know that the graph 2, for example, is the parabola in Figure 1. If we replace the equal sign by the symof bol, we obtain the inequality y x FIGURE 1 y x 2 SE CTI O N 6.5 \| Systems of Inequalities 475 Its graph consists of not just the parabola in Figure 1, but also every point whose y-coordinate is larger than x 2. We indicate the solution in Figure 2(a) by shading the points above the parabola. Similarly, the graph of y x 2 parabola. However, the graphs of parabola itself, as indicated by the dashed curves in Figures 2(c) and 2(d). in Figure 2(b) consists of all points on and below the y x do not include the points on the y x and a) y≥≈ (b) y≤≈ (c) y>≈ (d) y<≈ FIGURE 2 The graph of an inequality, in general, consists of a region in the plane whose bound,,, or ary is the graph of the equation obtained by replacing the inequality sign 2 with an equal sign. To determine which side of the graph gives the solution set of the inequality, we need only check test points. 1 GRAPHING INEQUALITIES To graph an inequality, we carry out
the following steps. 1. Graph Equation. Graph the equation corresponding to the inequality. Use a dashed curve for or and a solid curve for or. 2. Test Points. Test one point in each region formed by the graph in Step 1. If the point satisﬁes the inequality, then all the points in that region satisfy the inequality. (In that case, shade the region to indicate that it is part of the graph.) If the test point does not satisfy the inequality, then the region isn’t part of the graph. E X AM P L E 1 \| Graphs of Inequalities Graph each inequality. x2 y2 25 (a) (b) x 2y 5 ▼ SO LUTI O N (a) The graph of x2 y2 25 is a circle of radius 5 centered at the origin. The points on the circle itself do not satisfy the inequality because it is of the form, so we graph the circle with a dashed curve, as shown in Figure 3. To determine whether the inside or the outside of the circle satisﬁes the inequality, we use the test points on the outside. To do this, we sub0, 0 1 stitute the coordinates of each point into the inequality and check whether the result satisﬁes the inequality. (Note that any point inside or outside the circle can serve as a test point. We have chosen these points for simplicity.) 6, 0 on the inside and 1 2 2 y 1 0 ≈+¥<25 (6, 0) x 1 FIGURE 3 Thus, the graph of x2 y2 25 is the set of all points inside the circle (see Figure 3). Test point x2 y2 25 Conclusion 0, 0 2 1 6, 0 2 1 02 02 0 25 62 02 36 25 Not part of graph Part of graph 476 CHAPTER 6 \| Systems of Equations and Inequalities y 1 0 (5, 5) x+2y≥5 1 x FIGURE 4 (b) The graph of x 2y 5 is the line shown in Figure 4. We use the test points 5, 5 and 1 2 on opposite sides of the line. 0, 0 1 2 Test point x 2y 5 Conclusion 0, 0 1 5 15 5 Not part of graph Part of graph Our check shows that the points above the line satisfy the inequality. Alternatively, we could put the inequality into slope-intercept form and graph it directly
: x 2y 5 2y x 5 y 1 2 x 5 2 From this form we see that the graph includes all points whose y-coordinates are greater than those on the line ; that is, the graph consists of the points on or above this line, as shown in Figure 4. y 1 2 x 5 2 ✎ Practice what you’ve learned: Do Exercises 7 and 15. ▲ ■ Systems of Inequalities We now consider systems of inequalities. The solution of such a system is the set of all points in the coordinate plane that satisfy every inequality in the system. E X AM P L E 2 \| A System of Two Inequalities Graph the solution of the system of inequalities, and label its vertices. x2 y2 25 x 2y 5 x ▼ SO LUTI O N These are the two inequalities of Example 1. In this example we wish to graph only those points that simultaneously satisfy both inequalities. The solution consists of the intersection of the graphs in Example 1. In Figure 5(a) we show the two regions on the same coordinate plane (in different colors), and in Figure 5(b) we show their intersection. b 3, 4 Vertices The points They are obtained by solving the system of equations and 5, 0 2 2 1 1 in Figure 5(b) are the vertices