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Verify that each function in Exercises 64 - 69 is the inverse of the corresponding function in Exercises 58 - 63. (Match up #58 and #64, and so on.) In Exercises 71 - 74, find the inverse of the function from the ‘procedural perspective’ discussed in Example 6.1.5 and graph the function and its inverse on the same set ... |
the decibel scale can be used in many disciplines,13 we shall restrict our attention to its use in acoustics, specifically its use in measuring the intensity level of sound.14 The Sound Intensity Level L (measured in decibels) of a sound intensity I (measured in watts per square meter) is given by L(I) = 10 log I 10−12... |
to Exponential and Logarithmic Functions 433 6.1.2 Answers 1. log2(8) = 3 4. log 1 3 (9) = −2 7. ln(1) = 0 10. 3−4 = 1 81 13. 10−1 = 0.1 2. log5 1 125 = −3 5. log 4 25 5 2 = − 1 2 8. 52 = 25 11. 4 3 −1 = 3 4 14. e1 = e 3. log4(32) = 5 2 6. log(0.001) = −3 9. (25) 1 2 = 5 12. 102 = 100 15. e− 1 2 = 1√ e 16. log3(27) = ... |
5, ∞) 51. (−∞, ∞) 53. (−∞, −7) ∪ (1, ∞) 54. (13, ∞) 55. (0, 125) ∪ (125, ∞) 56. No domain 57. (−∞, −3) ∪ 1 2, 2 434 Exponential and Logarithmic Functions 58. Domain of g: (−∞, ∞) Range of g: (−1, ∞) 59. Domain of g: (−∞, ∞) Range of g: (0, ∞) 3−2−1 x 1 2 3 H.A. y = −1 y = g(x) = 2x − 1 60. Domain of g: (−∞, ∞) Range of... |
�) Range of g: (−∞, ∞) 67. Domain of g: (−20, ∞) Range of g: (−∞, ∞) y 3 2 1 −1 −2 − 10 11 x V.A. x = 2 y = g(x) = − log3(x − 2) y 3 2 1 −10 10 20 30 40 50 60 70 80 90 100 x −2 −3 V.A. x = −20 y = g(x) = 2 log(x + 20) − 1 68. Domain of g: (−∞, 8) Range of g:(−∞, ∞) y 3 2 1 −1 −2 −.A. x = 8 69. Domain of g: (0, ∞) Range... |
x −2 −3 −4 y = f (x) = 5 log(x) − 2 x+2 y = f −1(x) = 10 5 75. 76. 77. (a) M (0.001) = log 0.001 0.001 80,000 0.001 (b) M (80, 000) = log = log(1) = 0. = log(80, 000, 000) ≈ 7.9. (a) L(10−6) = 60 decibels. (b) I = 10−.5 ≈ 0.316 watts per square meter. (c) Since L(1) = 120 decibels and L(100) = 140 decibels, a sound wi... |
for exponential and logarithmic functions to be one-to-one. Theorem 6.4. (One-to-one Properties of Exponential and Logarithmic Functions) Let f (x) = bx and g(x) = logb(x) where b > 0, b = 1. Then f and g are one-to-one and bu = bw if and only if u = w for all real numbers u and w. logb(u) = logb(w) if and only if u =... |
) be a logarithmic function (b > 0, b = 1) and let u > 0 and w > 0 be real numbers. Product Rule: g(uw) = g(u) + g(w). In other words, logb(uw) = logb(u) + logb(w) Quotient Rule: g u w = g(u) − g(w). In other words, logb u w = logb(u) − logb(w) Power Rule: g (uw) = wg(u). In other words, logb (uw) = w logb(u) There are... |
�rst example, we are asked to ‘expand’ the logarithms. This means that we read the properties in Theorem 6.6 from left to right and rewrite products inside the log as sums outside the log, quotients inside the log as differences outside the log, and powers inside the log as factors outside the log.1 Example 6.2.1. Expan... |
Quotient Rule is applicable, with u = 3 and w = ex, so we replace is being multiplied by 2, the entire ln 3 ex quantity ln(3) − ln(ex) is multiplied by 2. Finally, we apply the Product Rule with u = e and w = x, and replace ln(ex) with the quantity ln(e) + ln(x), and simplify, keeping in mind that the natural log is l... |
we have no means of simplifying log117 x2 − 4, since none of the properties of logs addresses the issue of expanding a difference inside the logarithm. However, we may factor x2 − 4 = (x + 2)(x − 2) thereby introducing a product which gives us license to use the Product Rule. log117 x2 − 4 = log117 [(x + 2)(x − 2)] Fac... |
are well aware of the propensity for some students to become overexcited and invent their own properties of logs x2 − log117(4), which simply isn’t true, in general. The unwritten3 like log117 property of logarithms is that if it isn’t written in a textbook, it probably isn’t true. x2 − 4 = log117 Example 6.2.2. Use t... |
