Datasets:
jacobcd52
/
hs_math_cleaned

Dataset card Data Studio Files
xet
 Community
1
text
stringlengths
235
3.08k
table of points as in Table 5. x x x about the x-axis. State its domain, range, about the x-axis, we multiply f (x) by βˆ’1 to get, x βˆ’3 βˆ’2 1 ) g(x) = βˆ’ ( __ 4 x βˆ’64 βˆ’16 βˆ’1 βˆ’4 Table 5 0 1 2 3 βˆ’1 βˆ’0.25 βˆ’0.0625 βˆ’0.0156 Plot the y-intercept, (0, βˆ’1), along with two other points. We can use (βˆ’1, βˆ’4) and (1, βˆ’0.25). Draw a smooth curve connecting the points: g(x) 10 8 6 4 2 –1–2 –4 –6 –8 –10 y = 0 –2 –5 –3 –4 (βˆ’1, βˆ’4) (1, βˆ’0.25) 5 3 21 4 (0, βˆ’1) x x g(x) = βˆ’ 1 4 The domain is (βˆ’βˆž, ∞); the range is (βˆ’βˆž, 0); the horizontal asymptote is y = 0. Figure 11 Try It #5 Find and graph the equation for a function, g(x), that reflects f (x) = 1.25x about the y-axis. State its domain, range, and asymptote. Summarizing Translations of the Exponential Function Now that we have worked with each type of translation for the exponential function, we can summarize them in Table 6 to arrive at the general equation for translating exponential functions. Translations of the Parent Function f (x) = b x Translation Form Shift β€’ Horizontally c units to the left β€’ Vertically d units up Stretch and Compress β€’ Stretch if | a | > 1 β€’ Compression if 0 < | a | < 1 Reflect about the x-axis Reflect about the y-axis General equation for all translations Table 6 f (xx) = ab x f (x) = βˆ’b x 1 ) f (x) = bβˆ’x = ( _ b x f (x) = ab x + c + d Download the OpenStax text for free at http://cnx.org/content/col11759/latest. SECTION 6.2 GRAPHS OF EXPONENTIAL FUNCTIONS 487 translations of exponential functions A translation of an exponential function has the form f (x) = ab x + c + d Where the parent function, y = b x, b >
1, is β€’ shifted horizontally c units to the left. β€’ stretched vertically by a factor of ∣ a ∣ if ∣ a ∣ > 0. β€’ compressed vertically by a factor of ∣ a ∣ if 0 < ∣ a ∣ < 1. β€’ shifted vertically d units. β€’ reflected about the x-axis when a < 0. Note the order of the shifts, transformations, and reflections follow the order of operations. Example 6 Writing a Function from a Description Write the equation for the function described below. Give the horizontal asymptote, the domain, and the range. β€’ f (x) = e x is vertically stretched by a factor of 2, reflected across the y-axis, and then shifted up 4 units. Solution We want to find an equation of the general form f (x) = ab x + c + d. We use the description provided to find a, b, c, and d. β€’ We are given the parent function f (x) = e x, so b = e. β€’ The function is stretched by a factor of 2, so a = 2. β€’ The function is reflected about the y-axis. We replace x with βˆ’x to get: eβˆ’x. β€’ The graph is shifted vertically 4 units, so d = 4. Substituting in the general form we get, f (x) = ab x + c + d = 2eβˆ’x + 0 + 4 = 2eβˆ’x + 4 The domain is (βˆ’βˆž, ∞); the range is (4, ∞); the horizontal asymptote is y = 4. Try It #6 Write the equation for function described below. Give the horizontal asymptote, the domain, and the range. 1 _ β€’ f (x) = e x is compressed vertically by a factor of, reflected across the x-axis and then shifted down 2 units. 3 Access this online resource for additional instruction and practice with graphing exponential functions. β€’ Graph Exponential Functions (http://openstaxcollege.org/l/graphexpfunc) Download the OpenStax text for free at http://cnx.org/content/col11759/latest. SECTION 6.2 SECTION EXERCISES 489 Figure 13 fx=ab x y B C D A E F x Figure 13 19. b 21. a 20. b 22. a flx 23. fx= x 24. fx=xβˆ’ 25
. fx=βˆ’x+ f x=x 26. fx=βˆ’x 27. hx=x+ 28. fx=xβˆ’ 29. fx=βˆ’xβˆ’ ) 30. fx=( x βˆ’ 31. fx=βˆ’x+ f x=xTh 32. fx 35. fx 33. fx 36. fxx 34. fxts left 37. fxy y =x 38. y –– – – – – – – – – 39. x – – – – y –– – – – – 40. x – – – – y –– – – – – x Download the OpenStax text for free at http://cnx.org/content/col11759/latest. 49 0 CHAPTER 6 EXPONENTIAL AND LOGARITHMIC FUNCTIONS, fi 41. y 42. x – – – – – – – – –– – – – – NUMERIC y –– – – – – x x 43. gx= xβˆ’g 44. f x=xβˆ’βˆ’f 45. hx=βˆ’ ) ( x +hβˆ’ TECHNOLOGY f x=ab x +d 46. βˆ’=βˆ’ ( ) 47. = ) ( ) 49. =( 48. =x+ 50. βˆ’=βˆ’x++ βˆ’ xβˆ’ βˆ’x x EXTENSIONS 51. fx=bx x. Th gx> b 52. fx=x gx=xβˆ’hx=( bn ) b xn b x( b> )x. Th 53. 54. REAL-WORLD APPLICATIONS 55. =+/) = 56. a. b. Download the OpenStax text for free at http://cnx.org/content/col11759/latest. SECTION 6.3 LOGARITHMIC FUNCTIONS 491 LEARNING OBJECTIVES In this section, you will: β€’ efine a logarithmic function emphasiing it as an inerse of an eponential function. β€’ Conert a logarithm to eponential form. β€’ Conert from eponential to logarithmic form. β€’ aluate logarithms. β€’ aluate logarithms using natural and common logarithms. 6.3 LOGARITHMIC FUNCTIONS Figure 1 Devastation of March 11, 2011 earthquake in Honshu, Japan. (credit: Daniel Pierce) In 2010, a
major earthquake struck Haiti, destroying or damaging over 285,000 homes[19]. One year later, another, stronger earthquake devastated Honshu, Japan, destroying or damaging over 332,000 buildings,[20] like those shown in Figure 1. Even though both caused substantial damage, the earthquake in 2011 was 100 times stronger than the earthquake in Haiti. How do we know? The magnitudes of earthquakes are measured on a scale known as the Richter Scale. The Haitian earthquake registered a 7.0 on the Richter Scale[21] whereas the Japanese earthquake registered a 9.0.[22] The Richter Scale is a base-ten logarithmic scale. In other words, an earthquake of magnitude 8 is not twice as great as an earthquake of magnitude 4. It is 108 βˆ’ 4 = 104 = 10,000 times as great! In this lesson, we will investigate the nature of the Richter Scale and the base-ten function upon which it depends. Converting from Logarithmic to Exponential Form In order to analyze the magnitude of earthquakes or compare the magnitudes of two different earthquakes, we need to be able to convert between logarithmic and exponential form. For example, suppose the amount of energy released from one earthquake were 500 times greater than the amount of energy released from another. We want to calculate the difference in magnitude. The equation that represents this problem is 10 x = 500, where x represents the difference in magnitudes on the Richter Scale. How would we solve for x? We have not yet learned a method for solving exponential equations. None of the algebraic tools discussed so far is sufficient to solve 10 x = 500. We know that 102 = 100 and 103 = 1000, so it is clear that x must be some value between 2 and 3, since y = 10x is increasing. We can examine a graph, as in Figure 2, to better estimate the solution. y 1,000 800 600 400 200 –3 –2 –1 –200 1 2 3 Figure 2 y = 10x y = 500 x 19 http://earthquake.usgs.gov/earthquakes/eqinthenews/2010/us2010rja6/#summary. Accessed 3/4/2013. 20 http://earthquake.usgs.gov/earthquakes/eqinthenews/2011/usc001xgp/#summary. Accessed 3/4/2013. 21 http://earthquake.usgs.gov/earthquakes/eqint
henews/2010/us2010rja6/. Accessed 3/4/2013. 22 http://earthquake.usgs.gov/earthquakes/eqinthenews/2011/usc001xgp/#details. Accessed 3/4/2013. Download the OpenStax text for free at http://cnx.org/content/col11759/latest. 49 2 CHAPTER 6 EXPONENTIAL AND LOGARITHMIC FUNCTIONS Estimating from a graph, however, is imprecise. To find an algebraic solution, we must introduce a new function. Observe that the graph in Figure 2 passes the horizontal line test. The exponential function y = b x is one-to-one, so its inverse, x = b y is also a function. As is the case with all inverse functions, we simply interchange x and y and solve for y to find the inverse function. To represent y as a function of x, we use a logarithmic function of the form y = logb(x). The base b logarithm of a number is the exponent by which we must raise b to get that number. We read a logarithmic expression as, β€œThe logarithm with base b of x is equal to y,” or, simplified, β€œlog base b of x is y.” We can also say, β€œb raised to the power of y is x,” because logs are exponents. For example, the base 2 logarithm of 32 is 5, because 5 is the exponent we must apply to 2 to get 32. Since 25 = 32, we can write log2 32 = 5. We read this as β€œlog base 2 of 32 is 5.” We can express the relationship between logarithmic form and its corresponding exponential form as follows: Note that the base b is always positive. logb(x) = y ⇔ b y = x, b > 0, b β‰  1 = logb(x) = y to Think b to the y = x Because logarithm is a function, it is most correctly written as logb(x), using parentheses to denote function evaluation, just as we would with f (x). However, when the input is a single variable or number, it is common to see the parentheses dropped and the expression written without parentheses, as logb x. Note that many calculators require parentheses around
the x. We can illustrate the notation of logarithms as follows: = logb(c) = a means ba = c to Notice that, comparing the logarithm function and the exponential function, the input and the output are switched. This means y = logb (x) and y = b x are inverse functions. definition of the logarithmic function A logarithm base b of a positive number x satisfies the following definition. For x > 0, b > 0, b β‰  1, where, β€’ we read logb (x) as, β€œthe logarithm with base b of x” or the β€œlog base b of x.” β€’ the logarithm y is the exponent to which b must be raised to get x. y = logb(x) is equivalent to b y = x Also, since the logarithmic and exponential functions switch the x and y values, the domain and range of the exponential function are interchanged for the logarithmic function. Therefore, β€’ the domain of the logarithm function with base b is (0, ∞). β€’ the range of the logarithm function with base b is ( βˆ’βˆž, ∞). Q & A… Can we take the logarithm of a negative number? No. Because the base of an exponential function is always positive, no power of that base can ever be negative. We can never take the logarithm of a negative number. Also, we cannot take the logarithm of zero. Calculators may output a log of a negative number when in complex mode, but the log of a negative number is not a real number. How To… Given an equation in logarithmic form logb(x) = y, convert it to exponential form. 1. Examine the equation y = logb(x) and identify b, y, and x. 2. Rewrite logb(x) = y as b y = x. Download the OpenStax text for free at http://cnx.org/content/col11759/latest. SECTION 6.3 LOGARITHMIC FUNCTIONS 493 Example 1 Converting from Logarithmic Form to Exponential Form Write the following logarithmic equations in exponential form. a. log6. log3(9) = 2 β€” Solution First, identify the values of b, y, and x
. Then, write the equation in the form b y = x. 6. Therefore, the equation log6( √ a. log6( √ 1 _ Here, b = 6, y =, and x = √ 2 1 __ 6 = √ 6. 2 Here, b = 3, y = 2, and x = 9. Therefore, the equation log3(9) = 2 is equivalent to 32 = 9. 1 6 ) = _ is equivalent to 2 b. log3(9) = 2 1 6 ) = _ 2 β€” β€” β€” Try It #1 Write the following logarithmic equations in exponential form. a. log10(1,000,000) = 6 b. log5(25) = 2 Converting from Exponential to Logarithmic Form To convert from exponents to logarithms, we follow the same steps in reverse. We identify the base b, exponent x, and output y. Then we write x = logb(y). Example 2 Converting from Exponential Form to Logarithmic Form Write the following exponential equations in logarithmic form. 1 ______ 10,000 b. 52 = 25 c. 10βˆ’4 = a. 23 = 8 Solution First, identify the values of b, y, and x. Then, write the equation in the form x = logb(y). a. 23 = 8 Here, b = 2, x = 3, and y = 8. Therefore, the equation 23 = 8 is equivalent to log2(8) = 3. b. 52 = 25 Here, b = 5, x = 2, and y = 25. Therefore, the equation 52 = 25 is equivalent to log5(25) = 2. c. 10βˆ’4 = 1 ______ 10,000 Here, b = 10, x = βˆ’4, and y = ) = βˆ’4. log10 ( 1 _ 10,000 1 _ 10,000. Therefore, the equation 10βˆ’4 = is equivalent to 1 _ 10,000 Try It #20 Write the following exponential equations in logarithmic form. a. 32 = 9 b. 53 = 125 1 __ c. 2βˆ’1 = 2 Evaluating Logarithms Knowing the squares, cubes, and roots of numbers allows us to evaluate many logarithms mentally. For example, consider log2(8). We ask, β€œTo what exponent must 2 be raised in order to get 8?οΏ½
οΏ½ Because we already know 23 = 8, it follows that log2(8) = 3. Now consider solving log7(49) and log3(27) mentally. β€’ We ask, β€œTo what exponent must 7 be raised in order to get 49?” We know 72 = 49. Therefore, log7(49) = 2 β€’ We ask, β€œTo what exponent must 3 be raised in order to get 27?” We know 33 = 27. Therefore, log3(27) = 3 Even some seemingly more complicated logarithms can be evaluated without a calculator. For example, let’s evaluate log 2 _ 4 ) mentally. 3 ( _ 9 2 2 4 )?” We know 22 = 4 and 32 = 9, so ( _ _ _ β€’ We ask, β€œTo what exponent must be raised in order to get 3 9 3 4 ) = 2. 3 ( _ 9 Therefore, log 2 _ 2 4 _ =. 9 Download the OpenStax text for free at http://cnx.org/content/col11759/latest. 49 4 CHAPTER 6 EXPONENTIAL AND LOGARITHMIC FUNCTIONS How To… Given a logarithm of the form y = logb(x), evaluate it mentally. 1. Rewrite the argument x as a power of b : b y = x. 2. Use previous knowledge of powers of b identify y by asking, β€œTo what exponent should b be raised in order to get x?” Example 3 Solving Logarithms Mentally Solve y = log4(64) without using a calculator. Solution First we rewrite the logarithm in exponential form: 4y = 64. Ne xt, we ask, β€œTo what exponent must 4 be raised in order to get 64?” We know 43 = 64 therefore, log4(64) = 3. Try It #3 Solve y = log121(11) without using a calculator. Example 4 Evaluating the Logarithm of a Reciprocal Evaluate y = log3 ( ) without using a calculator. 1 _ 27 y Solution First we rewrite the logarithm in exponential form: 3 raised in order to get?” 1 _ 27 We know 33 = 27, but what must we do to get the reciprocal, We use this information to write 1 _ 27 = 1 _ 27. Ne xt, we ask, β€œ
To what exponent must 3 be? Recall from working with exponents that bβˆ’a = 1 _ ba. Therefore, log3 ( 1 _ 27 ) = βˆ’3. 3βˆ’3 = 1 __ 33 1 __ 27 = Try It #4 Evaluate y = log2 ( 1 _ 32 ) without using a calculator. Using Common Logarithms Sometimes we may see a logarithm written without a base. In this case, we assume that the base is 10. In other words, the expression log(x) means log10(x). We call a base-10 logarithm a common logarithm. Common logarithms are used to measure the Richter Scale mentioned at the beginning of the section. Scales for measuring the brightness of stars and the pH of acids and bases also use common logarithms. definition of the common logarithm A common logarithm is a logarithm with base 10. We write log10(x) simply as log(x). The common logarithm of a positive number x satisfies the following definition. For x > 0, y = log(x) is equivalent to 10 y = x We read log(x) as, β€œthe logarithm with base 10 of x” or β€œlog base 10 of x.” The logarithm y is the exponent to which 10 must be raised to get x. Download the OpenStax text for free at http://cnx.org/content/col11759/latest. SECTION 6.3 LOGARITHMIC FUNCTIONS 495 How To… Given a common logarithm of the form y = log(x), evaluate it mentally. 1. Rewrite the argument x as a power of 10: 10 y = x. 2. Use previous knowledge of powers of 10 to identify y by asking, β€œTo what exponent must 10 be raised in order to get x?” Example 5 Finding the Value of a Common Logarithm Mentally Evaluate y = log(1,000) without using a calculator. Solution First we rewrite the logarithm in exponential form: 10y = 1,000. Ne xt, we ask, β€œTo what exponent must 10 be raised in order to get 1,000?” We know 103 = 1,000 therefore, log(1,000) = 3. Try It #5 Evaluate y = log(
1,000,000). How To… Given a common logarithm with the form y = log(x), evaluate it using a calculator. 1. Press [LOG]. 2. Enter the value given for x, followed by [ ) ]. 3. Press [ENTER]. Example 6 Finding the Value of a Common Logarithm Using a Calculator Evaluate y = log(321) to four decimal places using a calculator. Solution β€’ Press [LOG]. β€’ Enter 321, followed by [ ) ]. β€’ Press [ENTER]. Rounding to four decimal places, log(321) β‰ˆ 2.5065. Analysis Note that 102 = 100 and that 103 = 1000. Since 321 is between 100 and 1000, we know that log(321) must be between log(100) and log(1000). This gives us the following: 100 < 321 < 1000 2 < 2.5065 < 3 Try It #6 Evaluate y = log(123) to four decimal places using a calculator. Example 7 Rewriting and Solving a Real-World Exponential Model The amount of energy released from one earthquake was 500 times greater than the amount of energy released from another. The equation 10x = 500 represents this situation, where x is the difference in magnitudes on the Richter Scale. To the nearest thousandth, what was the difference in magnitudes? Solution We begin by rewriting the exponential equation in logarithmic form. 10x = 500 log(500) = x Use the definition of the common log. Download the OpenStax text for free at http://cnx.org/content/col11759/latest. 49 6 CHAPTER 6 EXPONENTIAL AND LOGARITHMIC FUNCTIONS Ne xt we evaluate the logarithm using a calculator: β€’ Press [LOG]. β€’ Enter 500, followed by [ ) ]. β€’ Press [ENTER]. β€’ To the nearest thousandth, log(500) β‰ˆ 2.699. The difference in magnitudes was about 2.699. Try It #7 The amount of energy released from one earthquake was 8,500 times greater than the amount of energy released from another. The equation 10x = 8500 represents this situation, where x is the difference in magnitudes on the Richter Scale. To the nearest thousandth, what was the difference in magnitudes? Using Natural Logarithms The most frequently used base for logarithms is e. Base e logarithms are important in calculus
and some scientific applications; they are called natural logarithms. The base e logarithm, loge(x), has its own notation, ln(x). Most values of ln(x) can be found only using a calculator. The major exception is that, because the logarithm of 1 is always 0 in any base, ln1 = 0. For other natural logarithms, we can use the ln key that can be found on most scientific calculators. We can also find the natural logarithm of any power of e using the inverse property of logarithms. definition of the natural logarithm A natural logarithm is a logarithm with base e. We write loge(x) simply as ln(x). The natural logarithm of a positive number x satisfies the following definition. For x > 0, y = ln(x) is equivalent to e y = x We read ln(x) as, β€œthe logarithm with base e of x” or β€œthe natural logarithm of x.” The logarithm y is the exponent to which e must be raised to get x. Since the functions y = e and y = ln(x) are inverse functions, ln(e x) = x for all x and e = x for x > 0. How To… Given a natural logarithm with the form y = ln(x), evaluate it using a calculator. 1. Press [LN]. 2. Enter the value given for x, followed by [ ) ]. 3. Press [ENTER]. Example 8 Evaluating a Natural Logarithm Using a Calculator Evaluate y = ln(500) to four decimal places using a calculator. Solution β€’ Press [LN]. β€’ Enter 500, followed by [ ) ]. β€’ Press [ENTER]. Rounding to four decimal places, ln(500) β‰ˆ 6.2146 Try It #8 Evaluate ln(βˆ’500). Access this online resource for additional instruction and practice with logarithms. β€’ Introduction to Logarithms (http://openstaxcollege.org/l/intrologarithms) Download the OpenStax text for free at http://cnx.org/content/col11759/latest. SECTION 6.3 SECTION EXERCISES 497 6.
3 SECTION EXERCISES VERBAL 1. b b y= x bx = y b > b β‰  2. f x=bx gx=b x 3. b x = y x 5. b n diff 4. b n diff ALGEBRAIC 6. q=m 10. yx=βˆ’ 14. v=t 7. b=c 11. a=b 15. w=n 8. y=x 9. x=y 12. y=x 13. =a 16. x=y 18. mβˆ’=n 22. ( ) 20. x βˆ’ =y 21. n= 17. c d=k m =n 24. a=b 25. e k=h 19. x=y 23. y x= x 26. x= 28. x= 29. x= 32. x= 33. x=βˆ’ 30. x= 34. x= 27. x=βˆ’ 31. x= 35. x= fi 36. 40. e βˆ’ NUMERIC 37. 41. e + 38. 39. e b 42. 43. 44. 46. 47. 48. 50. p 54. () 4p 51. βœ“ β—† 55. 52. βœ“ β—† 45. 49. 53. Download the OpenStax text for free at http://cnx.org/content/col11759/latest. 49 8 CHAPTER 6 EXPONENTIAL AND LOGARITHMIC FUNCTIONS 56. ) ( 57. β€” √ 58. )+ ( 59. 60. 61. 62. + 63. βˆ’ 64. ( e ) 65. 66. eβˆ’βˆ’ 67. ( e ) TECHNOLOGY 68. 69. 70. ) ( 71. √ β€” 72. √ β€” EXTENSIONS 73. x =f(x=x x = 74. f(x=fx=x x 75. x x= 76. =βˆ’ ) ( 77. e = EIβˆ’ REAL-WORLD APPLICATIONS 78. ThEI 79. f ) EI=( t ft 80. ThI I =Mβˆ’M I M Download the OpenStax text for free at http://cnx.org/content/col11759/latest. SECTION 6.4 GRAPHS OF LOGARITHMIC FUNCTIONS 499 LEARNING OBJECTIVES In this section, you will: β€’ dentify the domain of a logarithmic function.