of the solution set. (5, 0) x x2 y2 25 x 2y 5 We solve this system of equations by substitution. Solving for x in the second equation gives, and substituting this into the ﬁrst equation gives x 5 2y b 5 2y 2 1 25 20y 4y2 1 2 y2 25 Substitute x 5 2y y2 25 Expand 2 20y 5y2 0 5y 1 4 y 2 0 Simplify Factor Thus, y 0 or y 4. When y 0, we have x 5 2 x 5 2 3. So the points of intersection of these curves are 0 2 1 5, and when y 4, we have 5, 0 1 2 and 1 3a) y 0 (_3, 4) (b) x2 y2 25 x 2y 5 FIGURE 5 b SE CTI O N 6.5 \| Systems of Inequalities 477 Note that in this case the vertices are not part of the solution set, since they don’t satisfy the inequality x 2 y 2 25 (so they are graphed as open circles in the �
��gure). They simply show where the “corners” of the solution set lie. ✎ Practice what you’ve learned: Do Exercise 33. ▲ ■ Systems of Linear Inequalities An inequality is linear if it can be put into one of the following forms: ax by c ax by c ax by c ax by c In the next example we graph the solution set of a system of linear inequalities. E X AM P L E 3 \| A System of Four Linear Inequalities Graph the solution set of the system, and label its vertices. x 3y 12 x y 8 x 0 y 0 ▼ SO LUTI O N In Figure 6 we ﬁrst graph the lines given by the equations that correspond to each inequality. To determine the graphs of the linear inequalities, we need to check only one test point. For simplicity let’s use the point d 0, 0 1. 2 Inequality Test point (0, 0) Conclusion x 3y 12 x y 8 0 12 0 3 0 2 1 0 0 0 8 Satisﬁes inequality Satisﬁes inequality 2 is below the line x 3y 12, our check shows that the region on or below the Since 0, 0 1 is below the line x y 8, our check line must satisfy the inequality. Likewise, since 0, 0 1 shows that the region on or below this line must satisfy the inequality. The inequalities x 0 and y 0 say that x and y are nonnegative. These regions are sketched in Figure 6(a), and the intersection—the solution set—is sketched in Figure 6(b). 2 Vertices The coordinates of each vertex are obtained by simultaneously solving the equations of the lines that intersect at that vertex. From the system x 3y 12 x y 8. The origin (0, 0) is also clearly a vertex. The other two vertices are 2. In this case all the ver2 and 1 0, 4 8, 0 6, 2 we get the vertex 1 at the x- and y-intercepts of the corresponding lines: tices are part of the solution set. ✎ Practice what you’ve learned: Do Exercise 39. E X AM P L E 4 \| A System of Linear Inequalities b 2 1 ▲ Graph the solution set of the system. x 2y 8 x 2y 4 3x 2y 8 ▼ SO LUTI O N We must graph
the lines that correspond to these inequalities and then shade the appropriate regions, as in Example 3. We will use a graphing calculator, so we must ﬁrst isolate y on the left-hand side of each inequality+y=8 x+3y=12 x 8 12 (a) (0, 4) (6, 2) (8, 0) (b) FIGURE 6 x 12 478 CHAPTER 6 \| Systems of Equations and Inequalities 8 8 _2 _2 FIGURE 7 y 1 0 (2, 3) 1 (6, 5) (4, 2) x FIGURE Using the shading feature of the calculator, we obtain the graph in Figure 7. The solution c set is the triangular region that is shaded in all three patterns. We then use or the Intersect command to ﬁnd the vertices of the region. The solution set is graphed in Figure 8. ✎ Practice what you’ve learned: Do Exercise 47. TRACE ▲ When a region in the plane can be covered by a (sufﬁciently large) circle, it is said to be bounded. A region that is not bounded is called unbounded. For example, the regions graphed in Figures 3, 5(b), 6(b), and 8 are bounded, whereas those in Figures 2 and 4 are unbounded. An unbounded region cannot be “fenced in”—it extends inﬁnitely far in at least one direction. ■ Application: Feasible Regions Many applied problems involve constraints on the variables. For instance, a factory manager has only a certain number of workers that can be assigned to perform jobs on the factory ﬂoor. A farmer deciding what crops to cultivate has only a certain amount of land that can be seeded. Such constraints or limitations can usually be expressed as systems of inequalities. When dealing with applied inequalities, we usually refer to the solution set of a system as a feasible region, because the points in the solution set represent feasible (or possible) values for the quantities being studied. E X AM P L E 5 \| Restricting Pollutant Outputs A factory produces two agricultural pesticides, A and B. For every barrel of A, the factory emits 0.25 kg of carbon monoxide (CO) and 0.60 kg of sulfur dioxide (SO2); and for every barrel of B, it emits 0.50 kg of CO and 0.20 kg