2 = log2 = log2 x4 + 3 x4 + log2 x4 + log2(8) 8x4 23 Since 3 = log2 Power Rule 23 Product Rule 3The authors relish the irony involved in writing what follows. 442 Exponential and Logarithmic Functions 4. To get started with − ln(x) − 1 2, we rewrite − ln(x) as (−1) ln(x). We can then use the Power Rule to obtain (−1) l... |
� buttons which correspond to the common and natural logs, respectively. Suppose we wanted an approximation to log2(7). The answer should be a little less than 3, (Can you explain why?) but how do we coerce the calculator into telling us a more accurate answer? We need the following theorem. Theorem 6.7. (Change of Bas... |
the following expressions to ones with the indicated base. Verify your answers using a calculator, as appropriate. 1. 32 to base 10 3. log4(5) to base e Solution. 2. 2x to base e 4. ln(x) to base 10 1. We apply the Change of Base formula with a = 3 and b = 10 to obtain 32 = 102 log(3). Typing the latter in the calcula... |
1 Exercises In Exercises 1 - 15, expand the given logarithm and simplify. Assume when necessary that all quantities represent positive real numbers. 1. ln(x3y2) 4. log(1.23 × 1037) 7. log√ 2 4x3 10. log3 x2 81y4 13. log √ 100x √ 3 10 y 2. log2 128 x2 + 4 √ z xy 5. ln 8. log 1 3 (9x(y3 − 8)) 11. ln 4 xy ez 4 3√ √ y x2 z... |
(x2 + 1) to base e In Exercises 34 - 39, use the appropriate change of base formula to approximate the logarithm. 34. log3(12) 1 10 37. log4 35. log5(80) 38. log 3 5 (1000) 36. log6(72) 39. log 2 3 (50) 40. Compare and contrast the graphs of y = ln(x2) and y = 2 ln(x). 41. Prove the Quotient Rule and Power Rule for Log... |
(y) 7. 3 log√ 2(x) + 4 2. 7 − log2(x2 + 4) 4. log(1.23) + 37 6. log5(x − 5) + log5(x + 5) 8. −2 + log 1 3 (x) + log 1 3 (y − 2) + log 1 3 (y2 + 2y + 4) 9. 3 + 3 log(x) + 5 log(y) 10. 2 log3(x) − 4 − 4 log3(y) 11. 1 4 ln(x) + 1 4 ln(y) − 1 4 − 1 4 ln(z) 12. 12 − 12 log6(x) − 4 log6(y) 13. 5 3 + log(x) + 1 2 log(y) 14. −... |
≈ −9.64824 448 Exponential and Logarithmic Functions 6.3 Exponential Equations and Inequalities In this section we will develop techniques for solving equations involving exponential functions. Suppose, for instance, we wanted to solve the equation 2x = 128. After a moment’s calculation, we find 128 = 27, so we have 2x... |
the exponential function. 2. (a) If convenient, express both sides with a common base and equate the exponents. (b) Otherwise, take the natural log of both sides of the equation and use the Power Rule. Example 6.3.1. Solve the following equations. Check your answer graphically using a calculator. 1. 23x = 161−x 4. 75 ... |
(3) to get t = − ln(2) ln(3). On the calculator, we graph f (x) = 2000 and g(x) = 1000 · 3−0.1x and find that they intersect at x = − 10 ln(2) ln(3) ≈ −6.3093. 0.1 ln(3) = − 10 ln(2) y = f (x) = 23x and y = g(x) = 161−x y = f (x) = 2000 and y = g(x) = 1000 · 3−0.1x 3. We first note that we can rewrite the equation 9·3x =... |
the exponential. To that end, we clear denominators and get 75 1 + 3e−2t = 100. From this we get 75 + 225e−2t = 100, which leads to 225e−2t = 25, and finally, e−2t = 1 9. Taking the natural log of both sides 1+3e−2t 450 Exponential and Logarithmic Functions. Since natural log is log base e, ln e−2t = −2t. We can also u... |
= 3. Since it isn’t convenient to express 3 as a power of 5, we take natural logs and get ln (5x) = ln(3) so that x ln(5) = ln(3) or x = ln(3) ln(5). On the calculator, we see the graphs of f (x) = 25x and g(x) = 5x + 6 intersect at x = ln(3) ln(5) ≈ 0.6826. 2 6. At first, it’s unclear how to proceed with ex−e−x = 5, b... |