β€’ Graph logarithmic functions including using transformations. 6.4 GRAPHS OF LOGARITHMIC FUNCTIONS In Graphs of Exponential Functions, we saw how creating a graphical representation of an exponential model gives us another layer of insight for predicting future events. How do logarithmic graphs give us insight into situations? Because every logarithmic function is the inverse function of an exponential function, we can think of every output on a logarithmic graph as the input for the corresponding inverse exponential equation. In other words, logarithms give the cause for an effect. To illustrate, suppose we invest $2,500 in an account that offers an annual interest rate of 5%, compounded continuously. We already know that the balance in our account for any year t can be found with the equation A = 2500e0.05t. But what if we wanted to know the year for any balance? We would need to create a corresponding new function by interchanging the input and the output; thus we would need to create a logarithmic model for this situation. By graphing the model, we can see the output (year) for any input (account balance). For instance, what if we wanted to know how many years it would take for our initial investment to double? Figure 1 shows this point on the logarithmic graph. Logarithmic Model Showing Years as a Function of the Balance in the Account s r a e Y 20 18 16 14 12 10 8 6 4 2 0 The balance reaches $5,000 near year 14 500 1,000 1,500 2,000 2,500 3,000 3,500 4,000 4,500 5,000 5,500 6,000 Account balance Figure 1 In this section we will discuss the values for which a logarithmic function is defined, and then turn our attention to graphing the family of logarithmic functions. Finding the Domain of a Logarithmic Function Before working with graphs, we will take a look at the domain (the set of input values) for which the logarithmic function is defined. Recall that the exponential function is defined as y = b x for any real number x and constant b > 0, b β‰  1, where β€’ The domain of y is (βˆ’βˆž, ∞). β€’ The range of y is (0, ∞). In the last section we learned that the logarithmic function y = logb(x) is
the inverse of the exponential function y = b x. So, as inverse functions: β€’ The domain of y = logb(x) is the range of y = b x : (0, ∞). β€’ The range of y = logb(x) is the domain of y = b x : (βˆ’βˆž, ∞). Download the OpenStax text for free at http://cnx.org/content/col11759/latest. 500 CHAPTER 6 EXPONENTIAL AND LOGARITHMIC FUNCTIONS Transformations of the parent function y = logb(x) behave similarly to those of other functions. Just as with other parent functions, we can apply the four types of transformationsβ€”shifts, stretches, compressions, and reflectionsβ€”to the parent function without loss of shape. In Graphs of Exponential Functions we saw that certain transformations can change the range of y = b x. Similarly, applying transformations to the parent function y = logb(x) can change the domain. When finding the domain of a logarithmic function, therefore, it is important to remember that the domain consists only of positive real numbers. That is, the argument of the logarithmic function must be greater than zero. For example, consider f (x) = log4(2x βˆ’ 3). This function is defined for any values of x such that the argument, in this case 2x βˆ’ 3, is greater than zero. To find the domain, we set up an inequality and solve for x : 2x βˆ’ 3 > 0 Show the argument greater than zero. 2x > 3 Add 3. x > 1.5 Divide by 2. In interval notation, the domain of f (x) = log4(2x βˆ’ 3) is (1.5, ∞). How To… Given a logarithmic function, identify the domain. 1. Set up an inequality showing the argument greater than zero. 2. Solve for x. 3. Write the domain in interval notation. Example 1 Identifying the Domain of a Logarithmic Shift What is the domain of f (x) = log2(x + 3)? Solution The logarithmic function is defined only when the input is positive, so this function is defined when x + 3 > 0. Solving this inequality, x + 3 > 0 The input must be positive. x > βˆ’3 The domain of f (x) = log2(x + 3) is
(βˆ’3, ∞). Subtract 3. Try It #1 What is the domain of f (x) = log5(x βˆ’ 2) + 1? Example 2 Identifying the Domain of a Logarithmic Shift and Reflection What is the domain of f (x) = log(5 βˆ’ 2x)? Solution The logarithmic function is defined only when the input is positive, so this function is defined when 5 βˆ’ 2x > 0. Solving this inequality, 5 βˆ’ 2x > 0 The input must be positive. βˆ’2x > βˆ’5 Subtract 5. 5 __ x < 2 Divide by βˆ’2 and switch the inequality. 5 ). The domain of f (x) = log(5 βˆ’ 2x) is ( β€“βˆž, _ 2 Download the OpenStax text for free at http://cnx.org/content/col11759/latest. SECTION 6.4 GRAPHS OF LOGARITHMIC FUNCTIONS 501 Try It #2 What is the domain of f (x) = log(x βˆ’ 5) + 2? Graphing Logarithmic Functions Now that we have a feel for the set of values for which a logarithmic function is defined, we move on to graphing logarithmic functions. The family of logarithmic functions includes the parent function y = logb(x) along with all its transformations: shifts, stretches, compressions, and reflections. We begin with the parent function y = logb(x). Because every logarithmic function of this form is the inverse of an exponential function with the form y = b x, their graphs will be reflections of each other across the line y = x. To illustrate this, we can observe the relationship between the input and output values of y = 2x and its equivalent x = log2(y) in Table 1. x 2x = y log2(y) = x βˆ’3 1 _ 8 βˆ’3 βˆ’2 1 _ 4 βˆ’2 0 1 0 βˆ’1 1 _ 2 βˆ’1 Table Using the inputs and outputs from Table 1, we can build another table to observe the relationship between points on the graphs of the inverse functions f (x) = 2x and g(x) = log2(x). See Table 2. f (x) = 2x g(x) = log2(x) 1 ) ( βˆ’3, _ 8 1, βˆ’3 )
( _ 8 1 ) ( βˆ’2, _ 4 1, βˆ’2 ) ( _ 4 1 ) ( βˆ’1, _ 2 1, βˆ’1 ) ( _ 2 (0, 1) (1, 2) (2, 4) (3, 8) (1, 0) (2, 1) (4, 2) (8, 3) Table 2 As we’d expect, the x- and y-coordinates are reversed for the inverse functions. Figure 2 shows the graph of f and g. f (x) = 2 x –5 –4 –3 –2 y = x g(x) = log2 (x) 321 1 –1 –2 –3 –4 –5 Figure 2 Notice that the graphs of f (x) = 2x and g(x) = log2(x) are reflections about the line y = x. Observe the following from the graph: β€’ f (x) = 2x has a y-intercept at (0, 1) and g(x) = log2(x) has an x-intercept at (1, 0). β€’ The domain of f (x) = 2x, (βˆ’βˆž, ∞), is the same as the range of g(x) = log2(x). β€’ The range of f (x) = 2x, (0, ∞), is the same as the domain of g(x) = log2(x). Download the OpenStax text for free at http://cnx.org/content/col11759/latest. 502 CHAPTER 6 EXPONENTIAL AND LOGARITHMIC FUNCTIONS characteristics of the graph of the parent function, f (x) = logb(x) For any real number x and constant b > 0, b β‰  1, we can see the following characteristics in the graph of f (x) = logb(x): β€’ one-to-one function β€’ vertical asymptote: x = 0 β€’ domain: (0, ∞) β€’ range: (βˆ’βˆž, ∞) β€’ x-intercept: (1, 0) f (x) = logb(x) 0 < b < 1 f(x) = logb(x) b > 1 (b, 1) (b, 1) (1, 0) f (x) f(x) x x (1, 0) x
= 0 and key point (b, 1) β€’ y-intercept: none β€’ increasing if b > 1 β€’ decreasing if 0 < b < 1 See Figure 3. Figure 4 shows how changing the base b in f (x) = logb(x) can affect the graphs. Observe that the graphs compress vertically as the value of the base increases. (Note: recall that the function ln(x) has base e β‰ˆ 2.718.) x = 0 Figure 3 x = 0 642 8 10 12 log2(x) ln(x) log(x) x y 5 4 3 2 1 –2 –1 –2 –3 –4 –5 –12 –10 –8 –6 –4 Figure 4 The graphs of three logarithmic functions with different bases, all greater than 1. How To… Given a logarithmic function with the form f (x) = logb(x), graph the function. 1. Draw and label the vertical asymptote, x = 0. 2. Plot the x-intercept, (1, 0). 3. Plot the key point (b, 1). 4. Draw a smooth curve through the points. 5. State the domain, (0, ∞), the range, (βˆ’βˆž, ∞), and the vertical asymptote, x = 0. Example 3 Graphing a Logarithmic Function with the Form f ( x) = logb( x). Graph f (x) = log5(x). State the domain, range, and asymptote. Solution Before graphing, identify the behavior and key points for the graph. β€’ Since b = 5 is greater than one, we know the function is increasing. The left tail of the graph will approach the vertical asymptote x = 0, and the right tail will increase slowly without bound. β€’ The x-intercept is (1, 0). β€’ The key point (5, 1) is on the graph. β€’ We draw and label the asymptote, plot and label the points, and draw a smooth curve through the points (see Figure 5). Download the OpenStax text for free at http://cnx.org/content/col11759/latest. SECTION 6.4 GRAPHS OF LOGARITHMIC FUNCTIONS 503 –10 –8 –6 –4 f (x) x = 0 (5, 1) 642 8 10
(1, 0) 5 4 3 2 1 –2 –1 –2 –3 –4 –5 Figure 5 f (x) = log5(x) x The domain is (0, ∞), the range is (βˆ’βˆž, ∞), and the vertical asymptote is x = 0. Try It #3 Graph f (x) = log 1 _ 5 (x). State the domain, range, and asymptote. Graphing Transformations of Logarithmic Functions As we mentioned in the beginning of the section, transformations of logarithmic graphs behave similarly to those of other parent functions. We can shift, stretch, compress, and reflect the parent function y = logb(x) without loss of shape. Graphing a Horizontal Shift of f (x ) = logb(x ) When a constant c is added to the input of the parent function f (x) = logb(x), the result is a horizontal shift c units in the opposite direction of the sign on c. To visualize horizontal shifts, we can observe the general graph of the parent function f (x) = logb(x) and for c > 0 alongside the shift left, g(x ) = logb(x + c), and the shift right, h(x) = logb(x βˆ’ c). See Figure 6. Shift left g (x) = logb(x + c) x = –c y x = 0 g(x) = logb(x + c) y x = 0 Shift right h(x) = logb(x βˆ’ c) x = c f (x) = logb(x) (b βˆ’ c, 1) (1 βˆ’ c, 0) f (x) = logb(x) x (b, 1) (1, 0) (b, 1) (b + c, 1) (1, 0) (1 + c, 0) x h(x) = logb(x βˆ’ c) β€’ The asymptote changes to x = βˆ’c. β€’ The domain changes to (βˆ’c, ∞). β€’ The range remains (βˆ’βˆž, ∞). β€’ The asymptote changes to x = c. β€’ The domain changes to (c, ∞). β€’ The range remains (βˆ’βˆž, ∞). Figure 6 horizontal shifts of the parent function y = logb(x) For any constant c, the
function f (x) = logb (x + c) β€’ shifts the parent function y = logb(x) left c units if c > 0. β€’ shifts the parent function y = logb(x) right c units if c < 0. β€’ has the vertical asymptote x = βˆ’c. β€’ has domain (βˆ’c, ∞). β€’ has range (βˆ’βˆž, ∞). Download the OpenStax text for free at http://cnx.org/content/col11759/latest. 504 CHAPTER 6 EXPONENTIAL AND LOGARITHMIC FUNCTIONS How To… Given a logarithmic function with the form f (x) = logb(x + c), graph the translation. 1. Identify the horizontal shift: a. If c > 0, shift the graph of f (x) = logb(x) left c units. b. If c < 0, shift the graph of f (x) = logb(x) right c units. 2. Draw the vertical asymptote x = βˆ’c. 3. Identify three key points from the parent function. Find new coordinates for the shifted functions by subtracting c from the x coordinate. 4. Label the three points. 5. The domain is (βˆ’c, ∞), the range is (βˆ’βˆž, ∞), and the vertical asymptote is x = βˆ’c. Example 4 Graphing a Horizontal Shift of the Parent Function y = logb( x) Sketch the horizontal shift f (x) = log3(x βˆ’ 2) alongside its parent function. Include the key points and asymptotes on the graph. State the domain, range, and asymptote. Solution Since the function is f (x) = log3(x βˆ’ 2), we notice x + (βˆ’2) = x βˆ’ 2. Thus c = βˆ’2, so c < 0. This means we will shift the function f (x) = log3(x) right 2 units. The vertical asymptote is x = βˆ’(βˆ’2) or x = 2. 1, βˆ’1 ), (1, 0), and (3, 1). Consider the three key points from the parent function, ( _ 3 The new coordinates are found by adding 2 to the x coordinates. 7, βˆ’1 ), (3, 0), and (5, 1). Label the points ( _ 3 The domain is (
2, ∞), the range is (βˆ’βˆž, ∞), and the vertical asymptote is x = 2. y 5 4 3 2 1 –1–1 –2 –3 –4 –5 (1, 0) y = log3(x) f (x) = log3(x βˆ’ 2) x (3, 1) (5, 1) 321 4 5 6 7 8 9 (3, 0) x = 2 x = 0 Figure 7 Try It #4 Sketch a graph of f (x) = log3(x + 4) alongside its parent function. Include the key points and asymptotes on the graph. State the domain, range, and asymptote. Graphing a Vertical Shift of y = logb(x ) When a constant d is added to the parent function f (x) = logb(x), the result is a vertical shift d units in the direction of the sign on d. To visualize vertical shifts, we can observe the general graph of the parent function f (x) = logb(x) alongside the shift up, g (x) = logb(x) + d and the shift down, h(x) = logb(x) βˆ’ d. See Figure 8. Download the OpenStax text for free at http://cnx.org/content/col11759/latest. SECTION 6.4 GRAPHS OF LOGARITHMIC FUNCTIONS 505 Shift up g (x) = logb(x) + d Shift down h(x) = logb(x) βˆ’ d y x = 0 (b1 βˆ’ d, 1) βˆ’d, 0) (b, 1) (b (1, 0) g(x) = logb(x) + d f (x) = logb(x) x y x = 0 f (x) = logb(x) (b, 1) (b1+d, 1) h(x) = logb(x) βˆ’ d (1, 0) (bd, 0) x β€’ The asymptote remains x = 0. β€’ The domain remains to (0, ∞). β€’ The range remains (βˆ’βˆž, ∞). β€’ The asymptote remains x = 0. β€’ The domain remains to (0, ∞). β€’ The range remains (βˆ’βˆž, ∞). Figure 8 vertical shifts of the parent function y =
logb(x) For any constant d, the function f (x) = logb(x) + d β€’ shifts the parent function y = logb(x) up d units if d > 0. β€’ shifts the parent function y = logb(x) down d units if d < 0. β€’ has the vertical asymptote x = 0. β€’ has domain (0, ∞). β€’ has range (βˆ’βˆž, ∞). How To… Given a logarithmic function with the form f (x) = logb(x) + d, graph the translation. 1. Identify the vertical shift: a. If d > 0, shift the graph of f (x) = logb(x) up d units. b. If d < 0, shift the graph of f (x) = logb(x) down d units. 2. Draw the vertical asymptote x = 0. 3. Identify three key points from the parent function. Find new coordinates for the shifted functions by adding d to the y coordinate. 4. Label the three points. 5. The domain is (0, ∞), the range is (βˆ’βˆž, ∞), and the vertical asymptote is x = 0. Example 5 Graphing a Vertical Shift of the Parent Function y = logb(x) Sketch a graph of f (x) = log3(x) βˆ’ 2 alongside its parent function. Include the key points and asymptote on the graph. State the domain, range, and asymptote. Solution Since the function is f (x) = log3(x) βˆ’ 2, we will notice d = βˆ’2. Thus d < 0. This means we will shift the function f (x) = log3(x) down 2 units. The vertical asymptote is x = 0. Download the OpenStax text for free at http://cnx.org/content/col11759/latest. 506 CHAPTER 6 EXPONENTIAL AND LOGARITHMIC FUNCTIONS 1, βˆ’1 ), (1, 0), and (3, 1). Consider the three key points from the parent function, ( _ 3 The new coordinates are found by subtracting 2 from the y coordinates. 1 Label the points (, βˆ’3 ), (1, βˆ’2), and (3, βˆ’1). _ 3 The domain is (0, ∞), the
range is (βˆ’βˆž, ∞), and the vertical asymptote is x = 0. y 5 4 3 2 1 (1, 0) –1–1 –2 –3 –4 –5 (3, 1) 321 4 5 6 7 8 9 (3, βˆ’1) (1, βˆ’2) y = log3(x) x f (x) = log3(x βˆ’ 2) x = 0 Figure 9 The domain is (0, ∞), the range is (βˆ’βˆž, ∞), and the vertical asymptote is x = 0. Try It #5 Sketch a graph of f (x) = log2(x) + 2 alongside its parent function. Include the key points and asymptote on the graph. State the domain, range, and asymptote. Graphing Stretches and Compressions of y = logb(x ) When the parent function f (x) = logb(x) is multiplied by a constant a > 0, the result is a vertical stretch or compression of the original graph. To visualize stretches and compressions, we set a > 1 and observe the general graph of the parent 1 _ a logb(x). function f (x) = logb(x) alongside the vertical stretch, g (x) = alogb(x) and the vertical compression, h(x) = See Figure 10. Vertical Stretch g (x) = alogb(x), a > 1 y x = 0 g(x) = alogb(x) 1/a (b, 1) f(x) = logb(x) (b, 1) (1, 0) x Vertical Compression 1 _ a logb(x), a > 1 h(x) = y x = 0 (b, 1) (1, 0) f(x) = logb(x) 1 a h(x) = logb(x) (ba, 1) x β€’ The asymptote remains x = 0. β€’ The x-intercept remains (1, 0). β€’ The domain remains (0, ∞). β€’ The range remains (βˆ’βˆž, ∞). β€’ The asymptote remains x = 0. β€’ The x-intercept remains (1, 0). β€’ The domain remains (0, ∞). β€’ The range remains (βˆ’βˆž, ∞). Figure 10 Download the Open
Stax text for free at http://cnx.org/content/col11759/latest. SECTION 6.4 GRAPHS OF LOGARITHMIC FUNCTIONS 507 vertical stretches and compressions of the parent function y = logb(x) For any constant a > 1, the function f (x) = alogb(x) β€’ stretches the parent function y = logb(x) vertically by a factor of a if a > 1. β€’ compresses the parent function y = logb(x) vertically by a factor of a if 0 < a < 1. β€’ has the vertical asymptote x = 0. β€’ has the x-intercept (1, 0). β€’ has domain (0, ∞). β€’ has range (βˆ’βˆž, ∞). How To… Given a logarithmic function with the form f (x) = alogb(x), a > 0, graph the translation. 1. Identify the vertical stretch or compressions: a. If ∣ a ∣ > 1, the graph of f (x) = logb(x) is stretched by a factor of a units. b. If ∣ a ∣ < 1, the graph of f (x) = logb(x) is compressed by a factor of a units. 2. Draw the vertical asymptote x = 0. 3. Identify three key points from the parent function. Find new coordinates for the shifted functions by multiplying the y coordinates by a. 4. Label the three points. 5. The domain is (0, ∞), the range is (βˆ’βˆž, ∞), and the vertical asymptote is x = 0. Example 6 Graphing a Stretch or Compression of the Parent Function y = logb( x ) Sketch a graph of f (x) = 2log4(x) alongside its parent function. Include the key points and asymptote on the graph. State the domain, range, and asymptote. Solution Since the function is f (x) = 2log4(x), we will notice a = 2. This means we will stretch the function f (x) = log4(x) by a factor of 2. The vertical asymptote is x = 0. 1, βˆ’1 ), (1, 0), and (4, 1). Consider the three key points from the parent function, ( _ 4 The new coordinates
are found by multiplying the y coordinates by 2. 1 Label the points (, βˆ’2 ), (1, 0), and (4, 2). _ 4 The domain is (0, ∞), the range is (βˆ’βˆž, ∞), and the vertical asymptote is x = 0. See Figure 11. y 5 4 3 2 1 –1–1 –2 –3 –4 –5 (4, 2) (2, 1) (4, 1) 4 5 6 7 8 9 321 (1, 0) f (x) = 2log4(x) y = log4(x) x x = 0 Figure 11 Download the OpenStax text for free at http://cnx.org/content/col11759/latest. 508 CHAPTER 6 EXPONENTIAL AND LOGARITHMIC FUNCTIONS Try It #6 1 _ Sketch a graph of f (x) = log4(x) alongside its parent function. Include the key points and asymptote on the graph. State 2 the domain, range, and asymptote. Example 7 Combining a Shift and a Stretch Sketch a graph of f (x) = 5log(x + 2). State the domain, range, and asymptote. Solution Remember: what happens inside parentheses happens first. First, we move the graph left 2 units, then stretch the function vertically by a factor of 5, as in Figure 12. The vertical asymptote will be shifted to x = βˆ’2. The x-intercept will be (βˆ’1, 0). The domain will be (βˆ’2, ∞). Two points will help give the shape of the graph: (βˆ’1, 0) and (8, 5). We chose x = 8 as the x-coordinate of one point to graph because when x = 8, x + 2 = 10, the base of the common logarithm. y y = 5 log(x + 2) y = log(x + 2) 321 4 5 x y = log(x) 5 4 3 2 1 –5 –4 –2 –3 –2 –1 –2 –3 –4 –5 x = βˆ’2 The domain is (βˆ’2, ∞), the range is (βˆ’βˆž, ∞), and the vertical asymptote is x = βˆ’2. Figure 12 Try It #7 Sketch a graph of the function f (x) = 3log(x
βˆ’ 2) + 1. State the domain, range, and asymptote. Graphing Reflections of f (x ) = logb(x ) When the parent function f (x) = logb(x) is multiplied by βˆ’1, the result is a reflection about the x-axis. When the input is multiplied by βˆ’1, the result is a reflection about the y-axis. To visualize reflections, we restrict b > 1, and observe the general graph of the parent function f (x) = logb(x) alongside the reflection about the x-axis, g(x) = βˆ’logb(x) and the reflection about the y-axis, h(x) = logb(βˆ’x). Reflection about the x-axis g (x) = logb(x), b > 1 Reflection about the y-axis h(x) = logb(βˆ’x), x) = logb(x) βˆ’1, 1) (b (b, 1) (1, 0) g(x) = βˆ’logb(x) h(x) = logb(βˆ’x) f (x) = logb(x) x (βˆ’b, 1) (βˆ’1, 0) (b, 1) (1, 0) x β€’ The reflected function is decreasing as x moves from zero to infinity. β€’ The asymptote remains x = 0. β€’ The x-intercept remains (1, 0). β€’ The key point changes to (b βˆ’ 1, 1). β€’ The domain remains (0, ∞). β€’ The range remains (βˆ’βˆž, ∞). β€’ The reflected function is decreasing as x moves from infinity to zero. β€’ The asymptote remains x = 0. β€’ The x-intercept remains (βˆ’1, 0). β€’ The key point changes to (βˆ’b, 1). β€’ The domain changes to (βˆ’βˆž, 0). β€’ The range remains (βˆ’βˆž, ∞). Figure 13 Download the OpenStax text for free at http://cnx.org/content/col11759/latest. SECTION 6.4 GRAPHS OF LOGARITHMIC FUNCTIONS 509 reflections of the parent function y = logb(x) The function f (x) = βˆ’logb(x) β€’ reflects the parent function y = logb(x) about the x-axis. β€’ has domain, (0, ∞
), range, (βˆ’βˆž, ∞), and vertical asymptote, x = 0, which are unchanged from the parent function. The function f (x) = logb(βˆ’x) β€’ reflects the parent function y = logb(x) about the y-axis. β€’ has domain (βˆ’βˆž, 0). β€’ has range, (βˆ’βˆž, ∞), and vertical asymptote, x = 0, which are unchanged from the parent function. How To… Given a logarithmic function with the parent function f (x) = logb(x), graph a translation. If f (x) = βˆ’logb(x) 1. Draw the vertical asymptote, x = 0. 2. Plot the x-intercept, (1, 0). 3. Reflect the graph of the parent function f (x) = logb(x) If f (x) = logb(βˆ’x) 1. Draw the vertical asymptote, x = 0. 2. Plot the x-intercept, (1, 0). 3. Reflect the graph of the parent function f (x) = logb(x) about the x-axis. about the y-axis. 4. Draw a smooth curve through the points. 5. State the domain, (0, ∞), the range, (βˆ’βˆž, ∞), and 4. Draw a smooth curve through the points. 5. State the domain, (βˆ’βˆž, 0), the range, (βˆ’βˆž, ∞), and the vertical asymptote x = 0. the vertical asymptote x = 0. Table 3 Example 8 Graphing a Reflection of a Logarithmic Function Sketch a graph of f (x) = log(βˆ’x) alongside its parent function. Include the key points and asymptote on the graph. State the domain, range, and asymptote. Solution Before graphing f (x) = log(βˆ’x), identify the behavior and key points for the graph. β€’ Since b = 10 is greater than one, we know that the parent function is increasing. Since the input value is multiplied by βˆ’1, f is a reflection of the parent graph about the y-axis. Thus, f (x) = log(βˆ’x) will be decreasing as x moves from negative infinity to zero, and the right tail of the graph will approach the vertical asym