of SO2. Pollution laws restrict the factory’s output of CO to a maximum of 75 kg and SO2 to a maximum of 90 kg per day. (a) Find a system of inequalities that describes the number of barrels of each pesticide the factory can produce and still satisfy the pollution laws. Graph the feasible region. (b) Would it be legal for the factory to produce 100 barrels of A and 80 barrels of B per day? (c) Would it be legal for the factory to produce 60 barrels of A and 160 barrels of B per day? ▼ SO LUTI O N (a) To set up the required inequalities, it is helpful to organize the given information into a table. A B Maximum CO (kg) SO2 (kg) 0.25 0.60 0.50 0.20 75 90 We let x number of barrels of A produced per day y number of barrels of B produced per day y 400 300 200 100 3x+y=450 (60, 160) (100, 80) x+2y=300 0 100 200 300 x FIGURE 9 SE CTI O N 6.5 \| Systems of Inequalities 479 From the data in the table and the fact that x and y can’t be negative, we obtain the following inequalities. 0.25x 0.50y 75 0.60x 0.20y 90 x 0, y 0 CO inequality SO2 inequality Multiplying the ﬁrst inequality by 4 and the second by 5 simpliﬁes this to c x 2y 300 3x y 450 x 0, y 0 The feasible region is the solution of this system of inequalities, shown in Figure 9. 100, 80 1 2 lies inside the feasible region, this production plan is legal c lies outside the feasible region, this production plan is not legal. It violates the CO restriction, although it does not violate the SO2 restriction (see Figure 9). 60, 160 1 2 (b) Since the point (see Figure 9). (c) Since the point ✎ Practice what you’ve learned: Do Exercise 51. ▲ 6. ▼ CONCE PTS 1. To graph an inequality, we ﬁrst graph the corresponding (b) x y 0 x y 2. So to graph y x 1, we ﬁrst graph the equation. To decide which side of the graph of the equation b is the graph of the inequality, we
use 0, 0 1 appropriate region. 2 as such a point, graph the inequality by shading the points. Using y 1 x-y=0 x+y=2 1 x y=x+1 (c) x y 0 x y 2 (d. Shade the solution of each system of inequalities on the given graph. (a-y=0 x+y=-y=0 x+y=2 x-y=0 x+y=2 1 x 1 x ▼ SKI LLS 3–16 ■ Graph the inequality. 3. x 3 5. y x 4. y 2 6. y x 2 y 1 0 480 CHAPTER 6 \| Systems of Equations and Inequalities ✎ ✎ 7. y 2x 2 9. 2 x y 8 11. 4x 5y 20 13. y x2 1 15. x2 y2 25 8. y x 5 10. 3x 4y 12 0 12. x2 y 10 14. x2 y 2 9 y 1 x2 16. 2 1 1 2 17–20 ■ An equation and its graph are given. Find an inequality whose solution is the shaded region. 17. y 1 2 x 1 18. y x2 19. x 2 y 2 4 20. y x 3 4x ✎ 33. x2 y2 4 x y 0 35. b x2 y 0 2x2 y 12 37. b x 2y 14 3x y 0 x y 2 ✎ 39 41. y x 1 d x 2y 12 x 1 0 c 43. c 45. x2 y2 8 x 2 y 0 x2 y2 9 x y 0 x 0 34. x 0 y 0 x y 10 x2 y2 9 36. x2 y2 9 d 2x y2 1 38. b y x 6 3x 2y 12 x 2y 2 c x 0 y 0 y 4 2x y 8 x y 12 d y 1 3x y 6 2 x 6 c x2 y 0 x y 6 x y 6 y x3 c y 2x 4 x y 0 40. 42. 44. 46. c 47–50 ■ Use a graphing calculator to graph the solution of the system of inequalities. Find the coordinates of all vertices, correct to one decimal place. c 21–46 ■ Graph the solution of the system of inequalities. Find the coordinates of all vertices, and determine whether the solution set is bounded. 49. c y