tically. For example, to verify our solution to 2000 = 1000 · 3−0.1t, we substitute t = − 10 ln(2) ln(3) and obtain 2000 2000 2000 2000 2000 − 10 ln(2) ln(3) −0.1 ln(2) ln(3)? = 1000 · 3? = 1000 · 3? = 1000 · 3log3(2)? = 1000 · 2 = 2000 Change of Base Inverse Property The other solutions can be verified by using a combi... |
side of the inequality. To that end, we subtract 3 from both sides and get a common denominator ex ex − 4 ≤ 3 ex ex − 4 − − 3 ≤ 0 ex ex − 4 3 (ex − 4) ex − 4 12 − 2ex ex − 4 ≤ 0 Common denomintors. ≤ 0 We set r(x) = 12−2ex ex−4 and we note that r is undefined when its denominator ex − 4 = 0, or when ex = 4. Solving thi... |
would that be? 6.3 Exponential Equations and Inequalities 453 (−) (+) 0 (−) ln(4) ln(6) y = f (x) = ex ex−4 y = g(x) = 3 3. As before, we start solving xe2x < 4x by getting 0 on one side of the inequality, xe2x − 4x < 0. We set r(x) = xe2x − 4x and since there are no denominators, even-indexed radicals, or logs, the d... |
Example 6.3.3. Recall from Example 6.1.2 that the temperature of coffee T (in degrees Fahrenheit) t minutes after it is served can be modeled by T (t) = 70 + 90e−0.1t. When will the coffee be warmer than 100◦F? 3 so that t = −10 ln 1 Solution. We need to find when T (t) > 100, or in other words, we need to solve the ineq... |
section by finding the inverse of a function which is a composition of a rational function with an exponential function. Example 6.3.4. The function f (x) = 5ex ex + 1 your answer graphically using your calculator. Solution. We start by writing y = f (x), and interchange the roles of x and y. To solve for y, we first cl... |
10 25. 3(x−1) = 2x 2. 3(x−1) = 27 5. 8x = 1 128 8. 9 · 37x = 1 9 2x 11. 5x = −2 14. e−5730k = 1 2 3. 52x−1 = 125 6. 2(x3−x) = 1 9. 32x = 5 12. 3(x−1) = 29 15. 2000e0.1t = 4000 17. 70 + 90e−0.1t = 75 18. 30 − 6e−0.1x = 20 20. 5000 1 + 2e−3t = 2500 23. e2x = 2ex 26. 3(x−1) = 1 2 (x+5) 21. 150 1 + 29e−0.8t = 75 24. 7e2x ... |
the inverse of f (x) = ex − e−x 2. State the domain and range of both f and f −1. 6.3 Exponential Equations and Inequalities 457 48. In Example 6.3.4, we found that the inverse of f (x) = we left a few loose ends for you to tie up. 5ex ex + 1 was f −1(x) = ln x 5 − x but (a) Show that f −1 ◦ f (x) = x for all x in the... |
= ln( 1 29 ) −0.8 = 5 4 ln(29) 23. x = ln(2) 25. x = ln(3) ln(3)−ln(2) 27. x = 4 ln(3)−3 ln(7) 7 ln(7)+2 ln(3) 16. x = 1 2 ln 1 2 = − 1 2 ln(2) 18. x = −10 ln 5 3 = 10 ln 3 5 20. t = 1 3 ln(2) 22. x = = ln(2)−ln(5) ln(4)−ln(5) 24. x = − 1 = 1 4 ln(2) ln( 2 5 ) ln( 4 5 ) 8 ln 1 ln(3)+5 ln( 1 2 ) ln(3)−ln( 1 2 ) 4 26. x... |
) 46. x > 1 3 (ln(6) + 1) √ 47. f −1 = ln x + x2 + 1. Both f and f −1 have domain (−∞, ∞) and range (−∞, ∞). 6.4 Logarithmic Equations and Inequalities 459 6.4 Logarithmic Equations and Inequalities In Section 6.3 we solved equations and inequalities involving exponential functions using one of two basic strategies. We... |
same base on both sides of the equation log117(1 − 3x) = log117 x2 − 3, we equate what’s inside the logs to get 1 − 3x = x2 − 3. Solving x2 + 3x − 4 = 0 gives x = −4 and x = 1. To check these answers using the calculator, we make use of the change ln(x2−3) of base formula and graph f (x) = ln(1−3x) ln(117) and we see ... |
= f (x) = ln(x+4) and y = g(x) = 1, we see they intersect twice, at x = −3 and x = 2. ln(6) + ln(3−x) ln(6) y = f (x) = log6(x + 4) + log6(3 − x) and y = g(x) = 1 4. Taking a cue from the previous problem, we begin solving log7(1 − 2x) = 1 − log7(3 − x) by first collecting the logarithms on the same side, log7(1 − 2x) ... |
(x) = log7(1 − 2x) and y = g(x) = 1 − log7(3 − x) y = f (x) = log2(x + 3) and y = g(x) = log2(6 − x) + 3 6. Starting with 1 + 2 log4(x + 1) = 2 log2(x), we gather the logs to one side to get the equation 1 = 2 log2(x) − 2 log4(x + 1). Before we can combine the logarithms, however, we need a common base. Since 4 is a p... |