ptote x = 0. β€’ The x-intercept is (βˆ’1, 0). β€’ We draw and label the asymptote, plot and label the points, and draw a smooth curve through the points. y x = 0 (–10, 0) f (x) = log(βˆ’x) y = log(x) (10, 0) (–1, 0) (1, 0) x The domain is (βˆ’βˆž, 0), the range is (βˆ’βˆž, ∞), and the vertical asymptote is x = 0. Figure 14 Try It #8 Graph f (x) = βˆ’log(βˆ’x). State the domain, range, and asymptote. Download the OpenStax text for free at http://cnx.org/content/col11759/latest. 510 CHAPTER 6 EXPONENTIAL AND LOGARITHMIC FUNCTIONS How To… Given a logarithmic equation, use a graphing calculator to approximate solutions. Press [Y=]. Enter the given logarithm equation or equations as Y1= and, if needed, Y2=. 1. 2. Press [GRAPH] to observe the graphs of the curves and use [WINDOW] to find an appropriate view of the graphs, including their point(s) of intersection. 3. To fi nd the value of x, we compute the point of intersection. Press [2ND] then [CALC]. Select β€œintersect” and press [ENTER] three times. The point of intersection gives the value of x, for the point(s) of intersection. Example 9 Approximating the Solution of a Logarithmic Equation Solve 4ln(x) + 1 = βˆ’2ln(x βˆ’ 1) graphically. Round to the nearest thousandth. Solution Press [Y=] and enter 4ln(x) + 1 next to Y1=. Then enter βˆ’2ln(x βˆ’ 1) next to Y2=. For a window, use the values 0 to 5 for x and –10 to 10 for y. Press [GRAPH]. The graphs should intersect somewhere a little to right of x = 1. For a better approximation, press [2ND] then [CALC]. Select [5: intersect] and press [ENTER] three times. The x-coordinate of the point of intersection is displayed as 1.3385297. (Your answer
may be different if you use a different window or use a different value for Guess?) So, to the nearest thousandth, x β‰ˆ 1.339. Try It #9 Solve 5log(x + 2) = 4 βˆ’ log(x) graphically. Round to the nearest thousandth. Summarizing Translations of the Logarithmic Function Now that we have worked with each type of translation for the logarithmic function, we can summarize each in Table 4 to arrive at the general equation for translating exponential functions. Translations of the Parent Function y = logb(x) Translation Form Shift β€’ Horizontally c units to the left β€’ Vertically d units up Stretch and Compress β€’ Stretch if ∣ a ∣ > 1 β€’ Compression if ∣ a ∣ < 1 Reflect about the x-axis Reflect about the y-axis General equation for all translations y = logb (x + c) + d y = alogb(x) y = βˆ’logb(x) y = logb(βˆ’x) y = alogb(x + c) + d Table 4 translations of logarithmic functions All translations of the parent logarithmic function, y = logb(x), have the form f (x) = alogb(x + c) + d where the parent function, y = logb(x), b > 1, is β€’ shifted vertically up d units. β€’ shifted horizontally to the left c units. β€’ stretched vertically by a factor of ∣ a ∣ if ∣ a ∣ > 0. β€’ compressed vertically by a factor of ∣ a ∣ if 0 < ∣ a ∣ < 1. β€’ reflected about the x-axis when a < 0. For f (x) = log(βˆ’x), the graph of the parent function is reflected about the y-axis. Download the OpenStax text for free at http://cnx.org/content/col11759/latest. SECTION 6.4 GRAPHS OF LOGARITHMIC FUNCTIONS 511 Example 10 Finding the Vertical Asymptote of a Logarithm Graph What is the vertical asymptote of f (x) = βˆ’2log3(x + 4) + 5? Solution The vertical asymptote is at x = βˆ’4. Analysis The coefficient, the base, and the upward translation do not affect the asymptote. The shift of the curve
4 units to the left shifts the vertical asymptote to x = βˆ’4. Try It #10 What is the vertical asymptote of f (x) = 3 + ln(x βˆ’ 1)? Example 11 Finding the Equation from a Graph Find a possible equation for the common logarithmic function graphed in Figure 15. f(x) 5 4 3 2 1 –1 –1 –2 –3 –5 –4 –3 –2 21 3 4 5 6 7 x Figure 15 Solution This graph has a vertical asymptote at x = βˆ’2 and has been vertically reflected. We do not know yet the vertical shift or the vertical stretch. We know so far that the equation will have form: It appears the graph passes through the points (βˆ’1, 1) and (2, βˆ’1). Substituting (βˆ’1, 1), f (x) = βˆ’alog(x + 2) + k Ne xt, substituting in (2, –1), 1 = βˆ’alog(βˆ’1 + 2) + k Substitute (βˆ’1, 1). 1 = βˆ’alog(1) + k 1 = k Arithmetic. log(1) = 0. βˆ’1 = βˆ’alog(2 + 2) + 1 Plug in (2, βˆ’1). βˆ’2 = βˆ’alog(4) a = 2 _____ log(4) Arithmetic. Solve for a. This gives us the equation f (x) = βˆ’ 2_ log(x + 2) + 1. log(4) Analysis Figure 15. You can verify this answer by comparing the function values in Table 5 with the points on the graph in x f (x) x f (x) βˆ’1 1 4 βˆ’1.5850 0 0 5 βˆ’1.8074 1 βˆ’0.58496 6 βˆ’2 2 βˆ’1 7 βˆ’2.1699 3 βˆ’1.3219 8 βˆ’2.3219 Table 5 Download the OpenStax text for free at http://cnx.org/content/col11759/latest. 512 CHAPTER 6 EXPONENTIAL AND LOGARITHMIC FUNCTIONS Try It #11 Give the equation of the natural logarithm graphed in Figure 16. f(x) –5 –4 –3 –2 4 3 2 1 –1 –1 –2 –3 –4 –5 321 4 5 x Figure 16 Q & A… Is it possible to tell the domain and range and describe the
end behavior of a function just by looking at the graph? Yes, if we know the function is a general logarithmic function. For example, look at the graph in Figure 16. The graph approaches x = βˆ’3 (or thereabouts) more and more closely, so x = βˆ’3 is, or is very close to, the vertical asymptote. It approaches from the right, so the domain is all points to the right, {x | x > βˆ’3}. The range, as with all general logarithmic functions, is all real numbers. And we can see the end behavior because the graph goes down as it goes left and up as it goes right. The end behavior is that as x β†’ βˆ’3+, f (x) β†’ βˆ’βˆž and as x β†’ ∞, f (x) β†’ ∞. Access these online resources for additional instruction and practice with graphing logarithms. β€’ Graph an Exponential Function and Logarithmic Function (http://openstaxcollege.org/l/graphexplog) β€’ Match Graphs with Exponential and Logarithmic Functions (http://openstaxcollege.org/l/matchexplog) β€’ Find the Domain of Logarithmic Functions (http://openstaxcollege.org/l/domainlog) Download the OpenStax text for free at http://cnx.org/content/col11759/latest. SECTION 6.4 SECTION EXERCISES 513 6.4 SECTION EXERCISES VERBAL 1. Th 2. ff 3. ff 4. f (x) = bx. x 5. ALGEBRAIC 6. f x=x+ 9. hx=x+βˆ’ βˆ’x ) 7. hx=( 10. fx=βˆ’xβˆ’ 8. gx=x+βˆ’ 11. fx=bxβˆ’ 12. gx=βˆ’x 13. fx=x+ 14. fx=βˆ’x+ 15. gx=βˆ’x+βˆ’ 17. fx=( xβˆ’ 16. fx=βˆ’x ) 20. fx=βˆ’x+ 19. gx=x+βˆ’ 18. hx= βˆ’xβˆ’+ xy 21. hx=xβˆ’+ 22. fx=x++ 23. gx=βˆ’xβˆ’ 24. fx=x+βˆ’ 25. hx=xβˆ’ GRAPH
ICAL Figure 17 A B C x 26. fx=x 27. gx=x 28. hx=x y – – Figure 17 Download the OpenStax text for free at http://cnx.org/content/col11759/latest. 514 CHAPTER 6 EXPONENTIAL AND LOGARITHMIC FUNCTIONS Figure 18 – – – – Figure 18 29. fx= x 30. gx=x 31. hx= x 32. fx=xgx=x 34. fx=xgx=x 33. fx=xgx= x 35. fx=e xgx=x Figure 19 Figure 19 36. fx=βˆ’x+ 37. gx=βˆ’x+ 38. hx=x+ 39. fx=x+ 40. fx=x 41. fx=βˆ’x 42. gx=x++ 43. gx=βˆ’x+ 44. hx=βˆ’ x+βˆ’ 45. y=x 46. fx= Download the OpenStax text for free at http://cnx.org/content/col11759/latest. SECTION 6.4 SECTION EXERCISES 515 47. fx=x 48. fx= – – – – – – – – – – TECHNOLOGY o fi 49. xβˆ’+=xβˆ’+ 50. xβˆ’+=βˆ’xβˆ’+ 51. xβˆ’=βˆ’x+ 52. x+= βˆ’x+ 53. βˆ’x=x++ EXTENSIONS 54. bbβ‰  b 55. fx= x gx) = βˆ’x 56. 57. fx=( x+ xβˆ’ ) 58. x fx=x +x+ Download the OpenStax text for free at http://cnx.org/content/col11759/latest. 516 CHAPTER 6 EXPONENTIAL AND LOGARITHMIC FUNCTIONS LEARNING OBJECTIVES In this section, you will: β€’ Use the product power andor quotient rule to epand logarithmic epressions. β€’ Use the product power andor quotient rule to condense logarithmic epressions. β€’ aluate logarithms. β€’ Use the change-of-base formula for logarithms. 6.5 LOGARITHMIC PROPERTIES Figure 1 The pH of hydrochloric acid is tested with litmus paper. (credit
: David Berardan) In chemistry, pH is used as a measure of the acidity or alkalinity of a substance. The pH scale runs from 0 to 14. Substances with a pH less than 7 are considered acidic, and substances with a pH greater than 7 are said to be alkaline. Our bodies, for instance, must maintain a pH close to 7.35 in order for enzymes to work properly. To get a feel for what is acidic and what is alkaline, consider the following pH levels of some common substances: β€’ Battery acid: 0.8 β€’ Stomach acid: 2.7 β€’ Orange juice: 3.3 β€’ Pure water: 7 (at 25Β° C) β€’ Human blood: 7.35 β€’ Fresh coconut: 7.8 β€’ Sodium hydroxide (lye): 14 To determine whether a solution is acidic or alkaline, we find its pH, which is a measure of the number of active positive hydrogen ions in the solution. The pH is defined by the following formula, where a is the concentration of hydrogen ion in the solution pH = βˆ’log([H+]) = log ( 1 _____ ) ([H+] The equivalence of βˆ’log ([H+]) and log ( ) is one of the logarithm properties we will examine in this section. 1 _ [H+] Using the Product Rule for Logarithms Recall that the logarithmic and exponential functions β€œundo” each other. This means that logarithms have similar properties to exponents. Some important properties of logarithms are given here. First, the following properties are easy to prove. logb(1) = 0 logb(b) = 1 For example, log5 1 = 0 since 50 = 1. And log5 5 = 1 since 51 = 5. Download the OpenStax text for free at http://cnx.org/content/col11759/latest. SECTION 6.5 LOGARITHMIC PROPERTIES 517 Ne xt, we have the inverse property. logb(b x) = x b logb(x) = x, x > 0 For example, to evaluate log(100), we can rewrite the logarithm as log10(102), and then apply the inverse property logb (b x) = x to get log10(102) = 2. To evaluate e ln(7), we can rewrite the logarithm as e loge(
7), and then apply the inverse property b log (x) = x to get eloge(7) = 7. b Finally, we have the one-to-one property. logbM = logbN if and only if M = N We can use the one-to-one property to solve the equation log3(3x) = log3(2x + 5) for x. Since the bases are the same, we can apply the one-to-one property by setting the arguments equal and solving for x : 3x = 2x + 5 x = 5 Set the arguments equal. Subtract 2x. But what about the equation log3(3x) + log3(2x + 5) = 2? The one-to-one property does not help us in this instance. Before we can solve an equation like this, we need a method for combining terms on the left side of the equation. Recall that we use the product rule of exponents to combine the product of exponents by adding. We have a similar property for logarithms, called the product rule for logarithms, which says that the logarithm of a product is equal to a sum of logarithms. Because logs are exponents, and we multiply like bases, we can add the exponents. We will use the inverse property to derive the product rule below. Given any real number x and positive real numbers M, N, and b, where b β‰  1, we will show logb(MN) = logb(M) + logb(N). Let m = logb(M) and n = logb(N). In exponential form, these equations are b m = M and b n = N. It follows that logb(MN) = logb(b mb n) = logb(b m + n) = m + n = logb(M) + logb(N) Substitute for M and N. Apply the product rule for exponents. Apply the inverse property of logs. Substitute for m and n. Note that repeated applications of the product rule for logarithms allow us to simplify the logarithm of the product of any number of factors. For example, consider logb(wxyz). Using the product rule for logarithms, we can rewrite this logarithm of a product as the sum of logarithms of its factors:
logb(wxyz) = logb(w) + logb(x) + logb(y) + logb(z) the product rule for logarithms The product rule for logarithms can be used to simplify a logarithm of a product by rewriting it as a sum of individual logarithms. logb(MN) = logb(M) + logb(N) for b > 0 How To… Given the logarithm of a product, use the product rule of logarithms to write an equivalent sum of logarithms. 1. Factor the argument completely, expressing each whole number factor as a product of primes. 2. Write the equivalent expression by summing the logarithms of each factor. Download the OpenStax text for free at http://cnx.org/content/col11759/latest. 518 CHAPTER 6 EXPONENTIAL AND LOGARITHMIC FUNCTIONS Example 1 Using the Product Rule for Logarithms Expand log3(30x(3x + 4)). Solution We begin by factoring the argument completely, expressing 30 as a product of primes. Ne xt we write the equivalent equation by summing the logarithms of each factor. log3(30x(3x + 4)) = log3(2 Β· 3 Β· 5 Β· x Β· (3x +4)) log3(30x(3x + 4)) = log3(2) + log3(3) + log3(5) + log3(x) + log3(3x + 4) Try It #1 Expand logb(8k). Using the Quotient Rule for Logarithms For quotients, we have a similar rule for logarithms. Recall that we use the quotient rule of exponents to combine a __ the quotient of exponents by subtracting: x = x aβˆ’b. The quotient rule for logarithms says that the logarithm of a b quotient is equal to a difference of logarithms. Just as with the product rule, we can use the inverse property to derive the quotient rule. Given any real number x and positive real numbers M, N, and b, where b β‰  1, we will show M ) = logb(M) βˆ’ logb(N). logb ( __ N Let m = logb(
M) and n = logb(N). In exponential form, these equations are bm = M and bn = N. It follows that bm bn ) ) = logb ( M logb ( __ __ N = logb(b m βˆ’ n) = m βˆ’ n = logb(M) βˆ’ logb(N) Substitute for M and N. Substitute for m and n. Apply the inverse property of logs. Apply the quotient rule for exponents. For example, to expand log ( 2x2 + 6x _ 3x + 9 we get, ), we must first express the quotient in lowest terms. Factoring and canceling log ( Factor the numerator and denominator. 2x(x + 3) ) ________ 3(x + 3) 2x2 + 6x _______ 3x + 9 ) = log ( 2x __ ) 3 = log ( Cancel the common factors. Ne xt we apply the quotient rule by subtracting the logarithm of the denominator from the logarithm of the numerator. Then we apply the product rule. log ( 2x __ ) = log(2x) βˆ’ log(3) 3 = log(2) + log(x) βˆ’ log(3) the quotient rule for logarithms The quotient rule for logarithms can be used to simplify a logarithm or a quotient by rewriting it as the difference of individual logarithms. M ) = logb(M) βˆ’ logb(N) logb ( __ N How To… Given the logarithm of a quotient, use the quotient rule of logarithms to write an equivalent difference of logarithms. 1. Express the argument in lowest terms by factoring the numerator and denominator and canceling common terms. 2. Write the equivalent expression by subtracting the logarithm of the denominator from the logarithm of the numerator. 3. Check to see that each term is fully expanded. If not, apply the product rule for logarithms to expand completely. Download the OpenStax text for free at http://cnx.org/content/col11759/latest. SECTION 6.5 LOGARITHMIC PROPERTIES 519 Example 2 Using the Quotient Rule for Logarithms 15x(x βˆ’ 1) ). __ (3x
+ 4)(2 βˆ’ x) Expand log 2 ( Solution First we note that the quotient is factored and in lowest terms, so we apply the quotient rule. log2 ( 15x(x βˆ’ 1)__ (3x + 4)(2 βˆ’ x)) = log2(15x(xβˆ’1))βˆ’ log2((3x + 4)(2 βˆ’ x)) Notice that the resulting terms are logarithms of products. To expand completely, we apply the product rule, noting that the prime factors of the factor 15 are 3 and 5. log2(15x(x βˆ’ 1)) βˆ’ log2((3x + 4)(2 βˆ’ x)) = [log2(3) + log2(5) + log2(x) + log2(x βˆ’ 1)] βˆ’ [log2(3x + 4) + log2(2 βˆ’ x)] = log2(3) + log2(5) + log2(x) + log2(x βˆ’ 1) βˆ’ log2(3x + 4) βˆ’ log2(2 βˆ’ x) Analysis There are exceptions to consider in this and later examples. First, because denominators must never be zero, this expression is not defined for x = βˆ’ 4 _ and x = 2. Also, since the argument of a logarithm must be positive, we note 3 as we observe the expanded logarithm, that x > 0, x > 1, x > βˆ’ 4 _, and x < 2. Combining these conditions is beyond the 3 scope of this section, and we will not consider them here or in subsequent exercises. Try It #2 Expand log3 ( 7x2 + 21x__ ). 7x(x βˆ’ 1)(x βˆ’ 2) Using the Power Rule for Logarithms We’ve explored the product rule and the quotient rule, but how can we take the logarithm of a power, such as x2? One method is as follows: logb(x2) = logb(x β‹… x) = logb (x) + logb (x) = 2log b (x) Notice that we used the product rule for logarithms to find a solution for the example above. By doing so, we have derived the power rule for logarithms, which says that the log of a power is equal to the exponent times the log of the base. Keep in mind that
, although the input to a logarithm may not be written as a power, we may be able to change it to a power. For example, 100 = 102 β€” 1 __ βˆ’1 the power rule for logarithms The power rule for logarithms can be used to simplify the logarithm of a power by rewriting it as the product of the exponent times the logarithm of the base. logb(Mn) = nlogb(M) How To… Given the logarithm of a power, use the power rule of logarithms to write an equivalent product of a factor and a logarithm. 1. Express the argument as a power, if needed. 2. Write the equivalent expression by multiplying the exponent times the logarithm of the base. Download the OpenStax text for free at http://cnx.org/content/col11759/latest. 520 CHAPTER 6 EXPONENTIAL AND LOGARITHMIC FUNCTIONS Example 3 Expanding a Logarithm with Powers Expand log2(x5). Solution The argument is already written as a power, so we identify the exponent, 5, and the base, x, and rewrite the equivalent expression by multiplying the exponent times the logarithm of the base. log2(x 5) = 5log2(x) Try It #3 Expand ln(x2). Example 4 Rewriting an Expression as a Power before Using the Power Rule Expand log3(25) using the power rule for logs. Solution Expressing the argument as a power, we get log3(25) = log3(52). Ne xt we identify the exponent, 2, and the base, 5, and rewrite the equivalent expression by multiplying the exponent times the logarithm of the base. log3(5 2) = 2log3(5) Try It #4 1 x2 ). Expand ln ( _ Example 5 Using the Power Rule in Reverse Rewrite 4ln(x) using the power rule for logs to a single logarithm with a leading coefficient of 1. Solution Because the logarithm of a power is the product of the exponent times the logarithm of the base, it follows that the product of a number and a logarithm can be written as a power. For the expression 4ln(x), we identify the factor, 4, as the exponent and the argument
, x, as the base, and rewrite the product as a logarithm of a power: 4ln(x) = ln(x4). Try It #5 Rewrite 2log3(4) using the power rule for logs to a single logarithm with a leading coefficient of 1. Expanding Logarithmic Expressions Taken together, the product rule, quotient rule, and power rule are often called β€œlaws of logs.” Sometimes we apply more than one rule in order to simplify an expression. For example: logb ( 6x _ y ) = logb(6x) βˆ’ logb(y) = logb(6) + logb(x) βˆ’ logb(y) We can use the power rule to expand logarithmic expressions involving negative and fractional exponents. Here is an alternate proof of the quotient rule for logarithms using the fact that a reciprocal is a negative power: A ) = logb(AC βˆ’1) logb ( __ C = logb(A) + logb(C βˆ’1) = logb(A) + (βˆ’1)logb(C) = logb(A) βˆ’ logb(C) We can also apply the product rule to express a sum or difference of logarithms as the logarithm of a product. With practice, we can look at a logarithmic expression and expand it mentally, writing the final answer. Remember, however, that we can only do this with products, quotients, powers, and rootsβ€”never with addition or subtraction inside the argument of the logarithm. Download the OpenStax text for free at http://cnx.org/content/col11759/latest. SECTION 6.5 LOGARITHMIC PROPERTIES 521 Example 6 Expanding Logarithms Using Product, Quotient, and Power Rules Rewrite ln ( x 4 y_ ) as a sum or difference of logs. 7 Solution First, because we have a quotient of two expressions, we can use the quotient rule: ln ( x 4y ) = ln(x 4y)βˆ’ ln(7) ___ 7 Then seeing the product in the first term, we use the product rule: ln(x 4y) βˆ’ ln(7) = ln(x 4) + ln(y) βˆ’ ln(7)
Finally, we use the power rule on the first term: ln(x4)+ ln(y) βˆ’ ln(7) = 4ln(x) + ln(y) βˆ’ ln(7) Try It #6 Expand log ( x2 y3 z4 ). _ Example 7 Using the Power Rule for Logarithms to Simplify the Logarithm of a Radical Expression Expand log( √ β€” x ). log(√ β€” 1 __ x ) = log (x) 2 = 1 __ log(x) 2 Solution Try It #7 Expand ln( 3 √ β€” x2 ). Q & A… Can we expand ln(x2 + y2)? No. There is no way to expand the logarithm of a sum or difference inside the argument of the logarithm. Example 8 Expanding Complex Logarithmic Expressions Expand log6 ( 64x3 (4x + 1) __ (2x βˆ’ 1) ). Solution We can expand by applying the Product and Quotient Rules. log6 ( 64x3(4x + 1) __ (2x βˆ’ 1) ) = log6(64) + log6(x3) + log6(4x + 1) βˆ’ log6(2x βˆ’ 1) = log6(26) + log6(x3) + log6(4x + 1) βˆ’ log6(2x βˆ’ 1) = 6log6(2) + 3log6(x) + log6(4x + 1) βˆ’ log6(2x βˆ’ 1) Apply the Quotient Rule. Simplify by writing 64 as 26. Apply the Power Rule. Try It #8 Expand ln ( β€” (x βˆ’ 1) (2x + 1) 2 √ ). __ x 2 βˆ’ 9 Condensing Logarithmic Expressions We can use the rules of logarithms we just learned to condense sums, differences, and products with the same base as a single logarithm. It is important to remember that the logarithms must have the same base to be combined. We will learn later how to change the base of any logarithm before condensing. Download the OpenStax text for free at http://cnx.org/content/col11759/latest. 522 CHAPTER 6 EXPONENTIAL AND LOGARITHMIC FUNCTIONS How
To… Given a sum, difference, or product of logarithms with the same base, write an equivalent expression as a single logarithm. 1. Apply the power property first. Identify terms that are products of factors and a logarithm, and rewrite each as the logarithm of a power. 2. Ne xt apply the product property. Rewrite sums of logarithms as the logarithm of a product. 3. Apply the quotient property last. Rewrite differences of logarithms as the logarithm of a quotient. Example 9 Using the Product and Quotient Rules to Combine Logarithms Write log3(5) + log3(8) βˆ’ log3(2) as a single logarithm. Solution Using the product and quotient rules This reduces our original expression to log3(5) + log3(8) = log3(5 Β· 8) = log3(40) Then, using the quotient rule log3(40) βˆ’ log3(2) log3(40) βˆ’ log3(2) = log3 ( 40 __ ) = log3(20) 2 Try It #9 Condense log(3) βˆ’ log(4) + log(5) βˆ’ log(6). Example 10 Condensing Complex Logarithmic Expressions Condense log2(x 2) + 1 __ log2(x βˆ’ 1) βˆ’ 3log2((x + 3)2). 2 Solution We apply the power rule first: log2(x 2) + 1 log2(x βˆ’ 1) βˆ’ 3log2((x + 3)2) = log2(x2) + log2( √ __ 2 Next we apply the product rule to the sum: β€” x βˆ’ 1 ) βˆ’ log2((x + 3)6) log2(x 2) + log2( √ β€” x βˆ’ 1 ) βˆ’ log2((x + 3)6) = log2(x2 √ β€” x βˆ’ 1 ) βˆ’ log2((x + 3)6) Finally, we apply the quotient rule to the difference: log2(x 2 √ β€” x βˆ’ 1 )βˆ’ log2((x + 3)6) = log2 ( β€” x βˆ’ 1 x2 √ ________ (x + 3)6 ) Try It #10 Rewrite log(5) + 0.5log(x)