6x x2 x y 4 ✎ 47. y x 3 y 2x 6 y 8 48. x y 12 2x y 24 x y 6 c 50. y x3 2x y 0 y 2x 6 21. 23 2x 5 b 25. b y 2x 8 y 1 2 x 5 x 0, y 0 27. c x 0 y 0 3x 5y 15 3x 2y 9 d 29. y 9 x2 x 0, y 0 31. b y 9 x2 y x 3 22. 24. 2x 3y 12 3x y 21 x y 0 4 y 2x b 26. b 4x 3y 18 2x y 8 x 0, y 0 28. 30. c x 2 y 12 2x 4y 8 c y x2 y 4 x 0 32. c y x2 x y 6 b b b c ✎ ▼ APPLICATIONS 51. Publishing Books A publishing company publishes a total of no more than 100 books every year. At least 20 of these are nonﬁction, but the company always publishes at least as much fiction as nonﬁction. Find a system of inequalities that describes the possible numbers of ﬁction and nonﬁction books that the company can produce each year consistent with these policies. Graph the solution set. 52. Furniture Manufacturing A man and his daughter manufacture unﬁnished tables and chairs. Each table requires 3 hours of sawing and 1 hour of assembly. Each chair requires 2 hours of sawing and 2 hours of assembly. Between the two of them, they can put in up to 12 hours of sawing and 8 hours of assembly work each day. Find a system of inequalities that describes all possible combinations of tables and chairs that they can make daily. Graph the solution set. 53. Coffee Blends A coffee merchant sells two different coffee blends. The Standard blend uses 4 oz of arabica and 12 oz of robusta beans per package; the Deluxe blend uses 10 oz of arabica and 6 oz of robusta beans per package. The merchant has 80 lb of arabica and 90 lb of robusta beans available. Find a system of inequalities that describes the possible number of Standard and Deluxe packages the merchant can make. Graph the solution set. 54. Nutrition A cat food manufacturer uses ﬁsh and beef byproducts. The ﬁsh contains 12 g of protein and 3 g of fat per ounce. The beef contains 6
g of protein and 9 g of fat per ounce. Each can of cat food must contain at least 60 g of protein and 45 g of fat. Find a system of inequalities that describes the possible number of ounces of ﬁsh and beef that can be used in each can to satisfy these minimum requirements. Graph the solution set. CHAPTER 6 \| REVIEW ▼ P R O P E RTI LAS Substitution Method (p. 443) To solve a pair of equations in two variables by substitution: CHAPTER 6 \| Review 481 ▼ DISCOVE RY • DISCUSSION • WRITI NG 55. Shading Unwanted Regions To graph the solution of a system of inequalities, we have shaded the solution of each inequality in a different color; the solution of the system is the region where all the shaded parts overlap. Here is a different method: For each inequality, shade the region that does not satisfy the inequality. Explain why the part of the plane that is left unshaded is the solution of the system. Solve the following system by both methods. Which do you prefer? Why? x 2y 4 x y 1 x 3y 9 x 3 d 1. Add a nonzero multiple of one equation to another. 2. Multiply an equation by a nonzero constant. 1. Solve for one variable in terms of the other variable in one 3. Interchange the position of two equations in the system. equation. 2. Substitute into the other equation to get an equation in one variable, and solve for this variable. 3. Substitute the value(s) of the variable you have found into either original equation, and solve for the remaining variable. Elimination Method (p. 444) To solve a pair of equations in two variables by elimination: 1. Multiply the equations by appropriate constants so that the term(s) involving one of the variables are of opposite sign in the equations. 2. Add the equations to eliminate that one variable; this gives an equation in the other variable. Solve for this variable. 3. Substitute the value(s) of the variable that you have found into either original equation, and solve for the remaining variable. Graphical Method (p. 446) To solve a pair of equations in two variables graphically: 1. Put each equation in function form, y f x 1 2 2. Use a graphing calculator to graph the equations on a common screen. 3. Find the points