all of the answers in Example 6.4.1 analytically. We leave this to the reader and turn our attention to inequalities involving logarithmic functions. Since logarithmic functions are continuous on their domains, we can use sign diagrams. Example 6.4.2. Solve the following inequalities. Check your answer graphically usi... |
, and that the graphs intersect at x = 1. e2 < 1 = ln e−2 = −2, and find r 1 e2 e, ∞. To find the zeros of r, we set r(x) = ln(x) e < 1 < e. To determine the sign of r 1 ∪ 1 e2 (+) (−) 0 (+) 0 1 e 1 2Recall that an extraneous solution is an answer obtained analytically which does not satisfy the original equation. y = f ... |
(x))2 and y = g(x) = 2 log2(x) + 3 3. We begin to solve x log(x+1) ≥ x by subtracting x from both sides to get x log(x+1)−x ≥ 0. We define r(x) = x log(x+1)−x and due to the presence of the logarithm, we require x+1 > 0, or x > −1. To find the zeros of r, we set r(x) = x log(x + 1) − x = 0. Factoring, we get x (log(x + 1... |
8 ≥ log[H+] and log[H+] ≥ −8.5 and take the intersection of the solution sets.3 The former inequality yields 0 < [H+] ≤ 10−7.8 and the latter yields [H+] ≥ 10−8.5. Taking the intersection gives us our final answer 10−8.5 ≤ [H+] ≤ 10−7.8. (Your Chemistry professor may want the answer written as 3.16 × 10−9 ≤ [H+] ≤ 1.58 ... |
ithmic Functions 6.4.1 Exercises In Exercises 1 - 24, solve the equation analytically. 1. log(3x − 1) = log(4 − x) 3. ln 8 − x2 = ln(2 − x) 5. log3(7 − 2x) = 2 7. ln x2 − 99 = 0 9. log125 3x − 2 2x + 3 = 1 3 11. − log(x) = 5.4 13. 6 − 3 log5(2x) = 0 2. log2 x3 = log2(x) 4. log5 18 − x2 = log5(6 − x) 6. log 1 2 (2x − 1)... |
. ln(x) = e−x 33. ln(x2 + 1) ≥ 5 √ 32. ln(x) = 4 x 34. ln(−2x3 − x2 + 13x − 6) < 0 6.4 Logarithmic Equations and Inequalities 467 35. Since f (x) = ex is a strictly increasing function, if a < b then ea < eb. Use this fact to solve the inequality ln(2x + 1) < 3 without a sign diagram. Use this technique to solve the in... |
in Section 6.1.) √ 42. With the help of your classmates, solve the inequality n x > ln(x) for a variety of natural numbers n. What might you conjecture about the “speed” at which f (x) = ln(x) grows versus any principal nth root function? 468 Exponential and Logarithmic Functions 6.4.2 Answers 1. x = 5 4 4. x = −3, 4 ... |
69 6.5 Applications of Exponential and Logarithmic Functions As we mentioned in Section 6.1, exponential and logarithmic functions are used to model a wide variety of behaviors in the real world. In the examples that follow, note that while the applications are drawn from many different disciplines, the mathematics rema... |
concept behind compound interest. In the previous discussion, we would say that the interest was compounded twice, or semiannually.4 If more money can be earned by earning interest on interest already earned, a natural question to ask is what happens if the interest is compounded more often, say 4 times a year, which ... |
the interest is compounded n times per year, the amount A in the account after t years is A(t) = P 1 + nt r n If we take P = 100, r = 0.05, and n = 4, Equation 6.2 becomes A(t) = 100 1 + 0.05 which 4 reduces to A(t) = 100(1.0125)4t. To check this new formula against our previous calculations, we find A 1 4 ≈ $103.79, a... |
59375)12t = 2. Taking natural 12 ln(1.0059375) ≈ 9.75. Hence, it takes approximately 9 years logs as in Section 6.3, we get t = 9 months for the investment to double. ln(2) 5−4 4. To find the average rate of change of A from the end of the fourth year to the end of the fifth year, we compute A(5)−A(4) ≈ 195.63. Similarly... |
= 1 + 1 n → e, where e is the natural base first presented in Section 6.1. Taking the number n of compoundings per year to infinity results in what is called continuously compounded interest. Theorem 6.8. If you invest $1 at 100% interest compounded continuously, then you will have $e at the end of one year. 6In fact, t... |