βˆ’ log(7x βˆ’ 1) + 3log(x βˆ’ 1) as a single logarithm. Example 11 Rewriting as a Single Logarithm Rewrite 2log(x) βˆ’ 4log(x + 5) + 1 _ x log(3x + 5) as a single logarithm. Solution We apply the power rule first: 2log(x) βˆ’ 4log(x + 5) + 1 _ x log(3x + 5) = log(x2) βˆ’ log((x + 5)4) + log ( (3x + 5) x βˆ’1 ) Ne xt we apply the product rule to the sum: log(x2)βˆ’ log((x + 5)4) + log ( (3x + 5) x βˆ’1 ) = log(x2)βˆ’ log ( (x + 5)4(3x + 5) x βˆ’1 ) Finally, we apply the quotient rule to the difference: log(x2) βˆ’ log ( (x + 5)4(3x + 5) x βˆ’1 ) = log ( x 2 ) __ (x + 5) 4 (3x + 5) x βˆ’1 Download the OpenStax text for free at http://cnx.org/content/col11759/latest. SECTION 6.5 LOGARITHMIC PROPERTIES 523 Try It #11 Condense 4(3log(x) + log(x + 5) βˆ’ log(2x + 3)). Example 12 Applying of the Laws of Logs Recall that, in chemistry, pH = βˆ’log[H+]. If the concentration of hydrogen ions in a liquid is doubled, what is the effect on pH? Solution Suppose C is the original concentration of hydrogen ions, and P is the original pH of the liquid. Then P = βˆ’log(C). If the concentration is doubled, the new concentration is 2C. Then the pH of the new liquid is Using the product rule of logs pH = βˆ’log(2C) pH = βˆ’log(2C) = βˆ’(log(2) + log(C)) = βˆ’log(2) βˆ’ log(C) Since P = βˆ’log(C), the new pH is pH = P βˆ’ log(2) β‰ˆ P βˆ’ 0.301 When the concentration of hydrogen ions is doubled, the pH decreases by about 0.301. Try It #12 How does the pH change when
the concentration of positive hydrogen ions is decreased by half? Using the Change-of-Base Formula for Logarithms Most calculators can evaluate only common and natural logs. In order to evaluate logarithms with a base other than 10 or e, we use the change-of-base formula to rewrite the logarithm as the quotient of logarithms of any other base; when using a calculator, we would change them to common or natural logs. To derive the change-of-base formula, we use the one-to-one property and power rule for logarithms. Given any positive real numbers M, b, and n, where n β‰  1 and b β‰  1, we show logb(M) = logn(M) _ logn(b) Let y = logb(M). By taking the log base n of both sides of the equation, we arrive at an exponential form, namely b y = M. It follows that logn(b y) = logn(M) Apply the one-to-one property. ylogn(b) = logn(M) Apply the power rule for logarithms. y = logn(M) _ logn(b) Isolate y. logb(M) = logn(M) _ logn(b) Substitute for y. For example, to evaluate log5(36) using a calculator, we must first rewrite the expression as a quotient of common or natural logs. We will use the common log. log5(36) = log(36) _ log(5) Apply the change of base formula using base 10. β‰ˆ 2.2266 Use a calculator to evaluate to 4 decimal places. Download the OpenStax text for free at http://cnx.org/content/col11759/latest. 524 CHAPTER 6 EXPONENTIAL AND LOGARITHMIC FUNCTIONS the change-of-base formula The change-of-base formula can be used to evaluate a logarithm with any base. For any positive real numbers M, b, and n, where n β‰  1 and b β‰  1, logn(M) _. logn(b) logb(M) = It follows that the change-of-base formula can be used to rewrite a logarithm with any base as the quotient of common or natural logs. logb(M
) = and logb(M) = ln(M)_ ln(b) logn(M) _ logn(b) How To… Given a logarithm with the form logb(M), use the change-of-base formula to rewrite it as a quotient of logs with any positive base n, where n β‰  1. 1. Determine the new base n, remembering that the common log, log(x), has base 10, and the natural log, ln(x), has base e. 2. Rewrite the log as a quotient using the change-of-base formula a. The numerator of the quotient will be a logarithm with base n and argument M. b. The denominator of the quotient will be a logarithm with base n and argument b. Example 13 Changing Logarithmic Expressions to Expressions Involving Only Natural Logs Change log5(3) to a quotient of natural logarithms. Solution Because we will be expressing log5(3) as a quotient of natural logarithms, the new base, n = e. We rewrite the log as a quotient using the change-of-base formula. The numerator of the quotient will be the natural log with argument 3. The denominator of the quotient will be the natural log with argument 5. logb(M) = ln(M)_ ln(b) log5(3) = ln(3)_ ln(5) Try It #13 Change log0.5(8) to a quotient of natural logarithms. Q & A… Can we change common logarithms to natural logarithms? Yes. Remember that log(9) means log10(9). So, log(9) = ln(9) _. ln(10) Example 14 Using the Change-of-Base Formula with a Calculator Evaluate log2(10) using the change-of-base formula with a calculator. Solution According to the change-of-base formula, we can rewrite the log base 2 as a logarithm of any other base. Since our calculators can evaluate the natural log, we might choose to use the natural logarithm, which is the log base e. log2(10) = ln(10)_ ln(2) β‰ˆ
3.3219 Apply the change of base formula using base e. Use a calculator to evaluate to 4 decimal places. Try It #14 Evaluate log5(100) using the change-of-base formula. Access this online resource for additional instruction and practice with laws of logarithms. β€’ The Properties of Logarithms (http://openstaxcollege.org/l/proplog) β€’ Expand Logarithmic Expressions (http://openstaxcollege.org/l/expandlog) β€’ Evaluate a Natural Logarithmic Expression (http://openstaxcollege.org/l/evaluatelog) Download the OpenStax text for free at http://cnx.org/content/col11759/latest. SECTION 6.5 SECTION EXERCISES 525 6.5 SECTION EXERCISES VERBAL 1. n √ β€” x 2. ALGEBRAIC ff 3. bxy 4. abc 5. b ( ) 6. ( x z w ) 7. ( ) 8. yx 9. +x+y 10. +a++b 12. aβˆ’dβˆ’c ) 13. βˆ’b ( 11. bβˆ’b 14. ff 15. ( xy z ) 19.! r 16. βœ“ β—† β—† 17. βœ“ 18. β€” √xyβˆ’ 20. p βœ“ β—† (√ y ) βˆ’y 21. 22. x+x 23. xβˆ’x 24. x+x+ 25. xβˆ’ y+z 26. c+ a + b 27. e 28. =a=b ab ( 31. 29. 30. ) NUMERIC 32. )βˆ’ ( 33. + 34. βˆ’+( ) o fi 35. 36. 37. 38. EXTENSIONS ( ) 39. 40. 42. x x++x+= nb bn= b>n> 41. x+βˆ’xβˆ’= 43. = Download the OpenStax text for free at http://cnx.org/content/col11759/latest. 526 CHAPTER 6 EXPONENTIAL AND LOGARITHMIC FUNCTIONS LEARNING OBJECTIVES In this section, you will: β€’ Use like bases to solve exponential equations. β€’ Use logarithms to solve exponential equations. β€’ Use the definition of a logarithm to solve logarithmic equations. β€’ Use the one-to-one property of log
arithms to solve logarithmic equations. 6.6 EXPONENTIAL AND LOGARITHMIC EQUATIONS Figure 1 Wild rabbits in Australia. The rabbit population grew so quickly in Australia that the event became known as the β€œrabbit plague.” (credit: Richard Taylor, Flickr) In 1859, an Australian landowner named Thomas Austin released 24 rabbits into the wild for hunting. Because Australia had few predators and ample food, the rabbit population exploded. In fewer than ten years, the rabbit population numbered in the millions. Uncontrolled population growth, as in the wild rabbits in Australia, can be modeled with exponential functions. Equations resulting from those exponential functions can be solved to analyze and make predictions about exponential growth. In this section, we will learn techniques for solving exponential functions. Using Like Bases to Solve Exponential Equations The first technique involves two functions with like bases. Recall that the one-to-one property of exponential functions tells us that, for any real numbers b, S, and T, where b > 0, b β‰  1, bS = bT if and only if S = T. In other words, when an exponential equation has the same base on each side, the exponents must be equal. This also applies when the exponents are algebraic expressions. Therefore, we can solve many exponential equations by using the rules of exponents to rewrite each side as a power with the same base. Then, we use the fact that exponential functions are one-to-one to set the exponents equal to one another, and solve for the unknown. For example, consider the equation 34x βˆ’ 7 =. To solve for x, we use the division property of exponents to rewrite the right side so that both sides have the common base, 3. Then we apply the one-to-one property of exponents by setting the exponents equal to one another and solving for x : 32x _ 3 34x βˆ’ 7 = 34x βˆ’ 7 = 32x ___ 3 32x ___ 31 Rewrite 3 as 31. 34x βˆ’ 7 = 32x βˆ’ 1 4x βˆ’ 7 = 2x βˆ’ 1 2x = 6 x = 3 Use the division property of exponents. Apply the one-to-one property of exponents. Subtract 2x and add 7 to both sides. Divide by 3. Download the OpenStax text for free at http://cnx.org/content/col11759/latest. SECTION 6.
6 EXPONENTIAL AND LOGARITHMIC EQUATIONS 527 using the one-to-one property of exponential functions to solve exponential equations For any algebraic expressions S and T, and any positive real number b β‰  1, bS = bT if and only if S = T How To… Given an exponential equation with the form bS = bT, where S and T are algebraic expressions with an unknown, solve for the unknown. 1. Use the rules of exponents to simplify, if necessary, so that the resulting equation has the form bS = bT. 2. Use the one-to-one property to set the exponents equal. 3. Solve the resulting equation, S = T, for the unknown. Example 1 Solving an Exponential Equation with a Common Base Solve 2x βˆ’ 1 = 22x βˆ’ 4. Solution Try It #1 Solve 52x = 53x + 2. 2x βˆ’ 1 = 22x βˆ’ 4 x βˆ’ 1 = 2x βˆ’ 4 x = 3 The common base is 2. By the one-to-one property the exponents must be equal. Solve for x. Rewriting Equations So All Powers Have the Same Base Sometimes the common base for an exponential equation is not explicitly shown. In these cases, we simply rewrite the terms in the equation as powers with a common base, and solve using the one-to-one property. For example, consider the equation 256 = 4x βˆ’ 5. We can rewrite both sides of this equation as a power of 2. Then we apply the rules of exponents, along with the one-to-one property, to solve for x : 256 = 4x βˆ’ 5 28 = (22)x βˆ’ 5 28 = 22x βˆ’ 10 8 = 2x βˆ’ 10 18 = 2x x = 9 Rewrite each side as a power with base 2. Use the one-to-one property of exponents. Apply the one-to-one property of exponents. Add 10 to both sides. Divide by 2. How To… Given an exponential equation with unlike bases, use the one-to-one property to solve it. 1. Rewrite each side in the equation as a power with a common base. 2. Use the rules of exponents to simplify, if necessary, so that the resulting equation has the form bS = bT. 3. Use the one-to-one property to set the exponents equal. 4.
Solve the resulting equation, S = T, for the unknown. Example 2 Solving Equations by Rewriting Them to Have a Common Base Solve 8x + 2 = 16x + 1. Solution 8x + 2 = 16x + 1 (23)x + 2 = (24)x + 1 23x + 6 = 24x + 4 3x + 6 = 4x + 4 x = 2 Write 8 and 16 as powers of 2. To take a power of a power, multiply exponents. Use the one-to-one property to set the exponents equal. Solve for x. Download the OpenStax text for free at http://cnx.org/content/col11759/latest. 528 CHAPTER 6 EXPONENTIAL AND LOGARITHMIC FUNCTIONS Try It #2 Solve 52x = 253x + 2. Solving Equations by Rewriting Roots with Fractional Exponents to Have a Common Base 1 __ 25x = 2 2 1 __ 5x = 2 1 __ 10 x = Write the square root of 2 as a power of 2. Use the one-to-one property. Solve for x. Example 3 Solve 25x = √ β€” 2. Solution Try It #3 Solve 5x = √ β€” 5. Q & A… Do all exponential equations have a solution? If not, how can we tell if there is a solution during the problemsolving process? No. Recall that the range of an exponential function is always positive. While solving the equation, we may obtain an expression that is undefined. Example 4 Solving an Equation with Positive and Negative Powers Solve 3x + 1 = βˆ’2. Solution This equation has no solution. There is no real value of x that will make the equation a true statement because any power of a positive number is positive. Analysis Figure 2 shows that the two graphs do not cross so the left side is never equal to the right side. Thus the equation has no solution. y y = 3 x + 1 –5 –4 –3 –2 5 4 3 2 1 –1–1 –2 –3 –4 –5 x y = βˆ’2 1 2 3 4 5 They do not cross. Figure 2 Try It #4 Solve 2x = βˆ’100. Solving Exponential Equations Using Logarithms Sometimes the terms of an exponential equation cannot be rewritten with a common base. In these cases, we solve by taking the
logarithm of each side. Recall, since log(a) = log(b) is equivalent to a = b, we may apply logarithms with the same base on both sides of an exponential equation. Download the OpenStax text for free at http://cnx.org/content/col11759/latest. SECTION 6.6 EXPONENTIAL AND LOGARITHMIC EQUATIONS 529 How To… Given an exponential equation in which a common base cannot be found, solve for the unknown. 1. Apply the logarithm of both sides of the equation. a. If one of the terms in the equation has base 10, use the common logarithm. b. If none of the terms in the equation has base 10, use the natural logarithm. 2. Use the rules of logarithms to solve for the unknown. Example 5 Solve 5x + 2 = 4x. Solution Solving an Equation Containing Powers of Different Bases 5x + 2 = 4x ln(5x + 2) = ln(4x) (x + 2)ln(5) = xln(4) xln(5) + 2ln(5) = xln(4) xln(5) βˆ’ xln(4) = βˆ’ 2ln(5) x(ln(5) βˆ’ ln(4)) = βˆ’ 2ln(5) 5 1 __ __ ) ) = ln ( xln ( 25 4 1 __ ln ( ) 25 _ 5 __ ) ln ( 4 x = There is no easy way to get the powers to have the same base. Take ln of both sides. Use laws of logs. Use the distributive law. Get terms containing x on one side, terms without x on the other. On the left hand side, factor out an x. Use the laws of logs. Divide by the coefficient of x. Try It #5 Solve 2x = 3x + 1. Q & A… Is there any way to solve 2x = 3x? Yes. The solution is 0. Equations Containing e One common type of exponential equations are those with base e. This constant occurs again and again in nature, in mathematics, in science, in engineering, and in finance. When we have an equation with a base e on either side, we can use the natural logarithm to solve it. How To… Given an equation
of the form y = Aekt, solve for t. 1. Divide both sides of the equation by A. 2. Apply the natural logarithm of both sides of the equation. 3. Divide both sides of the equation by k. Example 6 Solve an Equation of the Form y = Ae k t Solve 100 = 20e 2t. Solution 100 = 20e 2t 5 = e 2t ln(5) = 2t t = ln(5)___ 2 Divide by the coefficient of the power. Take ln of both sides. Use the fact that ln(x) and e x are inverse functions. Divide by the coefficient of t. Download the OpenStax text for free at http://cnx.org/content/col11759/latest. 530 CHAPTER 6 EXPONENTIAL AND LOGARITHMIC FUNCTIONS Analysis Using laws of logs, we can also write this answer in the form t = ln √ of the answer, we use a calculator. β€” 5. If we want a decimal approximation Try It #6 Solve 3e 0.5t = 11. Q & A… Does every equation of the form y = Aekt have a solution? No. There is a solution when k β‰  0, and when y and A are either both 0 or neither 0, and they have the same sign. An example of an equation with this form that has no solution is 2 = βˆ’3et. Example 7 Solving an Equation That Can Be Simplified to the Form y = Ae k t Solve 4e2x + 5 = 12. Solution 4e2x + 5 = 12 4e2x = 7 7 __ e2x = 4 7 __ ) 2x = ln ( 4 Combine like terms. Divide by the coefficient of the power. Take ln of both sides. 7 1 __ __ ) ln ( x = 4 2 Solve for x. Try It #7 Solve 3 + e2t = 7e2t. Extraneous Solutions Sometimes the methods used to solve an equation introduce an extraneous solution, which is a solution that is correct algebraically but does not satisfy the conditions of the original equation. One such situation arises in solving when the logarithm is taken on both sides of the equation. In such cases, remember that the argument of the logarithm must be positive. If the number we are evaluating in a
logarithm function is negative, there is no output. Example 8 Solving Exponential Functions in Quadratic Form Solve e2x βˆ’ e x = 56. Solution e 2x βˆ’ e x = 56 e 2x βˆ’ e x βˆ’ 56 = 0 (e x + 7)(e x βˆ’ 8) = 0 Get one side of the equation equal to zero. Factor by the FOIL method. e x + 7 = 0 or e x βˆ’ 8 = 0 If a product is zero, then one factor must be zero. e x = βˆ’7 or e x = 8 Isolate the exponentials. e x = 8 x = ln(8) Reject the equation in which the power equals a negative number. Solve the equation in which the power equals a positive number. Analysis When we plan to use factoring to solve a problem, we always get zero on one side of the equation, because zero has the unique property that when a product is zero, one or both of the factors must be zero. We reject the equation e x = βˆ’7 because a positive number never equals a negative number. The solution ln(βˆ’7) is not a real number, and in the real number system this solution is rejected as an extraneous solution. Download the OpenStax text for free at http://cnx.org/content/col11759/latest. SECTION 6.6 EXPONENTIAL AND LOGARITHMIC EQUATIONS 531 Try It #8 Solve e2x = e x + 2. Q & A… Does every logarithmic equation have a solution? No. Keep in mind that we can only apply the logarithm to a positive number. Always check for extraneous solutions. Using the Definition of a Logarithm to Solve Logarithmic Equations We have already seen that every logarithmic equation logb(x) = y is equivalent to the exponential equation b y = x. We can use this fact, along with the rules of logarithms, to solve logarithmic equations where the argument is an algebraic expression. For example, consider the equation log2(2) + log2(3x βˆ’ 5) = 3. To solve this equation, we can use rules of logarithms to rewrite the left side in compact form and then apply the definition of logs to solve for x: log2(2) + log2(3x
βˆ’ 5) = 3 log2(2(3x βˆ’ 5)) = 3 log2(6x βˆ’ 10) = 3 23 = 6x βˆ’ 10 8 = 6x βˆ’ 10 18 = 6x x = 3 Apply the product rule of logarithms. Distribute. Apply the definition of a logarithm. Calculate 23. Add 10 to both sides. Divide by 6. using the definition of a logarithm to solve logarithmic equations For any algebraic expression S and real numbers b and c, where b > 0, b β‰  1, logb(S) = c if and only if b c = S Example 9 Using Algebra to Solve a Logarithmic Equation Solve 2ln(x) + 3 = 7. Solution Try It #9 Solve 6 + ln(x) = 10. 2ln(x) + 3 = 7 2ln(x) = 4 ln(x) = 2 x = e2 Subtract 3. Divide by 2. Rewrite in exponential form. Example 10 Using Algebra Before and After Using the Definition of the Natural Logarithm Solve 2ln(6x) = 7. Solution 2ln(6x) = 7 7 __ ln(6x) = 2 7 __ 6x = e 2 7 1 __ __ e x = 2 6 Divide by 2. Use the definition of ln. Divide by 6. Download the OpenStax text for free at http://cnx.org/content/col11759/latest. 532 CHAPTER 6 EXPONENTIAL AND LOGARITHMIC FUNCTIONS Try It #10 Solve 2ln(x + 1) = 10. Example 11 Solve ln(x) = 3. Solution Using a Graph to Understand the Solution to a Logarithmic Equation ln(x) = 3 x = e 3 Use the definition of the natural logarithm. Figure 3 represents the graph of the equation. On the graph, the x-coordinate of the point at which the two graphs intersect is close to 20. In other words e3 β‰ˆ 20. A calculator gives a better approximation: e3 β‰ˆ 20.0855. y = 1n(x) y = 3 (e3, 3) β‰ˆ (20.0855, 3) 4 8 12 16 20 24 28 x y 4 2 1 –1
–2 Figure 3 The graphs of y = ln(x ) and y = 3 cross at the point (e 3, 3), which is approximately (20.0855, 3). Try It #11 Use a graphing calculator to estimate the approximate solution to the logarithmic equation 2x = 1000 to 2 decimal places. Using the One-to-One Property of Logarithms to Solve Logarithmic Equations As with exponential equations, we can use the one-to-one property to solve logarithmic equations. The one-to-one property of logarithmic functions tells us that, for any real numbers x > 0, S > 0, T > 0 and any positive real number b, where b β‰  1, logb(S) = logb(T) if and only if S = T. For example, If log2(x βˆ’ 1) = log2(8), then x βˆ’ 1 = 8. So, if x βˆ’ 1 = 8, then we can solve for x, and we get x = 9. To check, we can substitute x = 9 into the original equation: log2(9 βˆ’ 1) = log2(8) = 3. In other words, when a logarithmic equation has the same base on each side, the arguments must be equal. This also applies when the arguments are algebraic expressions. Therefore, when given an equation with logs of the same base on each side, we can use rules of logarithms to rewrite each side as a single logarithm. Then we use the fact that logarithmic functions are one-to-one to set the arguments equal to one another and solve for the unknown. For example, consider the equation log(3x βˆ’ 2) βˆ’ log(2) = log(x + 4). To solve this equation, we can use the rules of logarithms to rewrite the left side as a single logarithm, and then apply the one-to-one property to solve for x: log(3x βˆ’ 2) βˆ’ log(2) = log(x + 4) log ( ) = log(x + 4) 3x βˆ’ 2 ______ 2 3x βˆ’ 2 ______ 2 = x + 4 3x βˆ’ 2 = 2x + 8 Apply the quotient rule of logarithms. Apply the one to one property of a logarithm. Multiply both sides of
the equation by 2. x = 10 Subtract 2x and add 2. Download the OpenStax text for free at http://cnx.org/content/col11759/latest. SECTION 6.6 EXPONENTIAL AND LOGARITHMIC EQUATIONS 533 To check the result, substitute x = 10 into log(3x βˆ’ 2) βˆ’ log(2) = log(x + 4). log(3(10) βˆ’ 2) βˆ’ log(2) = log((10) + 4) log(28) βˆ’ log(2) = log(14) 28 __ ) = log(14) 2 log ( The solution checks. using the one-to-one property of logarithms to solve logarithmic equations For any algebraic expressions S and T and any positive real number b, where b β‰  1, logb(S) = logb(T) if and only if S = T Note, when solving an equation involving logarithms, always check to see if the answer is correct or if it is an extraneous solution. How To… Given an equation containing logarithms, solve it using the one-to-one property. 1. Use the rules of logarithms to combine like terms, if necessary, so that the resulting equation has the form logbS = logbT. 2. Use the one-to-one property to set the arguments equal. 3. Solve the resulting equation, S = T, for the unknown. Example 12 Solving an Equation Using the One-to-One Property of Logarithms Solve ln(x2) = ln(2x + 3). Solution ln(x2) = ln(2x + 3) x2 = 2x + 3 x2 βˆ’ 2x βˆ’ 3 = 0 (x βˆ’ 3)(x + 1) = 0 Use the one-to-one property of the logarithm. Get zero on one side before factoring. Factor using FOIL. x βˆ’ 3 = 0 or x + 1 = 0 If a product is zero, one of the factors must be zero. x = 3 or x = βˆ’1 Solve for x. Analysis There are two solutions: 3 or βˆ’1. The solution βˆ’1 is negative, but it checks when substituted into the original equation because the argument of the logarithm functions is still positive. Try It #12 Solve l