of intersection of the graphs. The solutions are the x- and y-coordinates of the points of intersection. Gaussian Elimination (p. 459) When we use Gaussian elimination to solve a system of linear equations, we use the following operations to change the system to an equivalent simpler system: Number of Solutions of a System of Linear Equations (pp. 451, 460) A system of linear equations can have: 1. A unique solution for each variable. 2. No solution, in which case the system is inconsistent. 3. Inﬁnitely many solutions, in which case the system is dependent. How to Determine the Number of Solutions of a Linear System (p. 460) When we use Gaussian elimination to solve a system of linear equations, then we can tell that the system has: 1. No solution (is inconsistent) if we arrive at a false equation of the form 0 c, where c is nonzero. 2. Inﬁnitely many solutions (is dependent) if the system is consistent but we end up with fewer equations than variables (after discarding redundant equations of the form 0 0). Partial Fractions (p. 469) The partial fraction decomposition of a rational function where the degree of P is less than the degree of Q) is a sum of r simpler fractional expressions that equal when brought to a common denominator. The denominator of each simpler fraction is either a linear or quadratic factor of Q(x) or a power of such a linear or quadratic factor. So to ﬁnd the terms of the partial fraction decomposition, we ﬁrst factor Q(x) into linear and irreducible x 1 2 Graphing Inequalities (p. 475) To graph an inequality: 1. Graph the equation that corresponds to the inequality. This “boundary curve” divides the coordinate plane into separate regions. 2. Use test points to determine which region(s) satisfy the inequality. 3. Shade the region(s) that satisfy the inequality, and use a solid line for the boundary curve if it satisﬁes the inequality ( or ) and a dashed line if it does not (< or >). Graphing Systems of Inequalities (p. 476) To graph the solution of a system of inequalities (or feasible region determined by the inequalities): 1. Graph all the inequalities on the same coordinate plane. 2. The solution is the
intersection of the solutions of all the in- equalities, so shade the region that satisﬁes all the inequalities. 3. Determine the coordinates of the intersection points of all the boundary curves that touch the solution set of the system. These points are the vertices of the solution. 482 CHAPTER 6 \| Systems of Equations and Inequalities quadratic factors. The terms then have the following forms, depending on the factors of Q(x). 1. For every distinct linear factor ax b, there is a term of the form A ax b 2. For every repeated linear factor the form ax b 1 2 m, there are terms of A1 ax b A2 ax b 1 2 2 p Am ax b m 1 2 ax2 bx c 3. For every distinct quadratic factor, there is a term of the form Ax B ax2 bx c 4. For every repeated quadratic factor are terms of the form ax2 bx c 1 m, there 2 A1x B1 ax2 bx c A2x B2 ax2 bx c 1 2 2 p Amx Bm ax2 bx c 1 m 2 ▼ CO N C E P T S U M MARY Section 6.1 ■ Solve a system of equations using the substitution method ■ Solve a system of equations using the elimination method ■ Solve a system of equations using the graphical method Section 6.2 ■ Solve a system of two linear equations in two variables Review Exercises 1–6, 11, 12 7–10, 13, 14 15–18 Review Exercises 5–8 5–8 27–28 Review Exercises 19–26 19–26 29–30 ■ Determine whether a system of two linear equations in two variables has one solution, inﬁnitely many solutions, or no solution ■ Model with linear systems in two variables Section 6.3 ■ Use Gaussian elimination to solve a system of three (or more) linear equations ■ Determine whether a system of (three or more) linear equations has one solution, inﬁnitely many solutions, or no solution ■ Model with linear systems in three (or more) variables Section 6.4 ■ Find the form of the partial fraction decomposition of a rational expression in the following cases: Review Exercises ■ Denominator contains distinct linear factors ■ Denominator contains repeated linear factors ■ Denominator contains distinct quadratic factors ■ Den