inhibited Growth, the number of organisms N at time t is given by the formula N (t) = N0ekt, where N (0) = N0 (read ‘N nought’) is the initial number of organisms and k > 0 is the constant of proportionality which satisfies the equation (instantaneous rate of change of N (t) at time t) = k N (t) It is worth taking some ... |
86.18% daily growth rate for the cells.. Hence, N (t) = 12e 7 ln 1250 7 ln( 1250 3 t Whereas Equations 6.3 and 6.4 model the growth of quantities, we can use equations like them to describe the decline of quantities. One example we’ve seen already is Example 6.1.1 in Section 6.1. There, the value of a car declined fro... |
8k = 2.5. Solving, we get k = 1 8 ≈ −0.08664, which we can interpret as a loss of material at a rate of 8.664% daily. Hence, A(t) = 5e− t ln(2) 8 ≈ 5e−0.08664t. = − ln(2) 2 We now turn our attention to some more mathematically sophisticated models. One such model is Newton’s Law of Cooling, which we first encountered in... |
of Cooling (Warming): The temperature T of an object at time t is given by the formula T (t) = Ta + (T0 − Ta) e−kt, where T (0) = T0 is the initial temperature of the object, Ta is the ambient temperaturea and k > 0 is the constant of proportionality which satisfies the equation (instantaneous rate of change of T (t) a... |
T (2) = 125. This gives rise to the equation 350 − 310e−2k = 125 which yields k = − 1 ≈ 0.1602. The temperature function is T (t) = 350 − 310e 2 ln 45 62 2 ln( 45 62 ) ≈ 350 − 310e−0.1602t. t 6.5 Applications of Exponential and Logarithmic Functions 475 2. To determine when the roast is done, we set T (t) = 165. This ... |
population, L is the limiting population,a C is a measure of how much room there is to grow given by C = L N0 − 1. and k > 0 is the constant of proportionality which satisfies the equation (instantaneous rate of change of N (t) at time t) = k N (t) (L − N (t)) aThat is, as t → ∞, N (t) → L The logistic function is used... |
1+very small (+) 3. To find how long it takes until 4200 people have heard the rumor, we set N (t) = 42. Solving 1+2799e−t = 42 gives t = ln(2799) ≈ 7.937. It takes around 8 days until 4200 people have heard the rumor. 84 4. We graph y = N (x) using the calculator and see that the line y = 84 is the horizontal asymptot... |
6.5.2 Applications of Logarithms Just as many physical phenomena can be modeled by exponential functions, the same is true of logarithmic functions. In Exercises 75, 76 and 77 of Section 6.1, we showed that logarithms are useful in measuring the intensities of earthquakes (the Richter scale), sound (decibels) and acid... |
buffer solution. When carbon dioxide is absorbed into the bloodstream it produces carbonic acid and lowers the pH. The body compensates by producing bicarbonate, a weak base to partially neutralize the acid. The equation18 which models blood pH in this situation is pH = 6.1 + log 800, where x is the partial pressure of... |
ms in an attempt to ‘linearize’ data sets - in other words, transform the data sets to produce ones which result in straight lines. To see how this could work, suppose we guessed the relationship between N and t was some kind of power function, not necessarily quadratic, say N = BtA. To try to determine the A and B, we... |
to your science faculty. 480 Exponential and Logarithmic Functions This is all well and good, but the quadratic model appears to fit the data better, and we’ve yet to mention any scientific principle which would lead us to believe the actual spread of the flu follows any kind of power function at all. If we are to attack... |
This appears to be an excellent fit, but there is no friendly coefficient of determination, R2, by which to judge this numerically. There are good reasons for this, but they are far beyond the scope of the text. Which of the models, quadratic, power, exponential, or logistic is the ‘best model’? If by ‘best’ we mean ‘fits... |