n(x2) = ln(1). Solving Applied Problems Using Exponential and Logarithmic Equations In previous sections, we learned the properties and rules for both exponential and logarithmic functions. We have seen that any exponential function can be written as a logarithmic function and vice versa. We have used exponents to solve logarithmic equations and logarithms to solve exponential equations. We are now ready to combine our skills to solve equations that model real-world situations, whether the unknown is in an exponent or in the argument of a logarithm. One such application is in science, in calculating the time it takes for half of the unstable material in a sample of a radioactive substance to decay, called its half-life. Table 1 lists the half-life for several of the more common radioactive substances. Download the OpenStax text for free at http://cnx.org/content/col11759/latest. 534 CHAPTER 6 EXPONENTIAL AND LOGARITHMIC FUNCTIONS Substance gallium-67 cobalt-60 Use nuclear medicine manufacturing technetium-99m nuclear medicine americium-241 construction Half-life 80 hours 5.3 years 6 hours 432 years carbon-14 uranium-235 archeological dating 5,715 years atomic power 703,800,000 years Table 1 We can see how widely the half-lives for these substances vary. Knowing the half-life of a substance allows us to calculate the amount remaining after a specified time. We can use the formula for radioactive decay: A(t) = A0 e ln(0.5) _____ T t t__ T A(t) = A0e ln(0.5) t _ A(t) = A0 (e ln(0.5)) T t _ 1 __ ) A(t) = A0 ( T 2 where β€’ A0 is the amount initially present β€’ T is the half-life of the substance β€’ t is the time period over which the substance is studied β€’ y is the amount of the substance present after time t Example 13 Using the Formula for Radioactive Decay to Find the Quantity of a Substance How long will it take for ten percent of a 1,000-gram sample of uranium-235 to decay? Solution t t y = 1000 e ln(0.5) __________ 703,800,000 900 = 1000 e ln(0.5)
__________ 703,800,000 0.9 = e ln(0.5) __________ 703,800,000 t ln(0.9) = ln ( e ln(0.5) __________ 703,800,000 t ) ln(0.9) = ln(0.5) __ t 703,800,000 t = 703,800,000 Γ— ln(0.9) _ ln(0.5) years After 10% decays, 900 grams are left. Divide by 1000. Take ln of both sides. ln(eM) = M Solve for t. t β‰ˆ 106,979,777 years Analysis Ten percent of 1,000 grams is 100 grams. If 100 grams decay, the amount of uranium-235 remaining is 900 grams. Try It #13 How long will it take before twenty percent of our 1,000-gram sample of uranium-235 has decayed? Access these online resources for additional instruction and practice with exponential and logarithmic equations. β€’ Solving Logarithmic Equations (http://openstaxcollege.org/l/solvelogeq) β€’ Solving Exponential Equations with Logarithms (http://openstaxcollege.org/l/solveexplog) Download the OpenStax text for free at http://cnx.org/content/col11759/latest. SECTION 6.6 SECTION EXERCISES 535 6.6 SECTION EXERCISES VERBAL 1. 2. 3. ALGEBRAIC 4. = 8. = = = 5. 9. 12. βˆ’vβˆ’= βˆ’v 13. β‹…x= 6. 10. = + = 7. 11. = + = 14. = 17. 20. 23. xβˆ’= = x+= xβˆ’ 15. 18. 21. 24. = ()= = x+= xβˆ’ 16. 19. 22. 25. = = += e r+βˆ’= βˆ’ fi 26. (+)= 27. (+)= 28. ( )= 29. 30. βˆ’x= βˆ’x )=( ( ) 32. βˆ’x= xβˆ’x x 31. (+)=(+) 35. (+)+( )= 33. 36. 39. 42. ()+( (/)+( )
= )= (+) (+)+( ()= )=() 34. 37. 40. 43. ()+(+)= )= ()+( (+) ( () ( )=() 38. () ) =() +(+) =() (+) (+)= ( 44. 41. )=() GRAPHICAL xTh 45. 48. xβˆ’= βˆ’ xβˆ’= 51. xβˆ’βˆ’= βˆ’ 54. x+= βˆ’x 57. βˆ’xβˆ’= 46. 49. 52. 55. 58. x+= +βˆ’x= βˆ’x= βˆ’x βˆ’x= xβˆ’ xβˆ’x+= 47. x= 50. 53. 56. βˆ’+βˆ’x= βˆ’ βˆ’xβˆ’x= xβˆ’ x+= x+ Download the OpenStax text for free at http://cnx.org/content/col11759/latest. 536 CHAPTER 6 EXPONENTIAL AND LOGARITHMIC FUNCTIONS 59. ft 60. Th I ) D D= ( I I I= βˆ’ β‹… 61. Th = e tt TECHNOLOGY Th 62. 64. 66. t= t= eβˆ’t= 63. 65. ex= xβˆ’= 67. exβˆ’+= 68. +x+= 69. βˆ’xβˆ’= + 70. P P= eβˆ’xx Hint 71. ThM E M= )E ( E E= Β· EXTENSIONS 72. x= x b b 73. 74. r ) kt A= a ( + k t 75. y= Ae kt tt T T= Ts+ Tβˆ’Tseβˆ’ktTs T k tt Download the OpenStax text for free at http://cnx.org/content/col11759/latest. SECTION 6.7 EXPONENTIAL AND LOGARITHMIC MODELS 537 LEARNING OBJECTIVES In this section, you will: β€’ odel eponential growth and decay. β€’ Use ewtons aw of Cooling. β€’ pply compound interest formulas and continuous growth formulas. β€’ Use logistic-growth models. 6. 7 EXPONENTIAL AND LOGARITHMIC MODELS Figure 1 A nuclear research reactor inside the Neely Nuclear Research Center on the Georgia Institute of Technology campus. (credit: Georgia Tech Research Institute) We have already explored some basic applications of exponential and logarithmic functions.
In this section, we explore some important applications in more depth, including radioactive isotopes and Newton’s Law of Cooling. Modeling Exponential Growth and Decay In real-world applications, we need to model the behavior of a function. In mathematical modeling, we choose a familiar general function with properties that suggest that it will model the real-world phenomenon we wish to analyze. In the case of rapid growth, we may choose the exponential growth function: y = A0e kt where A0 is equal to the value at time zero, e is Euler’s constant, and k is a positive constant that determines the rate (percentage) of growth. We may use the exponential growth function in applications involving doubling time, the time it takes for a quantity to double. Such phenomena as wildlife populations, financial investments, biological samples, and natural resources may exhibit growth based on a doubling time. In some applications, however, as we will see when we discuss the logistic equation, the logistic model sometimes fits the data better than the exponential model. On the other hand, if a quantity is falling rapidly toward zero, without ever reaching zero, then we should probably choose the exponential decay model. Again, we have the form y = A0e kt where A0 is the starting value, and e is Euler’s constant. Now k is a negative constant that determines the rate of decay. We may use the exponential decay model when we are calculating half-life, or the time it takes for a substance to exponentially decay to half of its original quantity. We use half-life in applications involving radioactive isotopes. Download the OpenStax text for free at http://cnx.org/content/col11759/latest. 538 CHAPTER 6 EXPONENTIAL AND LOGARITHMIC FUNCTIONS In our choice of a function to serve as a mathematical model, we often use data points gathered by careful observation and measurement to construct points on a graph and hope we can recognize the shape of the graph. Exponential growth and decay graphs have a distinctive shape, as we can see in Figure 2 and Figure 3. It is important to remember that, although parts of each of the two graphs seem to lie on the x-axis, they are really a tiny distance above the x-axis. y 6 5 4 3 2 1 y = 2e3 x,1 3 2e (0, 2) βˆ’ 1 3, 2 e –5 –4 –3 –
2 –1–1 –2 21 3 4 y = 0 5 x y = 3e βˆ’2 x y,βˆ’ 1 2 3e y = 0 –5 –4 –3 –2 10 8 6 4 2 –1 –2 –4 –6 –8 –10 (0, 3 Figure 2 A graph showing exponential growth. The equation is y = 2e 3x. Figure 3 A graph showing exponential decay. The equation is y = 3e βˆ’2x. Exponential growth and decay often involve very large or very small numbers. To describe these numbers, we often use orders of magnitude. The order of magnitude is the power of ten, when the number is expressed in scientific notation, with one digit to the left of the decimal. For example, the distance to the nearest star, Proxima Centauri, measured in kilometers, is 40,113,497,200,000 kilometers. Expressed in scientific notation, this is 4.01134972 Γ— 1013. So, we could describe this number as having order of magnitude 1013. characteristics of the exponential function, y = A0 e kt An exponential function with the form y = A0e kt has the following characteristics: β€’ one-to-one function β€’ horizontal asymptote: y = 0 β€’ domain: ( β€“βˆž, ∞) β€’ range: (0, ∞) β€’ x-intercept: none β€’ y-intercept: (0, A0) β€’ increasing if k > 0 (see Figure 4) β€’ decreasing if k < 0 (see Figure 4) ( ) 1 k, A0e y = A0ekt k > 0 ( ), – (0, A0) A0 e 1 k y ( ) 1 – k, A0e y = A0ekt k < 0 (0, A0) y = 0 t y = 0 y ( ), A0 e 1 k t Figure 4 An exponential function models exponential growth when k > 0 and exponential decay when k < 0. Example 1 Graphing Exponential Growth A population of bacteria doubles every hour. If the culture started with 10 bacteria, graph the population as a function of time. Solution When an amount grows at a fixed percent per unit time, the growth is exponential. To find A0 we use the fact that A0 is the amount at time zero, so A0 = 10. To find k, use the fact that after one hour (t = 1)
the population doubles from 10 to 20. The formula is derived as follows 20 = 10e k β‹… 1 2 = e k ln2 = k Divide by 10 Take the natural logarithm so k = ln(2). Thus the equation we want to graph is y = 10e(ln2)t = 10(eln2)t = 10 Β· 2t. The graph is shown in Figure 5. Download the OpenStax text for free at http://cnx.org/content/col11759/latest. SECTION 6.7 EXPONENTIAL AND LOGARITHMIC MODELS 539 y = 10e(ln 2)t y 320 280 240 200 160 120 80 40 1 2 3 4 5 6 t Figure 5 The graph of y = 10e (ln2)t. Analysis The population of bacteria after ten hours is 10,240. We could describe this amount is being of the order of magnitude 104. The population of bacteria after twenty hours is 10,485,760 which is of the order of magnitude 10 7, so we could say that the population has increased by three orders of magnitude in ten hours. Half-Life We now turn to exponential decay. One of the common terms associated with exponential decay, as stated above, is half-life, the length of time it takes an exponentially decaying quantity to decrease to half its original amount. Every radioactive isotope has a half-life, and the process describing the exponential decay of an isotope is called radioactive decay. To fi nd the half-life of a function describing exponential decay, solve the following equation: We find that the half-life depends only on the constant k and not on the starting quantity A0. The formula is derived as follows 1 __ A0 = A0e kt 2 1 __ A0 = A0e kt 2 1 __ = e kt 2 1 ) = kt ln ( __ 2 Divide by A0. Take the natural log. βˆ’ ln(2) = kt Apply laws of logarithms. βˆ’ ln(2) ____ k = t Divide by k. Since t, the time, is positive, k must, as expected, be negative. This gives us the half-life formula t = βˆ’ ln(2) ____ k How To… Given the half-life, find the decay rate. 1. Write A = A0 ekt. 1 _ 2. Replace A by A0 and replace
t by the given half-life. 2 3. Solve to find k. Express k as an exact value (do not round). ln(2) _. t Note: It is also possible to find the decay rate using k = βˆ’ Download the OpenStax text for free at http://cnx.org/content/col11759/latest. 540 CHAPTER 6 EXPONENTIAL AND LOGARITHMIC FUNCTIONS Example 2 Finding the Function that Describes Radioactive Decay The half-life of carbon-14 is 5,730 years. Express the amount of carbon-14 remaining as a function of time, t. Solution This formula is derived as follows. A = A0e kt 0.5A0 = A0e k β‹… 5730 0.5 = e5730k ln(0.5) = 5730k ln(0.5) ______ 5730 A = A0 e ( ln(0.5) ______ 5730 k = ) t The continuous growth formula. Substitute the half-life for t and 0.5A0 for f(t). Divide by A0. Take the natural log of both sides. Divide by the coefficient of k. Substitute for k in the continuous growth formula. ) t. We observe that the coefficient of t, ln(0.5) ______ 5730 The function that describes this continuous decay is f(t) = A0 e ( ln(0.5) _ 5730 β‰ˆ βˆ’1.2097 is negative, as expected in the case of exponential decay. Try It #14 The half-life of plutonium-244 is 80,000,000 years. Find function gives the amount of carbon-14 remaining as a function of time, measured in years. Radiocarbon Dating The formula for radioactive decay is important in radiocarbon dating, which is used to calculate the approximate date a plant or animal died. Radiocarbon dating was discovered in 1949 by Willard Libb y, who won a Nobel Prize for his discovery. It compares the difference between the ratio of two isotopes of carbon in an organic artifact or fossil to the ratio of those two isotopes in the air. It is believed to be accurate to within about 1% error for plants or animals that died within the last 60,000 years. Carbon-14 is a radioactive isotope of carbon that has a half-life of 5,730 years. It occurs
in small quantities in the carbon dioxide in the air we breathe. Most of the carbon on earth is carbon-12, which has an atomic weight of 12 and is not radioactive. Scientists have determined the ratio of carbon-14 to carbon-12 in the air for the last 60,000 years, using tree rings and other organic samples of known datesβ€”although the ratio has changed slightly over the centuries. As long as a plant or animal is alive, the ratio of the two isotopes of carbon in its body is close to the ratio in the atmosphere. When it dies, the carbon-14 in its body decays and is not replaced. By comparing the ratio of carbon-14 to carbon-12 in a decaying sample to the known ratio in the atmosphere, the date the plant or animal died can be approximated. Since the half-life of carbon-14 is 5,730 years, the formula for the amount of carbon-14 remaining after t years is ) t ln(0.5) ______ 5730 A β‰ˆ A0 e ( where β€’ A is the amount of carbon-14 remaining β€’ A0 is the amount of carbon-14 when the plant or animal began decaying. This formula is derived as follows: A = A0e kt 0.5A0 = A0e k β‹… 5730 0.5 = e5730k ln(0.5) = 5730k ln(0.5) ______ 5730 A = A0 e ( ln(0.5) ______ 5730 k = ) t To find the age of an object, we solve this equation for t: A ) ln ( _ A0 _ βˆ’ 0.000121 t = The continuous growth formula. Substitute the half-life for t and 0.5A0 for f (t). Divide by A0. Take the natural log of both sides. Divide by the coefficient of k. Substitute for r in the continuous growth formula. Download the OpenStax text for free at http://cnx.org/content/col11759/latest. SECTION 6.7 EXPONENTIAL AND LOGARITHMIC MODELS 541 Out of necessity, we neglect here the many details that a scientist takes into consideration when doing carbon-14 dating, and we only look at the basic formula. The ratio of carbon-14 to carbon-12 in the atmosphere is approximately 0.0000000001%. Let r be the ratio of carbon-14 to carbon-
12 in the organic artifact or fossil to be dated, determined by a method called liquid scintillation. From the equation A β‰ˆ A0eβˆ’0.000121t we know the ratio of the percentage of β‰ˆ eβˆ’0.000121t. We solve carbon-14 in the object we are dating to the percentage of carbon-14 in the atmosphere is r = this equation for t, to get A __ A0 t = ln(r) _ βˆ’0.000121 How To… Given the percentage of carbon-14 in an object, determine its age. 1. Express the given percentage of carbon-14 as an equivalent decimal, k. 2. Substitute for k in the equation t = and solve for the age, t. ln(r) _________ βˆ’0.000121 Example 3 Finding the Age of a Bone A bone fragment is found that contains 20% of its original carbon-14. To the nearest year, how old is the bone? Solution We substitute 20% = 0.20 for k in the equation and solve for t : t = = ln(r) _ βˆ’0.000121 ln(0.20) _ βˆ’0.000121 Use the general form of the equation. Substitute for r. β‰ˆ 13301 Round to the nearest year. The bone fragment is about 13,301 years old. Analysis The instruments that measure the percentage of carbon-14 are e xtremely sensitive and, as we mention above, a scientist will need to do much more work than we did in order to be satisfied. Even so, carbon dating is only accurate to about 1%, so this age should be given as 13,301 years Β± 1% or 13,301 years Β± 133 years. Try It #15 Cesium-137 has a half-life of about 30 years. If we begin with 200 mg of cesium-137, will it take more or less than 230 years until only 1 milligram remains? Calculating Doubling Time For decaying quantities, we determined how long it took for half of a substance to decay. For growing quantities, we might want to find out how long it takes for a quantity to double. As we mentioned above, the time it takes for a quantity to double is called the doubling time. Given the basic exponential growth equation A = A0e kt, doubling time can be found by solving for when the original quantity has doubled, that is, by solving 2A0 =
A0e kt. The formula is derived as follows: Thus the doubling time is 2A0 = A0e kt 2 = e kt ln(2) = kt t = ln(2) _ k t = ln(2) _ k Divide by A0. Take the natural logarithm. Divide by the coefficient of t. Download the OpenStax text for free at http://cnx.org/content/col11759/latest. 542 CHAPTER 6 EXPONENTIAL AND LOGARITHMIC FUNCTIONS Example 4 Finding a Function That Describes Exponential Growth According to Moore’s Law, the doubling time for the number of transistors that can be put on a computer chip is approximately two years. Give a function that describes this behavior. Solution The formula is derived as follows: t = 2 = ln(2) ___ k ln(2) ___ k ln(2) ___ 2 A = A0 e k = ln(2) _____ 2 t The doubling time formula. Use a doubling time of two years. Multiply by k and divide by 2. Substitute k into the continuous growth formula. The function is A0 e ln(2) t. _____ 2 Try It #16 Recent data suggests that, as of 2013, the rate of growth predicted by Moore’s Law no longer holds. Growth has slowed to a doubling time of approximately three years. Find the new function that takes that longer doubling time into account. Using Newton’s Law of Cooling Exponential decay can also be applied to temperature. When a hot object is left in surrounding air that is at a lower temperature, the object’s temperature will decrease exponentially, leveling off as it approaches the surrounding air temperature. On a graph of the temperature function, the leveling off will correspond to a horizontal asymptote at the temperature of the surrounding air. Unless the room temperature is zero, this will correspond to a vertical shift of the generic exponential decay function. This translation leads to Newton’s Law of Cooling, the scientific formula for temperature as a function of time as an object’s temperature is equalized with the ambient temperature This formula is derived as follows: T(t) = Ae kt + Ts T(t) = Ab ct + Ts T(t) = Ae ln(bct) + Ts T(t) = Ae ctln(b
) + Ts T(t) = Ae kt + Ts Laws of logarithms. Laws of logarithms. Rename the constant cln(b), calling it k. Newton’s law of cooling The temperature of an object, T, in surrounding air with temperature Ts will behave according to the formula T(t) = Ae kt + Ts where β€’ t is time β€’ A is the difference between the initial temperature of the object and the surroundings β€’ k is a constant, the continuous rate of cooling of the object How To… Given a set of conditions, apply Newton’s Law of Cooling. 1. Set Ts equal to the y-coordinate of the horizontal asymptote (usually the ambient temperature). 2. Substitute the given values into the continuous growth formula T(t) = Ae kt + Ts to find the parameters A and k. 3. Substitute in the desired time to find the temperature or the desired temperature to find the time. Download the OpenStax text for free at http://cnx.org/content/col11759/latest. SECTION 6.7 EXPONENTIAL AND LOGARITHMIC MODELS 543 Example 5 Using Newton’s Law of Cooling A cheesecake is taken out of the oven with an ideal internal temperature of 165Β°F, and is placed into a 35Β°F refrigerator. After 10 minutes, the cheesecake has cooled to 150Β°F. If we must wait until the cheesecake has cooled to 70Β°F before we eat it, how long will we have to wait? Solution Because the surrounding air temperature in the refrigerator is 35 degrees, the cheesecake’s temperature will decay exponentially toward 35, following the equation We know the initial temperature was 165, so T(0) = 165. T(t) = Ae kt + 35 165 = Ae k0 + 35 Substitute (0, 165). A = 130 Solve for A. We were given another data point, T(10) = 150, which we can use to solve for k. 150 = 130e k10 + 35 Substitute (10, 150). 115 = 130e k10 115 ___ 130 = e 10k Subtract 35. Divide by 130. 115 ___ 130 ln ( ) = 10k 115 ___ ln ( ) 130 _ 10 This gives us the equation for the cooling of the cheesecake: T(t) = 130e βˆ’0.0123t + 35
. β‰ˆ βˆ’0.0123 Divide by the coefficient of k. k = Take the natural log of both sides. Now we can solve for the time it will take for the temperature to cool to 70 degrees. 70 = 130eβˆ’0.0123t + 35 35 = 130eβˆ’0.0123t Substitute in 70 for T(t). Subtract 35. 35 ___ 130 = eβˆ’0.0123t Divide by 130. ln ( 35 ___ 130 Take the natural log of both sides ) = βˆ’0.0123t 35 ___ ) ln ( 130 _ β‰ˆ 106.68 Divide by the coefficient of t. βˆ’0.0123 t = It will take about 107 minutes, or one hour and 47 minutes, for the cheesecake to cool to 70Β°F. Try It #17 A pitcher of water at 40 degrees Fahrenheit is placed into a 70 degree room. One hour later, the temperature has risen to 45 degrees. How long will it take for the temperature to rise to 60 degrees? Using Logistic Growth Models Exponential growth cannot continue forever. Exponential models, while they may be useful in the short term, tend to fall apart the longer they continue. Consider an aspiring writer who writes a single line on day one and plans to double the number of lines she writes each day for a month. By the end of the month, she must write over 17 billion lines, or one-half-billion pages. It is impractical, if not impossible, for anyone to write that much in such a short period of time. Eventually, an exponential model must begin to approach some limiting value, and then the growth is forced to slow. For this reason, it is often better to use a model with an upper bound instead of an exponential growth model, though the exponential growth model is still useful over a short term, before approaching the limiting value. Download the OpenStax text for free at http://cnx.org/content/col11759/latest. 544 CHAPTER 6 EXPONENTIAL AND LOGARITHMIC FUNCTIONS The logistic growth model is approximately exponential at first, but it has a reduced rate of growth as the output approaches the model’s upper bound, called the carrying capacity. For constants a, b, and c, the logistic growth of a population over time x is represented by the model f (x) = c _______ 1 + aeβˆ’b x The graph in Figure 6 shows how