ominator contains repeated quadratic factors ■ Find the partial fraction decomposition of a rational expression in the above cases Section 6.5 ■ Graph the solution of an inequality ■ Graph the solution of a system of inequalities ■ Graph the solution of a system of linear inequalities 31–32 33–34 35–36 37–38 31–38 Review Exercises 39–44 45–46, 49–50 47–48, 51–52 ▼ E X E RC I S E S 1–4 ■ Two equations and their graphs are given. Find the intersection point(s) of the graphs by solving the system. 1. 2x 3y 7 x 2y 0 e 2. 3x y 8 y x2 5x e y 1 0 1 3. x2 y 2 x2 3x. x y 2 x2 y2 4y –10 ■ Solve the system of equations and graph the lines. 5. 7. 9. 3x y 5 2x y 5 2x 7y 28 y 2 7 x 4 e e 2x 0y 01 x 3y 10 3x 4y 15 y 2x 6 y x 3 6x 8y 15 2 x 2y 4 3 6. 8. e e 10. 2x 5y 09 x 3y 01 7x 2y 14 c 11–14 ■ Solve the system of equations. c 11. 13. y x2 2x y 6 x e 3x 4 y x 8 y 4 6 12. x2 y2 8 y x 2 e x2 y2 10 x2 2y2 7y 0 e 14. CHAPTER 6 \| Review 483 19–26 ■ Find the complete solution of the system, or show that the system has no solution. 19. x 0y 2z 06 2x 2y 5z 12 0x 2y 3z 09 21. x 2y 03z 1 c 2x 0y 05z 3 2x 7y 11z 2 20. 22. x 2y 3z 1 x 3y 0z 0 2x 3y 6z 6 c x 0y 0z 0„ 2 2x 2y 3z 4w 5 x 2y 2z 4„ 9 x 0y 2z 3„ 5 c e 23. 0x 3y 15z 4 4x 3y 15z 5 24. d 2x 3y 4z 03 4x 5y 9z 13 2x 3y 7z 60 25. x 4
y z 8 2x 6y z 9 x 6y 4z 15 26. x 0y z 0„ 02 c 2x 0y z 2„ 12 3y z 0„ 04 x 3y z 0w 10 27. Eleanor has two children, Kieran and Siobhan. Kieran is 4 d c years older than Siobhan, and the sum of their ages is 22. How old are the children? 28. A man invests his savings in two accounts, one paying 6% interest per year and the other paying 7%. He has twice as much invested in the 7% account as in the 6% account, and his annual interest income is $600. How much is invested in each account? 29. A piggy bank contains 50 coins, all of them nickels, dimes, or quarters. The total value of the coins is $5.60, and the value of the dimes is ﬁve times the value of the nickels. How many coins of each type are there? 30. Tornie is a commercial ﬁsherman who trolls for salmon on the British Columbia coast. One day he catches a total of 25 ﬁsh of three salmon species: coho, sockeye, and pink. He catches three more coho than the other two species combined; moreover, he catches twice as many coho as sockeye. How many ﬁsh of each species has he caught? 31–38 ■ Find the partial fraction decomposition of the rational function. 31. 3x 1 x2 2x 15 32. 8 x3 4x d 15–18 ■ Use a graphing device to solve the system, correct to the nearest hundredth. 33. x 1 15. 17. 0.32x 0.43y 000 0.07x 0.12y 341 x y2 10 22 y 12 x 1 e e 16. 18. 212 x 322 y 00,660 7137x 3931y 20,000 y 5x x y x5 5 e e 2x 4 x 1 2 2 34. 6x 4 x3 2x2 4x 8 36. 5x2 3x 10 x4 x2 2 35. 2x 1 x3 x 37. 3x2 x 6 x2 2 2 1 2 38. x2 x 1 x2 1 2 x 1 2 49–52 ■ Graph the solution set of the system of inequalities. Find the coordinates of all vert
ices, and determine whether the solution set is bounded or unbounded. 49. x2 y2 9 x y 0 51. b x 0, y 0 x 2y 12 y x 4 50. y x2 4 y 20 52. b x 4 x y 24 x 2y 12 53–54 ■ Solve for x, y, and z in terms of a, b, and c 53. c c 54. ax by cz a b c bx by cz c cx cy cz c 1 a b, b c, c 0 2 55. For what values of k do the following three lines have a c common point of intersection? x y 12 kx y 0 y x 2k 56. For what value of k does the following system have inﬁnitely many solutions? kx 0y z 0 x 2y kz 0 3z 0 x c 484 CHAPTER 6 \| Systems of Equations and Inequalities 39–40 ■ An equation and its graph are given. Find an inequality whose solution is the shaded region. 39. x y2 4 40. x2 y2 41–44 ■ Graph the inequality. 41. 43. 3x y 6 x2 y2 9 42. 44. y x2 3 x y2 4 45–48 ■ The ﬁgure shows the graphs of the equations corresponding to the given inequalities. Shade the solution set of the system of inequalities. 45. y x2 3x 3 x 1 y 1 46. y x 1 x2 y2 47. x y 2 y x 2 x 3 48. y 2x y 2x ■ CHAPTER 6 \| TEST 1–2 ■ A system of equations is given. (a) Determine whether the system is linear or nonlinear. (b) Find all solutions of the system. 1. 3x 5y 4 0x 4y 7 2. 10x y2 4 02x y 2 3. Use a graphing device to ﬁnd all solutions of the system correct to two decimal places. b b x 2y 1 y x3 2x2 e 4. In 21 hours an airplane travels 600 km against the wind. It takes 50 min to travel 300 km with 2 the wind. Find the speed of the wind and the speed of the airplane in still air. 5–8 ■ A system of linear equations is given. (a) Find the complete solution of the system, or show that there is no solution. (b)