the growth rate changes over time. The graph increases from left to right, but the growth rate only increases until it reaches its point of maximum growth rate, at which point the rate of increase decreases. f (x+a logistic growth The logistic growth model is where c _____ 1 + a β€’ is the initial value Carrying capacity f (x) = c 1 + ae–bx ln(a) b ( )c, 2 Point of maximum growth Initial value of population x Figure 6 f (x) = c _ 1 + aeβˆ’b x β€’ c is the carrying capacity, or limiting value β€’ b is a constant determined by the rate of growth. Example 6 Using the Logistic-Growth Model An influenza epidemic spreads through a population rapidly, at a rate that depends on two factors: The more people who have the flu, the more rapidly it spreads, and also the more uninfected people there are, the more rapidly it spreads. These two factors make the logistic model a good one to study the spread of communicable diseases. And, clearly, there is a maximum value for the number of people infected: the entire population. For example, at time t = 0 there is one person in a community of 1,000 people who has the flu. So, in that community, at most 1,000 people can have the flu. Researchers find that for this particular strain of the flu, the logistic growth constant is b = 0.6030. Estimate the number of people in this community who will have had this flu after ten days. Predict how many people in this community will have had this flu after a long period of time has passed. Solution We substitute the given data into the logistic growth model c _______ 1 + aeβˆ’b x f (x) = This model predicts that, after ten days, the number of people who have had the flu is f (x) = Because at most 1,000 people, the entire population of the community, can get the flu, we know the limiting value is c = 1000. To find a, we use the formula that the number of cases at time t = 0 is = 1, from which it follows that a = 999. 1000 ____________ 1 + 999eβˆ’0.6030x β‰ˆ 293.8. Because the actual number must be a whole number (a person has either had the flu or not) we round to 294. In the long term, the number of people
who will contract the flu is the limiting value, c = 1000. c _ 1 + a Analysis Remember that, because we are dealing with a virus, we cannot predict with certainty the number of people infected. The model only approximates the number of people infected and will not give us exact or actual values. The graph in Figure 7 gives a good picture of how this model fits the data. Download the OpenStax text for free at http://cnx.org/content/col11759/latest. SECTION 6.7 EXPONENTIAL AND LOGARITHMIC MODELS 545 y = 1000 1,000 cases on day 21 1,100 1,000 900 800 700 s e s a C 600 500 400 300 200 100 294 cases on day 10 1 case on day 0 20 cases on day 5 2 4 6 8 10 12 14 Days 16 18 20 22 24 26 Figure 7 The graph of f (x) = 1000 __ 1 + 999e βˆ’0.6030x Try It #18 Using the model in Example 6, estimate the number of cases of flu on day 15. Choosing an Appropriate Model for Data Now that we have discussed various mathematical models, we need to learn how to choose the appropriate model for the raw data we have. Many factors influence the choice of a mathematical model, among which are experience, scientific laws, and patterns in the data itself. Not all data can be described by elementary functions. Sometimes, a function is chosen that approximates the data over a given interval. For instance, suppose data were gathered on the number of homes bought in the United States from the years 1960 to 2013. After plotting these data in a scatter plot, we notice that the shape of the data from the years 2000 to 2013 follow a logarithmic curve. We could restrict the interval from 2000 to 2010, apply regression analysis using a logarithmic model, and use it to predict the number of home buyers for the year 2015. Three kinds of functions that are often useful in mathematical models are linear functions, exponential functions, and logarithmic functions. If the data lies on a straight line, or seems to lie approximately along a straight line, a linear model may be best. If the data is non-linear, we often consider an exponential or logarithmic model, though other models, such as quadratic models, may also be considered. In choosing between an exponential model and a logarithmic model, we look at the way the data curves. This is
called the concavity. If we draw a line between two data points, and all (or most) of the data between those two points lies above that line, we say the curve is concave down. We can think of it as a bowl that bends downward and therefore cannot hold water. If all (or most) of the data between those two points lies below the line, we say the curve is concave up. In this case, we can think of a bowl that bends upward and can therefore hold water. An exponential curve, whether rising or falling, whether representing growth or decay, is always concave up away from its horizontal asymptote. A logarithmic curve is always concave away from its vertical asymptote. In the case of positive data, which is the most common case, an exponential curve is always concave up, and a logarithmic curve always concave down. A logistic curve changes concavity. It starts out concave up and then changes to concave down beyond a certain point, called a point of inflection. After using the graph to help us choose a type of function to use as a model, we substitute points, and solve to find the parameters. We reduce round-off error by choosing points as far apart as possible. Download the OpenStax text for free at http://cnx.org/content/col11759/latest. 546 CHAPTER 6 EXPONENTIAL AND LOGARITHMIC FUNCTIONS Example 7 Choosing a Mathematical Model Does a linear, exponential, logarithmic, or logistic model best fit the values listed in Table 1? Find the model, and use a graph to check your choice.386 2.197 2.773 3.219 3.584 3.892 4.159 4.394 Table 1 Solution First, plot the data on a graph as in Figure 8. For the purpose of graphing, round the data to two significant digits. y 5.5 5 4.5 4 3.5 3 2.5 2 1.5 1 0. 10 Figure 8 x Clearly, the points do not lie on a straight line, so we reject a linear model. If we draw a line between any two of the points, most or all of the points between those two points lie above the line, so the graph is concave down, suggesting a logarithmic model. We can try y = aln(b x). Plugging in the
first point, (1,0), gives 0 = alnb. We reject the case that a = 0 (if it were, all outputs would be 0), so we know ln(b) = 0. Thus b = 1 and y = aln(x). Ne xt we can use the point (9,4.394) to solve for a: y = aln(x) 4.394 = aln(9) a = 4.394 _____ ln(9) Because a = β‰ˆ 2, an appropriate model for the data is y = 2ln(x). 4.394 _ ln(9) To check the accuracy of the model, we graph the function together with the given points as in Figure 9. y = 2 ln(x) x = 0 y 5.5 5 4.5 4 3.5 3 2.5 2 1.5 1 0. 10 Figure 9 The graph of y = 2lnx. x We can conclude that the model is a good fit to the data. Compare Figure 9 to the graph of y = ln(x2) shown in Figure 10. Download the OpenStax text for free at http://cnx.org/content/col11759/latest. SECTION 6.7 EXPONENTIAL AND LOGARITHMIC MODELS 547 y x = 0 5.5 5 4.5 4 3.5 3 2.5 2 1.5 1 0.5 y = ln(x2 10 Figure 10 The graph of y = ln(x 2) x The graphs appear to be identical when x > 0. A quick check confirms this conclusion: y = ln(x 2) = 2ln(x) for x > 0. However, if x < 0, the graph of y = ln(x 2) includes a β€œextra” branch, as shown in Figure 11. This occurs because, while y = 2ln(x) cannot have negative values in the domain (as such values would force the argument to be negative), the function y = ln(x 2) can have negative domain values. y y = ln(x2) 642 8 10 x –10 –8 –6 –4 10 8 6 4 2 –2 –2 –4 –6 –8 –10 Figure 11 Try It #19 Does a linear, exponential, or logarithmic model best fit the data in Table 2? Find the
model.297 5.437 8.963 14.778 24.365 40.172 66.231 109.196 180.034 Table 2 Expressing an Exponential Model in Base e While powers and logarithms of any base can be used in modeling, the two most common bases are 10 and e. In science and mathematics, the base e is often preferred. We can use laws of exponents and laws of logarithms to change any base to base e. How To… Given a model with the form y = ab x, change it to the form y = A0e kx. 1. Rewrite y = ab x as y = aeln(b x). 2. Use the power rule of logarithms to rewrite y as y = ae xln(b) = aeln(b)x. 3. Note that a = A0 and k = ln(b) in the equation y = A0e kx. Download the OpenStax text for free at http://cnx.org/content/col11759/latest. 548 CHAPTER 6 EXPONENTIAL AND LOGARITHMIC FUNCTIONS Example 8 Changing to base e Change the function y = 2.5(3.1)x so that this same function is written in the form y = A0e kx. Solution The formula is derived as follows y = 2.5(3.1)x = 2.5e ln(3.1x ) = 2.5e xln3.1 Insert exponential and its inverse. Laws of logs. = 2.5e (ln3.1)x Commutative law of multiplication Try It #20 Change the function y = 3(0.5)x to one having e as the base. Access these online resources for additional instruction and practice with exponential and logarithmic models. β€’ Logarithm Application – pH (http://openstaxcollege.org/l/logph) β€’ Exponential Model – Age Using Half-Life (http://openstaxcollege.org/l/expmodelhalf) β€’ Newton’s Law of Cooling (http://openstaxcollege.org/l/newtoncooling) β€’ Exponential Growth Given Doubling Time (http://openstaxcollege.org/l/expgrowthdbl) β€’ Exponential Growth – Find Initial Amount Given Doubling Time (http://
openstaxcollege.org/l/initialdouble) Download the OpenStax text for free at http://cnx.org/content/col11759/latest. SECTION 6.7 SECTION EXERCISES 549 6.7 SECTION EXERCISES VERBAL 1. halflife 2. 3. doubling time 4.. Th 5. NUMERIC 6. Thftt Tt=e βˆ’t+ft f x= +eβˆ’x 7. f 9. 11. hmic. Th 12. f(x=x eo fi 8. f 10. x f(x) TECHNOLOGY x f(x) x 13. 14. 15. 16. f(x) x f(x) x f(x) fit Pt= +eβˆ’t 17. 19. 18. f fi 20. e fi e fi ft 21. 22. e fi P Download the OpenStax text for free at http://cnx.org/content/col11759/latest. 550 CHAPTER 6 EXPONENTIAL AND LOGARITHMIC FUNCTIONS EXTENSIONS 23. 25. S M= ) ( S S 27. b x=e xbbβ‰  24. Th Pt=Pe rtP r>t M 26. y c +aeβˆ’rx y= REAL-WORLD APPLICATIONS For the following exercises, use this scenario: A doctor prescribes 125 milligrams of a therapeutic drug that decays by about 30% each hour. 28. To the nearest hour, what is the half-life of the drug? 29. Write an exponential model representing the amount of the drug remaining in the patient’s system after t hours. Then use the formula to find the amount of the drug that would remain in the patient’s system after 3 hours. Round to the nearest milligram. 30. Using the model found in the previous exercise, find f (10) and interpret the result. Round to the nearest hundredth. For the following exercises, use this scenario: A tumor is injected with 0.5 grams of Iodine-125, which has a decay rate of 1.15% per day. 31. To the nearest day, how long will it take for half of 32. Write an exponential model representing the amount of Iodine-125 remaining in the tumor after t days. Then use the formula to find the amount of Iodine-125 that would remain in the tumor after
60 days. Round to the nearest tenth of a gram. 34. The half-life of Radium-226 is 1590 years. What is the annual decay rate? Express the decimal result to four significant digits and the percentage to two significant digits. 36. A wooden artifact from an archeological dig contains 60 percent of the carbon-14 that is present in living trees. To the nearest year, about how many years old is the artifact? (The half-life of carbon-14 is 5730 years.) the Iodine-125 to decay? 33. A scientist begins with 250 grams of a radioactive substance. After 250 minutes, the sample has decayed to 32 grams. Rounding to five significant digits, write an exponential equation representing this situation. To the nearest minute, what is the half-life of this substance? 35. The half-life of Erbium-165 is 10.4 hours. What is the hourly decay rate? Express the decimal result to four significant digits and the percentage to two significant digits. 37. A research student is working with a culture of bacteria that doubles in size every twenty minutes. The initial population count was 1350 bacteria. Rounding to five significant digits, write an exponential equation representing this situation. To the nearest whole number, what is the population size after 3 hours? Download the OpenStax text for free at http://cnx.org/content/col11759/latest. SECTION 6.7 SECTION EXERCISES 551 ft ft 38. 39. 40. 41. 42. 43. 44. 45. Download the OpenStax text for free at http://cnx.org/content/col11759/latest. 552 CHAPTER 6 EXPONENTIAL AND LOGARITHMIC FUNCTIONS LEARNING OBJECTIVES In this section, you will: β€’ Build an exponential model from data. β€’ Build a logarithmic model from data. β€’ Build a logistic model from data. 6.8 FITTING EXPONENTIAL MODELS TO DATA In previous sections of this chapter, we were either given a function explicitly to graph or evaluate, or we were given a set of points that were guaranteed to lie on the curve. Then we used algebra to find the equation that fit the points exactly. In this section, we use a modeling technique called regression analysis to find a curve that models data collected from real-world observations. With regression analysis, we don
’t expect all the points to lie perfectly on the curve. The idea is to find a model that best fits the data. Then we use the model to make predictions about future events. Do not be confused by the word model. In mathematics, we often use the terms function, equation, and model interchangeably, even though they each have their own formal definition. The term model is typically used to indicate that the equation or function approximates a real-world situation. We will concentrate on three types of regression models in this section: exponential, logarithmic, and logistic. Having already worked with each of these functions gives us an advantage. Knowing their formal definitions, the behavior of their graphs, and some of their real-world applications gives us the opportunity to deepen our understanding. As each regression model is presented, key features and definitions of its associated function are included for review. Take a moment to rethink each of these functions, reflect on the work we’ve done so far, and then explore the ways regression is used to model real-world phenomena. Building an Exponential Model from Data As we’ve learned, there are a multitude of situations that can be modeled by exponential functions, such as investment growth, radioactive decay, atmospheric pressure changes, and temperatures of a cooling object. What do these phenomena have in common? For one thing, all the models either increase or decrease as time moves forward. But that’s not the whole story. It’s the way data increase or decrease that helps us determine whether it is best modeled by an exponential equation. Knowing the behavior of exponential functions in general allows us to recognize when to use exponential regression, so let’s review exponential growth and decay. Recall that exponential functions have the form y = ab x or y = A0e kx. When performing regression analysis, we use the form most commonly used on graphing utilities, y = ab x. Take a moment to reflect on the characteristics we’ve already learned about the exponential function y = ab x (assume a > 0): β€’ b must be greater than zero and not equal to one. β€’ The initial value of the model is y = a. β€’ If b > 1, the function models exponential growth. As x increases, the outputs of the model increase slowly at first, but then increase more and more rapidly, without bound. β€’ If 0 < b < 1, the function models exponential decay. As x increases, the outputs for the model decrease rapidly
at first and then level off to become asymptotic to the x-axis. In other words, the outputs never become equal to or less than zero. As part of the results, your calculator will display a number known as the correlation coefficient, labeled by the variable r, or r 2. (You may have to change the calculator’s settings for these to be shown.) The values are an indication of the β€œgoodness of fit” of the regression equation to the data. We more commonly use the value of r 2 instead of r, but the closer either value is to 1, the better the regression equation approximates the data. Download the OpenStax text for free at http://cnx.org/content/col11759/latest. SECTION 6.8 FITTING EXPONENTIAL MODELS TO DATA 553 exponential regression Exponential regression is used to model situations in which growth begins slowly and then accelerates rapidly without bound, or where decay begins rapidly and then slows down to get closer and closer to zero. We use the command β€œExpReg” on a graphing utility to fit an exponential function to a set of data points. This returns an equation of the form, y = ab x Note that: β€’ b must be non-negative. β€’ when b > 1, we have an exponential growth model. β€’ when 0 < b < 1, we have an exponential decay model. How To… Given a set of data, perform exponential regression using a graphing utility. 1. Use the STAT then EDIT menu to enter given data. a. Clear any existing data from the lists. b. List the input values in the L1 column. c. List the output values in the L2 column. 2. Graph and observe a scatter plot of the data using the STATPLOT feature. a. Use ZOOM [9] to adjust axes to fit the data. b. Verify the data follow an exponential pattern. 3. Find the equation that models the data. a. Select β€œExpReg” from the STAT then CALC menu. b. Use the values returned for a and b to record the model, y = ab x. 4. Graph the model in the same window as the scatterplot to verify it is a good fit for the data. Example 1 Using Exponential Regression to Fit a Model to Data In 2007, a university study was published investigating the crash risk of alcohol impaired driving. Data from 2,871 crashes
were used to measure the association of a person’s blood alcohol level (BAC) with the risk of being in an accident. Table 1 shows results from the study[24]. The relative risk is a measure of how many times more likely a person is to crash. So, for example, a person with a BAC of 0.09 is 3.54 times as likely to crash as a person who has not been drinking alcohol. BAC Relative Risk of Crashing BAC Relative Risk of Crashing 0 1 0.11 6.41 0.01 1.03 0.13 12.6 Table 1 0.03 1.06 0.15 22.1 0.05 1.38 0.17 0.07 2.09 0.19 0.09 3.54 0.21 39.05 65.32 99.78 a. Let x represent the BAC level, and let y represent the corresponding relative risk. Use exponential regression to fit a model to these data. b. After 6 drinks, a person weighing 160 pounds will have a BAC of about 0.16. How many times more likely is a person with this weight to crash if they drive after having a 6-pack of beer? Round to the nearest hundredth. 24 Source: Indiana University Center for Studies of Law in Action, 2007 Download the OpenStax text for free at http://cnx.org/content/col11759/latest. 554 CHAPTER 6 EXPONENTIAL AND LOGARITHMIC FUNCTIONS Solution a. Using the STAT then EDIT menu on a graphing utility, list the BAC values in L1 and the relative risk values in L2. Then use the STATPLOT feature to verify that the scatterplot follows the exponential pattern shown in Figure 1: y 110 100 90 80 70 60 50 40 30 20 10.02.04.06.08.10.12.14.16.18.20.22 x Figure 1 Use the β€œExpReg” command from the STAT then CALC menu to obtain the exponential model, y = 0.58304829(2.20720213E10)x Converting from scientific notation, we have: y = 0.58304829(22,072,021,300)x Notice that r 2 β‰ˆ 0.97 which indicates the model is a good fit to the data. To see this, graph the model in the same window as the scatterplot to verify
it is a good fit as shown in Figure 2: y 110 100 90 80 70 60 50 40 30 20 10.02.04.06.08.10.12.14.16.18.20.22 x Figure 2 b. Use the model to estimate the risk associated with a BAC of 0.16. Substitute 0.16 for x in the model and solve for y. y = 0.58304829(22,072,021,300)x Use the regression model found in part (a). = 0.58304829(22,072,021,300)0.16 Substitute 0.16 for x. β‰ˆ 26.35 Round to the nearest hundredth. If a 160-pound person drives after having 6 drinks, he or she is about 26.35 times more likely to crash than if driving while sober. Download the OpenStax text for free at http://cnx.org/content/col11759/latest. SECTION 6.8 FITTING EXPONENTIAL MODELS TO DATA 555 Try It #1 Table 2 shows a recent graduate’s credit card balance each month after graduation. Month 1 2 3 4 5 6 7 8 Debt ($) 620.00 761.88 899.80 1039.93 1270.63 1589.04 1851.31 2154.92 Table 2 a. Use exponential regression to fit a model to these data. b. If spending continues at this rate, what will the graduate’s credit card debt be one year after graduating? Q & A… Is it reasonable to assume that an exponential regression model will represent a situation indefinitely? No. Remember that models are formed by real-world data gathered for regression. It is usually reasonable to make estimates within the interval of original observation (interpolation). However, when a model is used to make predictions, it is important to use reasoning skills to determine whether the model makes sense for inputs far beyond the original observation interval (extrapolation). Building a Logarithmic Model from Data Just as with exponential functions, there are many real-world applications for logarithmic functions: intensity of sound, pH levels of solutions, yields of chemical reactions, production of goods, and growth of infants. As with exponential models, data modeled by logarithmic functions are either always increasing or always decreasing as time moves forward. Again, it is the way they increase or decrease that helps us determine whether a log
arithmic model is best. Recall that logarithmic functions increase or decrease rapidly at first, but then steadily slow as time moves on. By reflecting on the characteristics we’ve already learned about this function, we can better analyze real world situations that reflect this type of growth or decay. When performing logarithmic regression analysis, we use the form of the logarithmic function most commonly used on graphing utilities, y = a + bln(x). For this function β€’ All input values, x, must be greater than zero. β€’ The point (1, a) is on the graph of the model. β€’ If b > 0, the model is increasing. Growth increases rapidly at first and then steadily slows over time. β€’ If b < 0, the model is decreasing. Decay occurs rapidly at first and then steadily slows over time. logarithmic regression Logarithmic regression is used to model situations where growth or decay accelerates rapidly at first and then slows over time. We use the command β€œLnReg” on a graphing utility to fit a logarithmic function to a set of data points. This returns an equation of the form, y = a + bln(x) Note that: β€’ all input values, x, must be non-negative. β€’ when b > 0, the model is increasing. β€’ when b < 0, the model is decreasing. How To… Given a set of data, perform logarithmic regression using a graphing utility. 1. Use the STAT then EDIT menu to enter given data. a. Clear any existing data from the lists. b. List the input values in the L1 column. c. List the output values in the L2 column. Download the OpenStax text for free at http://cnx.org/content/col11759/latest. 556 CHAPTER 6 EXPONENTIAL AND LOGARITHMIC FUNCTIONS 2. Graph and observe a scatter plot of the data using the STATPLOT feature. a. Use ZOOM [9] to adjust axes to fit the data. b. Verify the data follow a logarithmic pattern. 3. Find the equation that models the data. a. Select β€œLnReg” from the STAT then CALC menu. b. Use the values returned for a and b to record the model, y = a + bln(x). 4. Graph the model in the same window as