State whether the system is inconsistent, dependent, or neither. 5. 6. 7. 8. x 2y 0z 3 x 3y 2z 3 2x 3y 0z 8 c x y 9z 8 4z 7 3x y 0z 5 2x 0y 0z 0 c 3x 2y 3z 1 x 4y 5z 1 c x y 2z 08 2x y 20 2x 2y 5z 15 9. Anne, Barry, and Cathy enter a coffee shop. Anne orders two coffees, one juice, and two c doughnuts and pays $6.25. Barry orders one coffee and three doughnuts and pays $3.75. Cathy orders three coffees, one juice, and four doughnuts and pays $9.25. Find the price of coffee, juice, and doughnuts at this coffee shop. 10–11 ■ Graph the solution set of the system of inequalities. Label the vertices with their coordinates. 10. 2x 0y 8 x 0y 2 x 2y 4 11. x2 y 5 y 2x 5 e 12–13 ■ Find the partial fraction decomposition of the rational function. c 4x 1 2 x 1 x 2 2 1 12. 1 2 13. 2x 3 x3 3x 485 LINEAR PROGRAMMING Linear programming is a modeling technique that is used to determine the optimal allocation of resources in business, the military, and other areas of human endeavor. For example, a manufacturer who makes several different products from the same raw materials can use linear programming to determine how much of each product should be produced to maximize the proﬁt. This modeling technique is probably the most important practical application of systems of linear inequalities. In 1975 Leonid Kantorovich and T. C. Koopmans won the Nobel Prize in economics for their work in the development of this technique. Although linear programming can be applied to very complex problems with hundreds or even thousands of variables, we consider only a few simple examples to which the graphical methods of Section 6.4 can be applied. (For large numbers of variables a linear programming method based on matrices is used.) Let’s examine a typical problem. E X AM P L E 1 \| Manufacturing for Maximum Proﬁt A small shoe manufacturer makes two styles of shoes: oxfords and loafers. Two machines are used in the process: a cutting machine and a sewing machine. Each type of shoe requires 15 min per pair on the cutting
machine. Oxfords require 10 min of sewing per pair, and loafers require 20 min of sewing per pair. Because the manufacturer can hire only one operator for each machine, each process is available for just 8 hours per day. If the proﬁt is $15 on each pair of oxfords and $20 on each pair of loafers, how many pairs of each type should be produced per day for maximum proﬁt? ▼ SO LUTI O N First we organize the given information into a table. To be consistent, let’s convert all times to hours. Oxfords Loafers Time available Time on cutting machine (h) Time on sewing machine (h Proﬁt $15 $20 We describe the model and solve the problem in four steps. Choose the variables. To make a mathematical model, we ﬁrst give names to the variable quantities. For this problem we let x number of pairs of oxfords made daily y number of pairs of loafers made daily Find the objective function. Our goal is to determine which values for x and y give maximum proﬁt. Since each pair of oxfords provides $15 proﬁt and each pair of loafers $20, the total proﬁt is given by P 15x 20y This function is called the objective function. Graph the feasible region. The larger x and y are, the greater is the proﬁt. But we cannot choose arbitrarily large values for these variables because of the restrictions, or constraints, in the problem. Each restriction is an inequality in the variables. Because loafers produce more proﬁt, it would seem best to manufacture only loafers. Surprisingly, this does not turn out to be the most proﬁtable solution. 486 Linear Programming 487 1 In this problem the total number of cutting hours needed is 4 x 1 4 y. Since only 8 hours are available on the cutting machine, we have 4 x 1 1 4 y 8 Similarly, by considering the amount of time needed and available on the sewing machine, we get 1 6 x 1 3 y 8 We cannot produce a negative number of shoes, so we also have x 0 and y 0 Thus, x and y must satisfy the constraints If we multiply the ﬁrst inequality by 4 and the second by 6, we obtain the simpliﬁed system d x 2y 32 x 2y 48 2

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