the scatterplot to verify it is a good fit for the data. Example 2 Using Logarithmic Regression to Fit a Model to Data Due to advances in medicine and higher standards of living, life expectancy has been increasing in most developed countries since the beginning of the 20th century. Table 3 shows the average life expectancies, in years, of Americans from 1900–2010[25]. Year Life Expectancy (Years) 1900 47.3 1910 50.0 1920 54.1 1930 59.7 1940 62.9 1950 68.2 Year Life Expectancy (Years) 1960 69.7 1970 70.8 1980 73.7 1990 75.4 2000 76.8 2010 78.7 Table 3 a. Let x represent time in decades starting with x = 1 for the year 1900, x = 2 for the year 1910, and so on. Let y represent the corresponding life expectancy. Use logarithmic regression to fit a model to these data. b. Use the model to predict the average American life expectancy for the year 2030. Solution a. Using the STAT then EDIT menu on a graphing utility, list the years using values 1–12 in L1 and the corresponding life expectancy in L2. Then use the STATPLOT feature to verify that the scatterplot follows a logarithmic pattern as shown in Figure 3: y 85 80 75 70 65 60 55 50 45 40 1 2 3 4 5 6 7 8 9 10 11 12 13 x Figure 3 Use the β€œLnReg” command from the STAT then CALC menu to obtain the logarithmic model, y = 42.52722583 + 13.85752327ln(x) Ne xt, graph the model in the same window as the scatterplot to verify it is a good fit as shown in Figure 4: 25 Source: Center for Disease Control and Prevention, 2013 Download the OpenStax text for free at http://cnx.org/content/col11759/latest. SECTION 6.8 FITTING EXPONENTIAL MODELS TO DATA 557 y 85 80 75 70 65 60 55 50 45 40 1 2 3 4 5 6 7 8 9 10 11 12 13 x Figure 4 b. To predict the life expectancy of an American in the year 2030, substitute x = 14 for the in the model and solve for y: y = 42.52722583 + 13.85752327ln(x) Use the regression model found in part ( a). =
42.52722583 + 13.85752327ln(14) Substitute 14 for x. β‰ˆ 79.1 Round to the nearest tenth If life expectancy continues to increase at this pace, the average life expectancy of an American will be 79.1 by the year 2030. Try It #2 Sales of a video game released in the year 2000 took off at first, but then steadily slowed as time moved on. Table 4 shows the number of games sold, in thousands, from the years 2000–2010. Year 2000 2001 2002 2003 2004 2005 2006 2007 2008 2009 2010 Number Sold (Thousands) 142 149 154 155 159 161 163 164 164 166 167 a. Let x represent time in years starting with x = 1 for the year 2000. Let y represent the number of games sold in thousands. Use logarithmic regression to fit a model to these data. b. If games continue to sell at this rate, how many games will sell in 2015? Round to the nearest thousand. Table 4 Download the OpenStax text for free at http://cnx.org/content/col11759/latest. 558 CHAPTER 6 EXPONENTIAL AND LOGARITHMIC FUNCTIONS Building a Logistic Model from Data Like exponential and logarithmic growth, logistic growth increases over time. One of the most notable differences with logistic growth models is that, at a certain point, growth steadily slows and the function approaches an upper bound, or limiting value. Because of this, logistic regression is best for modeling phenomena where there are limits in expansion, such as availability of living space or nutrients. It is worth pointing out that logistic functions actually model resource-limited exponential growth. There are many examples of this type of growth in real-world situations, including population growth and spread of disease, rumors, and even stains in fabric. When performing logistic regression analysis, we use the form most commonly used on graphing utilities: y = c _______ 1 + aeβˆ’b x Recall that: c _____ 1 + a is the initial value of the model. β€’ β€’ when b > 0, the model increases rapidly at first until it reaches its point of maximum growth rate, ( that point, growth steadily slows and the function becomes asymptotic to the upper bound y = c. ln(a) c ). At _ _, 2 b β€’ c is the limiting value, sometimes called the carrying capacity, of the model. logistic regression Logistic regression is
used to model situations where growth accelerates rapidly at first and then steadily slows to an upper limit. We use the command β€œLogistic” on a graphing utility to fit a logistic function to a set of data points. This returns an equation of the form Note that y = c _______ 1 + aeβˆ’b x β€’ The initial value of the model is c _. 1 + a β€’ Output values for the model grow closer and closer to y = c as time increases. How To… Given a set of data, perform logistic regression using a graphing utility. 1. Use the STAT then EDIT menu to enter given data. a. Clear any existing data from the lists. b. List the input values in the L1 column. c. List the output values in the L2 column. 2. Graph and observe a scatter plot of the data using the STATPLOT feature. a. Use ZOOM [9] to adjust axes to fit the data. b. Verify the data follow a logistic pattern. 3. Find the equation that models the data. a. Select β€œLogistic” from the STAT then CALC menu. b. Use the values returned for a, b, and c to record the model, y = c _ 1 + aeβˆ’b x. 4. Graph the model in the same window as the scatterplot to verify it is a good fit for the data. Example 3 Using Logistic Regression to Fit a Model to Data Mobile telephone service has increased rapidly in America since the mid 1990s. Today, almost all residents have cellular service. Table 5 shows the percentage of Americans with cellular service between the years 1995 and 2012[26]. 26 Source: The World Bankn, 2013 Download the OpenStax text for free at http://cnx.org/content/col11759/latest. SECTION 6.8 FITTING EXPONENTIAL MODELS TO DATA 559 Year 1995 1996 1997 1998 1999 2000 2001 2002 2003 Americans with Cellular Service (%) 12.69 16.35 20.29 25.08 30.81 38.75 45.00 49.16 55.15 Year 2004 2005 2006 2007 2008 2009 2010 2011 2012 Americans with Cellular Service (%) 62.852 68.63 76.64 82.47 85.68 89.14 91.86 95.28 98.17 Table 5 a. Let x represent time in years starting with x = 0 for the year 1995. Let y represent the corresponding percentage of
residents with cellular service. Use logistic regression to fit a model to these data. b. Use the model to calculate the percentage of Americans with cell service in the year 2013. Round to the nearest tenth of a percent. c. Discuss the value returned for the upper limit c. What does this tell you about the model? What would the limiting value be if the model were exact? Solution a. Using the STAT then EDIT menu on a graphing utility, list the years using values 0–15 in L1 and the corresponding percentage in L2. Then use the STATPLOT feature to verify that the scatterplot follows a logistic pattern as shown in Figure 5: y 110 100 90 80 70 60 50 40 30 20 10 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 x Figure 5 Use the β€œLogistic” command from the STAT then CALC menu to obtain the logistic model, 105.7379526 ___________________ 1 + 6.88328979eβˆ’0.2595440013x Ne xt, graph the model in the same window as shown in Figure 6 the scatterplot to verify it is a good fit: y = y 110 100 90 80 70 60 50 40 30 20 10 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 x Figure 6 Download the OpenStax text for free at http://cnx.org/content/col11759/latest. 560 CHAPTER 6 EXPONENTIAL AND LOGARITHMIC FUNCTIONS b. To approximate the percentage of Americans with cellular service in the year 2013, substitute x = 18 for the in the model and solve for y: y = = 105.7379526 ____________________ 1 + 6.88328979e βˆ’0.2595440013x 105.7379526 _____________________ 1 + 6.88328979e βˆ’0.2595440013(18) Substitute 18 for x. Use the regression model found in part ( a). β‰ˆ 99.3% Round to the nearest tenth According to the model, about 99.3% of Americans had cellular service in 2013. c. The model gives a limiting value of about 105. This means that the maximum possible percentage of Americans with cellular service would be 105%, which is impossible. (How could over 100% of a population have cellular service?) If the model were exact, the limiting value would
be c = 100 and the model’s outputs would get very close to, but never actually reach 100%. After all, there will always be someone out there without cellular service! Try It #3 Table 6 shows the population, in thousands, of harbor seals in the Wadden Sea over the years 1997 to 2012. Year 1997 1998 1999 2000 2001 2002 2003 2004 Seal Population (Thousands) 3.493 5.282 6.357 9.201 11.224 12.964 16.226 18.137 Year 2005 2006 2007 2008 2009 2010 2011 2012 Seal Population (Thousands) 19.590 21.955 22.862 23.869 24.243 24.344 24.919 25.108 Table 6 a. Let x represent time in years starting with x = 0 for the year 1997. Let y represent the number of seals in thousands. Use logistic regression to fit a model to these data. b. Use the model to predict the seal population for the year 2020. c. To the nearest whole number, what is the limiting value of this model? Access this online resource for additional instruction and practice with exponential function models. β€’ Exponential Regression on a Calculator (http://openstaxcollege.org/l/pregresscalc) Download the OpenStax text for free at http://cnx.org/content/col11759/latest. SECTION 6.8 SECTION EXERCISES 561 6.8 SECTION EXERCISES VERBAL 1. What situations are best modeled by a logistic equation? Give an example, and state a case for why the example is a good fit. 2. What is a carrying capacity? What kind of model has a carrying capacity built into its formula? Why does this make sense? 3. What is regression analysis? Describe the process of performing regression analysis on a graphing utility. 4. What might a scatterplot of data points look like if it were best described by a logarithmic model? 5. What does the y-intercept on the graph of a logistic equation correspond to for a population modeled by that equation? GRAPHICAL For the following exercises, match the given function of best fit with the appropriate scatterplot in Figure 7 through Figure 11. Answer using the letter beneath the matching graph. y y 16 14 12 10 8 6 4 2 16 14 12 10 8 6 4 2 y 16 14 12 10 8 6 4 2 y 16 14 12 10 a) 6 7 8 9 10
1 2 3 4 x 6 7 8 9 10 1 2 3 4 x 5 (b) 6 7 8 9 10 x 5 (c) Figure 7 Figure 8 Figure 9 y 16 14 12 10 d) Figure 10 x 7 8 9 10 1 2 3 4 x 6 7 8 9 10 5 (e) Figure 11 6. y = 10.209eβˆ’0.294x 7. y = 5.598 βˆ’ 1.912ln(x) 8. y = 2.104(1.479)x 9. y = 4.607 + 2.733ln(x) 10. y = 14.005 __________ 1 + 2.79eβˆ’0.812x Download the OpenStax text for free at http://cnx.org/content/col11759/latest. 562 CHAPTER 6 EXPONENTIAL AND LOGARITHMIC FUNCTIONS NUMERIC 11. To the nearest whole number, what is the initial value of a population modeled by the logistic equation P(t) = 175 ____________ 1 + 6.995eβˆ’0.68t? What is the carrying capacity? 12. Rewrite the exponential model A(t) = 1550(1.085)x as an equivalent model with base e. Express the exponent to four significant digits. 13. A logarithmic model is given by the equation h(p) = 67.682 βˆ’ 5.792ln(p). To the nearest hundredth, for what value of p does h(p) = 62? 15. What is the y-intercept on the graph of the logistic model given in the previous exercise? TECHNOLOGY 14. A logistic model is given by the equation 90 P(t) = ________ 1 + 5eβˆ’0.42t. To the nearest hundredth, for what value of t does P(t) = 45? For the following exercises, use this scenario: The population P of a koi pond over x months is modeled by the function P(x) = 68 __ 1 + 16eβˆ’0.28x. 16. Graph the population model to show the population 17. What was the initial population of koi? over a span of 3 years. 18. How many koi will the pond have after one and a 19. How many months will it take before there are 20 half years? koi in the pond? 20. Use the intersect feature to approximate the number
of months it will take before the population of the pond reaches half its carrying capacity. For the following exercises, use this scenario: The population P of an endangered species habitat for wolves 558 __ is modeled by the function P(x) = 1 + 54.8eβˆ’0.462x, where x is given in years. 21. Graph the population model to show the population 22. What was the initial population of wolves over a span of 10 years. transported to the habitat? 23. How many wolves will the habitat have after 3 years? 24. How many years will it take before there are 100 wolves in the habitat? 25. Use the intersect feature to approximate the number of years it will take before the population of the habitat reaches half its carrying capacity. For the following exercises, refer to Table 7. x f (x) 1 1125 2 1495 3 2310 Table 7 4 3294 5 4650 6 6361 26. Use a graphing calculator to create a scatter diagram 27. Use the regression feature to find an exponential of the data. function that best fits the data in the table. 28. Write the exponential function as an exponential 29. Graph the exponential equation on the scatter equation with base e. diagram. Download the OpenStax text for free at http://cnx.org/content/col11759/latest. SECTION 6.8 SECTION EXERCISES 563 30. Use the intersect feature to find the value of x for which f (x) = 4000. For the following exercises, refer to Table 8. x f (x) 1 555 2 383 3 307 Table 8 4 210 5 158 6 122 31. Use a graphing calculator to create a scatter diagram 32. Use the regression feature to find an exponential of the data. function that best fits the data in the table. 33. Write the exponential function as an exponential 34. Graph the exponential equation on the scatter equation with base e. diagram. 35. Use the intersect feature to find the value of x for which f (x) = 250. For the following exercises, refer to Table 9. x f (x) 1 5.1 2 6.3 3 7.3 Table 9 4 7.7 5 8.1 6 8.6 36. Use a graphing calculator to create a scatter diagram 37. Use the LOGarithm option of the REGression of the data. feature to find a logarithmic function of the form y = a + bln(x) that
best fits the data in the table. 38. Use the logarithmic function to find the value of the 39. Graph the logarithmic equation on the scatter function when x = 10. diagram. 40. Use the intersect feature to find the value of x for which f (x) = 7. For the following exercises, refer to Table 10. x f (x) 1 7.5 2 6 3 5.2 4 4.3 5 3.9 6 3.4 7 3.1 8 2.9 Table 10 41. Use a graphing calculator to create a scatter diagram 42. Use the LOGarithm option of the REGression of the data. feature to find a logarithmic function of the form y = a + bln(x) that best fits the data in the table. 43. Use the logarithmic function to find the value of the 44. Graph the logarithmic equation on the scatter function when x = 10. diagram. 45. Use the intersect feature to find the value of x for which f (x) = 8. For the following exercises, refer to Table 11. x f (x) 1 8.7 2 12.3 3 15.4 4 18.5 5 20.7 6 22.5 7 23.3 8 24 9 24.6 10 24.8 Table 11 Download the OpenStax text for free at http://cnx.org/content/col11759/latest. 564 CHAPTER 6 EXPONENTIAL AND LOGARITHMIC FUNCTIONS 46. Use a graphing calculator to create a scatter diagram of the data. 47. Use the LOGISTIC regression option to find a logistic growth model of the form y = best fits the data in the table. c _ 1 + aeβˆ’b x that 48. Graph the logistic equation on the scatter diagram. 49. To the nearest whole number, what is the predicted carrying capacity of the model? 50. Use the intersect feature to find the value of x for which the model reaches half its carrying capacity. For the following exercises, refer to Table 12. x f (x) 0 12 2 28.6 4 52.8 5 70.3 7 8 10 11 15 17 99.9 112.5 125.8 127.9 135.1 135.9 Table 12 51. Use a graphing calculator to create a scatter diagram of the data. 52. Use the LOGISTIC regression option to find
a logistic growth model of the form y = best fits the data in the table. c ________ 1 + aeβˆ’b x that 53. Graph the logistic equation on the scatter diagram. 54. To the nearest whole number, what is the predicted carrying capacity of the model? 55. Use the intersect feature to find the value of x for which the model reaches half its carrying capacity. EXTENSIONS 56. Recall that the general form of a logistic equation for a population is given by P(t) = c _ 1 + aeβˆ’bt, such that the initial population at time t = 0 is P(0) = P0. Show algebraically that c βˆ’ P0 _ c βˆ’ P(t) _ P0 P(t) eβˆ’bt. = 57. Use a graphing utility to find an exponential regression formula f (x) and a logarithmic regression formula g(x) for the points (1.5, 1.5) and (8.5, 8.5). Round all numbers to 6 decimal places. Graph the points and both formulas along with the line y = x on the same axis. Make a conjecture about the relationship of the regression formulas. 58. Verify the conjecture made in the previous exercise. Round all numbers to six decimal places when necessary. 59. Find the inverse function f βˆ’1 (x) for the logistic c _ 1 + aeβˆ’b x. Show all steps. function f (x) = the logistic model P(t) = 60. Use the result from the previous exercise to graph ________ 1 + 4eβˆ’0.5t along with its inverse on the same axis. What are the intercepts and asymptotes of each function? 20 Download the OpenStax text for free at http://cnx.org/content/col11759/latest. CHAPTER 6 REVIEW 565 CHAPTER 6 REVIEW Key Terms annual percentage rate (APR) the yearly interest rate earned by an investment account, also called nominal rate carrying capacity in a logistic model, the limiting value of the output change-of-base formula a formula for converting a logarithm with any base to a quotient of logarithms with any other base. common logarithm the exponent to which 10 must be raised to get x; log10(x) is written simply as log(x). compound interest interest earned on the total balance, not just the
principal doubling time the time it takes for a quantity to double exponential growth a model that grows by a rate proportional to the amount present extraneous solution a solution introduced while solving an equation that does not satisfy the conditions of the original equation half-life the length of time it takes for a substance to exponentially decay to half of its original quantity logarithm the exponent to which b must be raised to get x; written y = logb(x) c _ logistic growth model a function of the form f (x) = 1 + a c ________ 1+ aeβˆ’b x where limiting value, and b is a constant determined by the rate of growth is the initial value, c is the carrying capacity, or natural logarithm the exponent to which the number e must be raised to get x; loge(x) is written as ln(x). Newton’s Law of Cooling the scientific formula for temperature as a function of time as an object’s temperature is equalized with the ambient temperature nominal rate the yearly interest rate earned by an investment account, also called annual percentage rate order of magnitude the power of ten, when a number is expressed in scientific notation, with one non-zero digit to the left of the decimal power rule for logarithms a rule of logarithms that states that the log of a power is equal to the product of the exponent and the log of its base product rule for logarithms a rule of logarithms that states that the log of a product is equal to a sum of logarithms quotient rule for logarithms a rule of logarithms that states that the log of a quotient is equal to a difference of logarithms Key Equations definition of the exponential function f (x) = b x, where b > 0, b β‰  1 definition of exponential growth compound interest formula continuous growth formula General Form for the Translation of the Parent Function f (x) = b x f (x) = ab x, where a > 0, b > 0, b β‰  1 r ) kt, where _ A = a ( 1 + k A(t) is the account value at time t t is the number of years P is the initial investment, often called the principal r is the annual percentage rate (APR), or nominal rate n is the number of compounding periods in one year A(t) = ae rt, where t is the number
of unit time periods of growth a is the starting amount (in the continuous compounding formula a is replaced with P, the principal) e is the mathematical constant, e β‰ˆ 2.718282 f (x) = ab x + c + d Definition of the logarithmic function Definition of the common logarithm For x > 0, b > 0, b β‰  1, y = logb(x) if and only if b y = x. For x > 0, y = log(x) if and only if 10y = x. Download the OpenStax text for free at http://cnx.org/content/col11759/latest. 566 CHAPTER 6 EXPONENTIAL AND LOGARITHMIC FUNCTIONS Definition of the natural logarithm General Form for the Translation of the Parent Logarithmic Function f (x) = logb(x) The Product Rule for Logarithms The Quotient Rule for Logarithms The Power Rule for Logarithms The Change-of-Base Formula For x > 0, y = ln(x) if and only if ey = x. f (x) = alogb(x + c) + d logb(MN) = logb(M) + logb(N) M ) = logbM βˆ’ logbN logb ( __ N logb(Mn) = nlogbM logbM = lognM _ lognb n > 0, n β‰  1, b β‰  1 One-to-one property for exponential functions Definition of a logarithm One-to-one property for logarithmic functions For any algebraic expressions S and T and any positive real number b, where bS = bT if and only if S = T. For any algebraic expression S and positive real numbers b and c, where b β‰  1, logb(S) = c if and only if bc = S. For any algebraic expressions S and T and any positive real number b, where b β‰  1, logbS = logbT if and only if S = T. If A = A0e kt, k < 0, the half-life is t = βˆ’. t = ln(2) _ k A ln ( ) _ A0 _ βˆ’0.000121 is the amount of carbon-14 when the plant or animal died,
A0 A is the amount of carbon-14 remaining today, t is the age of the fossil in years If A = A0e kt, k > 0, the doubling time is t = T(t) = Ae kt + Ts, where Ts is the ambient temperature, A = T(0) βˆ’ Ts, and k is the continuous rate of cooling. ln(2) ___ k Half-life formula Carbon-14 dating Doubling time formula Newton’s Law of Cooling Key Concepts 6.1 Exponential Functions β€’ An exponential function is defined as a function with a positive constant other than 1 raised to a variable exponent. See Example 1. β€’ A function is evaluated by solving at a specific value. See Example 2 and Example 3. β€’ An exponential model can be found when the growth rate and initial value are known. See Example 4. β€’ An exponential model can be found when the two data points from the model are known. See Example 5. β€’ An exponential model can be found using two data points from the graph of the model. See Example 6. β€’ An exponential model can be found using two data points from the graph and a calculator. See Example 7. β€’ The value of an account at any time t can be calculated using the compound interest formula when the principal, annual interest rate, and compounding periods are known. See Example 8. β€’ The initial investment of an account can be found using the compound interest formula when the value of the account, annual interest rate, compounding periods, and life span of the account are known. See Example 9. β€’ The number e is a mathematical constant often used as the base of real world exponential growth and decay models. Its decimal approximation is e β‰ˆ 2.718282. β€’ Scientific and graphing calculators have the key [e x] or [exp(x)] for calculating powers of e. See Example 10. β€’ Continuous growth or decay models are exponential models that use e as the base. Continuous growth and decay models can be found when the initial value and growth or decay rate are known. See Example 11 and Example 12. Download the OpenStax text for free at http://cnx.org/content/col11759/latest. CHAPTER 6 REVIEW 567 6.2 Graphs of Exponential Functions β€’ The graph of the function f (x) = b x has a y-intercept at (0, 1), domain (βˆ’βˆž, ∞), range (0, ∞),
and horizontal asymptote y = 0. See Example 1. β€’ If b > 1, the function is increasing. The left tail of the graph will approach the asymptote y = 0, and the right tail will increase without bound. β€’ If 0 < b < 1, the function is decreasing. The left tail of the graph will increase without bound, and the right tail will approach the asymptote y = 0. β€’ The equation f (x) = b x + d represents a vertical shift of the parent function f (x) = b x. β€’ The equation f (x) = b x + c represents a horizontal shift of the parent function f (x) = b x. See Example 2. β€’ Approximate solutions of the equation f (x) = b x + c + d can be found using a graphing calculator. See Example 3. β€’ The equation f (x) = ab x, where a > 0, represents a vertical stretch if ∣ a ∣ > 1 or compression if 0 < ∣ a ∣ < 1 of the parent function f (x) = b x. See Example 4. β€’ When the parent function f (x) = b x is multiplied by βˆ’1, the result, f (x) = βˆ’b x, is a reflection about the x-axis. When the input is multiplied by βˆ’1, the result, f (x) = bβˆ’x, is a reflection about the y-axis. See Example 5. β€’ All translations of the exponential function can be summarized by the general equation f (x) = ab x + c + d. See Table 3. β€’ Using the general equation f (x) = ab x + c + d, we can write the equation of a function given its description. See Example 6. 6.3 Logarithmic Functions β€’ The inverse of an exponential function is a logarithmic function, and the inverse of a logarithmic function is an exponential function. β€’ Logarithmic equations can be written in an equivalent exponential form, using the definition of a logarithm. See Example 1. β€’ Exponential equations can be written in their equivalent logarithmic form using the definition of a logarithm See Example 2. β€’ Logarithmic functions with base b can be evaluated mentally using previous knowledge of powers of b. See Example 3 and Example 4. β€’ Common logarithms can be evaluated mentally using previous knowledge of
powers of 10. See Example 5. β€’ When common logarithms cannot be evaluated mentally, a calculator can be used. See Example 6. β€’ Real-world exponential problems with base 10 can be rewritten as a common logarithm and then evaluated using a calculator. See Example 7. β€’ Natural logarithms can be evaluated using a calculator Example 8. 6.4 Graphs of Logarithmic Functions β€’ To find the domain of a logarithmic function, set up an inequality showing the argument greater than zero, and solve for x. See Example 1 and Example 2. β€’ The graph of the parent function f (x) = logb(x) has an x-intercept at (1, 0), domain (0, ∞), range (βˆ’βˆž, ∞), vertical asymptote x = 0, and β€’ if b > 1, the function is increasing. β€’ if 0 < b < 1, the function is decreasing. See Example 3. β€’ The equation f (x) = logb(x + c) shifts the parent function y = logb(x) horizontally β€’ left c units if c > 0. β€’ right c units if c < 0. See Example 4. β€’ The equation f (x) = logb(x) + d shifts the parent function y = logb(x) vertically β€’ up d units if d > 0. β€’ down d units if d < 0. See Example 5. Download the OpenStax text for free at http://cnx.org/content/col11759/latest. 568 CHAPTER 6 EXPONENTIAL AND LOGARITHMIC FUNCTIONS β€’ For any constant a > 0, the equation f (x) = alogb(x) β€’ stretches the parent function y = logb(x) vertically by a factor of a if ∣ a ∣ > 1. β€’ compresses the parent function y = logb(x) vertically by a factor of a if ∣ a ∣ < 1. See Example 6 and Example 7. β€’ When the parent function y = logb(x) is multiplied by βˆ’1, the result is a reflection about the x-axis. When the input is multiplied by βˆ’1, the result is a reflection about the y-axis. β€’ The equation f (x) = βˆ’logb(x) represents a reflection of the parent function about the x-axis. β€’ The equation f (x) = log
b(βˆ’x) represents a reflection of the parent function about the y-axis. See Example 8. β€’ A graphing calculator may be used to approximate solutions to some logarithmic equations See Example 9. β€’ All translations of the logarithmic function can be summarized by the general equation f (x) = alogb(x + c) + d. See Table 4. β€’ Given an equation with the general form f (x) = alogb(x + c) + d, we can identify the vertical asymptote x = βˆ’c for the transformation. See Example 10. β€’ Using the general equation f (x) = alogb(x + c) + d, we can write the equation of a logarithmic function given its graph. See Example 11. 6.5 Logarithmic Properties β€’ We can use the product rule of logarithms to rewrite the log of a product as a sum of logarithms. See Example 1. β€’ We can use the quotient rule of logarithms to rewrite the log of a quotient as a difference of logarithms. See Example 2. β€’ We can use the power rule for logarithms to rewrite the log of a power as the product of the exponent and the log of its base. See Example 3, Example 4, and Example 5. β€’ We can use the product rule, the quotient rule, and the power rule together to combine or expand a logarithm with a complex input. See Example 6, Example 7, and Example 8. β€’ The rules of logarithms can also be used to condense sums, differences, and products with the same base as a single logarithm. See Example 9, Example 10, Example 11, and Example 12. β€’ We can convert a logarithm with any base to a quotient of logarithms with any other base using the change-of- base formula. See Example 13. β€’ The change-of-base formula is often used to rewrite a logarithm with a base other than 10 and e as the quotient of natural or common logs. That way a calculator can be used to evaluate. See Example 14. 6.6 Exponential and Logarithmic Equations β€’ We can solve many exponential equations by using the rules of exponents to rewrite each side as a power with the same base. Then we use the fact that exponential functions are one-
to-one to set the exponents equal to one another and solve for the unknown. β€’ When we are given an exponential equation where the bases are explicitly shown as being equal, set the exponents equal to one another and solve for the unknown. See Example 1. β€’ When we are given an exponential equation where the bases are not explicitly shown as being equal, rewrite each side of the equation as powers of the same base, then set the exponents equal to one another and solve for the unknown. See Example 2, Example 3, and Example 4. β€’ When an exponential equation cannot be rewritten with a common base, solve by taking the logarithm of each side. See Example 5. β€’ We can solve exponential equations with base e, by applying the natural logarithm of both sides because exponential and logarithmic functions are inverses of each other. See Example 6 and Example 7. β€’ After solving an exponential equation, check each solution in the original equation to find and eliminate any extraneous solutions. See Example 8. Download the OpenStax text for free at http://cnx.org/content/col11759/latest. CHAPTER 6 REVIEW 569 β€’ When given an equation of the form logb(S) = c, where S is an algebraic expression, we can use the definition of a logarithm to rewrite the equation as the equivalent exponential equation bc = S, and solve for the unknown. See Example 9 and Example 10. β€’ We can also use graphing to solve equations with the form logb(S) = c. We graph both equations y = logb(S) and y = c on the same coordinate plane and identify the solution as the x-value of the intersecting point. See Example 11. β€’ When given an equation of the form logbS = logbT, where S and T are algebraic expressions, we can use the one- to-one property of logarithms to solve the equation S = T for the unknown. See Example 12. β€’ Combining the skills learned in this and previous sections, we can solve equations that model real world situations, whether the unknown is in an exponent or in the argument of a logarithm. See Example 13. 6.7 Exponential and Logarithmic Models β€’ The basic exponential function is f (x) = ab x. If b > 1, we have exponential growth; if 0 < b < 1, we have exponential decay. β€’
We can also write this formula in terms of continuous growth as A = A0e kx, where A0 is the starting value. If A0 is positive, then we have exponential growth when k > 0 and exponential decay when k < 0. See Example 1. β€’ In general, we solve problems involving exponential growth or decay in two steps. First, we set up a model and use the model to find the parameters. Then we use the formula with these parameters to predict growth and decay. See Example 2. β€’ We can find the age, t, of an organic artifact by measuring the amount, k, of carbon-14 remaining in the artifact and using the formula t = to solve for t. See Example 3. ln(k) _ βˆ’0.000121 β€’ Given a substance’s doubling time or half-life we can find a function that represents its exponential growth or decay. See Example 4. β€’ We can use Newton’s Law of Cooling to find how long it will take for a cooling object to reach a desired temperature, or to find what temperature an object will be after a given time. See Example 5. β€’ We can use logistic growth functions to model real-world situations where the rate of growth changes over time, such as population growth, spread of disease, and spread of rumors. See Example 6. β€’ We can use real-world data gathered over time to observe trends. Knowledge of linear, exponential, logarithmic, and logistic graphs help us to develop models that best fit our data. See Example 7. β€’ Any exponential function with the form y = ab x can be rewritten as an equivalent exponential function with the form y = A0e kx where k = lnb. See Example 8. 6.8 Fitting Exponential Models to Data β€’ Exponential regression is used to model situations where growth begins slowly and then accelerates rapidly without bound, or where decay begins rapidly and then slows down to get closer and closer to zero. β€’ We use the command β€œExpReg” on a graphing utility to fit function of the form y = ab x to a set of data points. See Example 1. β€’ Logarithmic regression is used to model situations where growth or decay accelerates rapidly at first and then slows over time. β€’ We use the command β€œLnReg” on a graphing utility to fit a function of the form y = a + bln(x) to a set of data points
. See Example 2. β€’ Logistic regression is used to model situations where growth accelerates rapidly at first and then steadily slows as the function approaches an upper limit. β€’ We use the command β€œLogistic” on a graphing utility to fit a function of the form y = points. See Example 3. c _________ 1 + aeβˆ’b x to a set of data Download the OpenStax text for free at http://cnx.org/content/col11759/latest. 570 CHAPTER 6 EXPONENTIAL AND LOGARITHMIC FUNCTIONS CHAPTER 6 REVIEW EXERCISES EXPONENTIAL FUNCTIONS 1. Determine whether the function y = 156(0.825)t represents exponential growth, exponential decay, or neither. Explain 2. The population of a herd of deer is represented by the function A(t) = 205(1.13)t, where t is given in years. To the nearest whole number, what will the herd population be after 6 years? 3. Find an exponential equation that passes through the points (2, 2.25) and (5, 60.75). 4. Determine whether Table 1 could represent a function that is linear, exponential, or neither. If it appears to be exponential, find a function that passes through the points. x f (x) 1 3 2 0.9 Table 1 3 0.27 4 0.081 5. A retirement account is opened with an initial deposit of $8,500 and earns 8.12% interest compounded monthly. What will the account be worth in 20 years? 7. Does the equation y = 2.294eβˆ’0.654t represent continuous growth, continuous decay, or neither? Explain. 6. Hsu-Mei wants to save $5,000 for a down payment on a car. To the nearest dollar, how much will she need to invest in an account now with 7.5% APR, compounded daily, in order to reach her goal in 3 years? 8. Suppose an investment account is opened with an initial deposit of $10,500 earning 6.25% interest, compounded continuously. How much will the account be worth after 25 years? GRAPHS OF EXPONENTIAL FUNCTIONS 9. Graph the function f (x) = 3.5(2)x. State the domain and range and give the y-intercept. 1 ) 10. Graph the function f (x) =
4 ( __ 8 x and its reflection about the y-axis on the same axes, and give the y-intercept. 11. The graph of f (x) = 6.5x is reflected about the y-axis and stretched vertically by a factor of 7. What is the equation of the new function, g (x)? State its y-intercept, domain, and range. 12. The graph below shows transformations of the graph of f (x) = 2x. What is the equation for the transformation1–1 –2 –3 –6 –5 –4 –3 –2 21 3 4 5 6 x LOGARITHMIC FUNCTIONS 13. Rewrite log17(4913) = x as an equivalent exponential 14. Rewrite ln(s) = t as an equivalent exponential equation. equation. Figure 1 βˆ’ 2 __ = b as an equivalent logarithmic 5 15. Rewrite a equation. 1 ) to exponential form. 17. Solve for xlog64(x) = ( _ 3 19. Evaluate log(0.000001) without using a calculator. 16. Rewrite eβˆ’3.5 = h as an equivalent logarithmic equation. 18. Evaluate log5 ( 20. Evaluate log(4.005) using a calculator. Round to the ) without using a calculator. 1 _ 125 nearest thousandth. Download the OpenStax text for free at http://cnx.org/content/col11759/latest. CHAPTER 6 REVIEW 571 21. Evaluate ln(eβˆ’0.8648) without using a calculator. 22. Evaluate ln ( 3 √ β€” 18 ) using a calculator. Round to the GRAPHS OF LOGARITHMIC FUNCTIONS 23. Graph the function g(x) = log(7x + 21) βˆ’ 4. 25. State the domain, vertical asymptote, and end behavior of the function g (x) = ln(4x + 20) βˆ’ 17. LOGARITHMIC PROPERTIES nearest thousandth. 24. Graph the function h(x) = 2ln(9 βˆ’ 3x) + 1. 26. Rewrite ln(7r Β· 11st) in expanded form. 27. Rewrite log8(x) + log8(5) + log8(y) + log8(13) in ) in expanded form. 67 28.
Rewrite logm ( ___ 83 1 30. Rewrite ln ( x5 ) as a product. __ 32. Use properties of logarithms to expand log ( r 2s11 t14 ). _ compact form. 29. Rewrite ln(z) – ln(x) – ln(y) in compact form. 1 ) as a single logarithm. 31. Rewrite βˆ’logy ( __ 12 33. Use properties of logarithms to expand ln ( 2b √ ______ b + 1 ). _____ b βˆ’ 1 to a single logarithm. 37. Rewrite 512x βˆ’ 17 = 125 as a logarithm. Then apply the change of base formula to solve for x using the common log. Round to the nearest thousandth. 125 __ βˆ’x βˆ’ 3 = 53 by rewriting each side with a 39. Solve 1 ( ) _ 625 common base. 34. Condense the expression 5ln(b) + ln(c) + 35. Condense the expression 3log7v + 6log7w βˆ’ ln(4 βˆ’ a) _ 2 log 7 u _ 3 to a single logarithm. 36. Rewrite log3(12.75) to base e. EXPONENTIAL AND LOGARITHMIC EQUATIONS 38. Solve 2163x Β· 216x = 363x + 2 by rewriting each side with a common base. 40. Use logarithms to find the exact solution for 41. Use logarithms to find the exact solution for 7 Β· 17βˆ’9x βˆ’ 7 = 49. If there is no solution, write no solution. 3e6n βˆ’ 2 + 1 = βˆ’60. If there is no solution, write no solution. 42. Find the exact solution for 5e3x βˆ’ 4 = 6. If there is 43. Find the exact solution for 2e5x βˆ’ 2 βˆ’ 9 = βˆ’56. no solution, write no solution. If there is no solution, write no solution. 44. Find the exact solution for 52x βˆ’ 3 = 7x + 1. If there is 45. Find the exact solution for e 2x βˆ’ e x βˆ’ 110 = 0. If no solution, write no solution. there is no solution, write no solution. 46. Use the definition of a logarithm to solve. 47. Use the definition of a logarithm to find the exact βˆ’
5log7(10n) = 5. solution for 9 + 6ln(a + 3) = 33. 48. Use the one-to-one property of logarithms to find an exact solution for log8(7) + log8(βˆ’4x) = log8(5). If there is no solution, write no solution. 49. Use the one-to-one property of logarithms to find an exact solution for ln(5) + ln(5x2 βˆ’ 5) = ln(56). If there is no solution, write no solution. 50. The formula for measuring sound intensity in decibels D is defined by the equation D = 10log I _ ), where I is the intensity of the sound in watts ( I0 per square meter and I0 = 10βˆ’12 is the lowest level of sound that the average person can hear. How many decibels are emitted from a large orchestra with a sound intensity of 6.3 Β· 10βˆ’3 watts per square meter? 52. Find the inverse function f βˆ’1 for the exponential function f (x. 51. The population of a city is modeled by the equation P(t) = 256, 114e0.25t where t is measured in years. If the city continues to grow at this rate, how many years will it take for the population to reach one million? 53. Find the inverse function f βˆ’1 for the logarithmic function f (x) = 0.25 Β· log2(x3 + 1). Download the OpenStax text for free at http://cnx.org/content/col11759/latest. 572 CHAPTER 6 EXPONENTIAL AND LOGARITHMIC FUNCTIONS EXPONENTIAL AND LOGARITHMIC MODELS For the following exercises, use this scenario: A doctor prescribes 300 milligrams of a therapeutic drug that decays by about 17% each hour. 54. To the nearest minute, what is the half-life of the drug? 55. Write an exponential model representing the amount of the drug remaining in the patient’s system after t hours. Then use the formula to find the amount of the drug that would remain in the patient’s system after 24 hours. Round to the nearest hundredth of a gram. For the following exercises, use this scenario: A soup with an internal temperature of 350Β° Fahrenheit was taken off the stove to cool in a 71Β°
F room. After fifteen minutes, the internal temperature of the soup was 175Β°F. 56. Use Newton’s Law of Cooling to write a formula that 57. How many minutes will it take the soup to cool models this situation. to 85Β°F? For the following exercises, use this scenario: The equation N(t) = school who have heard a rumor after t days. 1200 __ 1 + 199eβˆ’0.625t models the number of people in a 58. How many people started the rumor? 59. To the nearest tenth, how many days will it be before the rumor spreads to half the carrying capacity? 60. What is the carrying capacity? For the following exercises, enter the data from each table into a graphing calculator and graph the resulting scatter plots. Determine whether the data from the table would likely represent a function that is linear, exponential, or logarithmic. 61. 62. x f (x) x f (x) 1 3.05 0.5 18.05 2 4.42 1 17 3 6.4 4 9.28 5 13.46 6 19.52 7 28.3 8 41.04 9 59.5 10 86.28 3 15.33 5 14.55 7 14.04 10 13.5 12 13.22 13 13.1 15 12.88 17 12.69 20 12.45 63. Find a formula for an exponential equation that goes through the points (βˆ’2, 100) and (0, 4). Then express the formula as an equivalent equation with base e. FITTING EXPONENTIAL MODELS TO DATA 64. What is the carrying capacity for a population modeled by the logistic equation P(t) = initial population for the model? 65. The population of a culture of bacteria is modeled by the logistic equation P(t) = 250, 000 ___________ 1 + 499eβˆ’0.45t? What is the 14, 250 __ 1 + 29eβˆ’0.62t, where t is in days. To the nearest tenth, how many days will it take the culture to reach 75% of its carrying capacity? For the following exercises, use a graphing utility to create a scatter diagram of the data given in the table. Observe the shape of the scatter diagram to determine whether the data is best described by an exponential, logarithmic, or logistic model. Then use the appropriate regression feature to find an equation that models the data. When
necessary, round values to five decimal places. 66. 67. 68. x f (x) x f (x) x f (x) 1 409.4 2 260.7 3 170.4 4 110.6 5 74 6 44.7 7 32.4 8 19.5 9 12.7 0.15 36.21 0.25 28.88 0.5 24.39 0.75 18.28 1 16.5 1.5 12.99 2 9.91 2.25 8.57 2.75 7.23 3 5.99 0 9 2 22.6 4 44.2 5 62.1 7 96.9 8 113.4 10 133.4 11 137.6 15 148.4 10 8.1 3.5 4.81 17 149.3 Download the OpenStax text for free at http://cnx.org/content/col11759/latest. CHAPTER 6 PRACTICE TEST 573 CHAPTER 6 PRACTICE TEST 1. The population of a pod of bottlenose dolphins is 2. Find an exponential equation that passes through modeled by the function A(t) = 8(1.17)t, where t is given in years. To the nearest whole number, what will the pod population be after 3 years? the points (0, 4) and (2, 9). 3. Drew wants to save $2,500 to go to the next 4. An investment account was opened with an World Cup. To the nearest dollar, how much will he need to invest in an account now with 6.25% APR, compounding daily, in order to reach his goal in 4 years? 5. Graph the function f (x) = 5(0.5)βˆ’x and its reflection across the y-axis on the same axes, and give the y-intercept. initial deposit of $9,600 and earns 7.4% interest, compounded continuously. How much will the account be worth after 15 years? 6. The graph below shows transformations of the 1 ) graph of f (x) = ( __ 2 transformation? x. What is the equation for the –4 –3 –2 y 4 3 2 1 0 –1–1 –2 –3 –4 –5 –6 –7 –8 21 3 4 5 6 7 8 x 7. Rewrite log8.5(614.125) = a as an equivalent exponential equation. 1 __ 8. Rewrite e =