an account which offers 0.75%, compounded continuously. 3. $1000 is invested in an account which offers 1.25%, compounded monthly. 4. $1000 is invested in an account which offers 1.25%, compounded continuously. 5. $5000 is invested in an account which offers 2.125%, compounded monthly. 6. $5000 is invested in an account w... |
compute the APY for this investment. 12. A finance company offers a promotion on $5000 loans. The borrower does not have to make any payments for the first three years, however interest will continue to be charged to the loan at 29.9% compounded continuously. What amount will be due at the end of the three year period, a... |
! 484 Exponential and Logarithmic Functions 19. With the help of your classmates, show that the time it takes for 90% of each isotope listed in Exercises 14 - 18 to decay does not depend on the initial amount of the substance, but rather, on only the decay constant k. Find a formula, in terms of k only, to determine ho... |
be 6 million organisms per cc. Let t be the time elapsed since the first observation, measured in hours. Assume that the yeast growth follows the Law of Uninhibited Growth N (t) = N0ekt. (a) Find the growth constant k. Round your answer to four decimal places. (b) Find a function which gives the number of yeast (in mil... |
to the nearest Sasquatch. (c) When will the population of Sasquatch in Bigfoot county reach 60? Round your answer to the nearest year. (d) Find and interpret the end behavior of the graph of y = P (t). Check your answer using a graphing utility. 28. The half-life of the radioactive isotope Carbon-14 is about 5730 year... |
F had it not been carved and eaten. 32. In reference to Exercise 44 in Section 5.3, if Fritzy the Fox’s speed is the same as Chewbacca the Bunny’s speed, Fritzy’s pursuit curve is given by y(x) = x2 − 1 4 1 4 ln(x) − 1 4 Use your calculator to graph this path for x > 0. Describe the behavior of y as x → 0+ and interpre... |
27511 (a) Use your calculator to fit a logistic model to these data, using x = 0 to represent the year 1860. (b) Graph these data and your logistic function on your calculator to judge the reasonable- ness of the fit. (c) Use this model to estimate the population of Lake County in 2010. (The 2010 census gave the populati... |
Exponential and Logarithmic Functions 489 Week x Total Headcount 17 18 19 20 21 22 23 24 7161 7772 8505 9256 10201 10743 11102 11181 With the help of your classmates, find a model for this data. Unlike most of the phenomena we have studied in this section, there is no single differential equation which governs the enrol... |
≈ $538.94, A(30) ≈ $626.16, A(35) ≈ $650.09 It will take approximately 92 years for the investment to double. The average rate of change from the end of the fourth year to the end of the fifth year is approximately 3.88. This means that the investment is growing at an average rate of $3.88 per year at this point. The a... |
125 A(5) ≈ $5559.98, A(10) ≈ $6182.67, A(30) ≈ $9453.40, A(35) ≈ $10512.13 It will take approximately 33 years for the investment to double. 12t 12 6.5 Applications of Exponential and Logarithmic Functions 491 The average rate of change from the end of the fourth year to the end of the fifth year is approximately 116.80... |
c) P = ≈ 30.83 years ln(2) 12 ln 1 + 0.0225 12 2000 1 + 0.0225 12 12 ≈ 1.0227 so the APY is 2.27% 36 ≈ $1869.57 (d) 1 + 0.0225 12 12. A(3) = 5000e0.299·3 ≈ $12, 226.18, A(6) = 5000e0.299·6 ≈ $30, 067.29 14. 5.27 ≈ −0.1315 k = ln(1/2) A(t) = 50e−0.1315t t = ln(0.1) −0.1315 ≈ 17.51 years. 15. 14 ≈ −0.0495 k = ln(1/2) A(t... |
, 963.24 billion in 2007 and $16, 291.25 billion in 2010. (a) D(0) = 15, so the tumor was 15 millimeters in diameter when it was first detected. (b) t = ln(2) 0.0277 ≈ 25 days. 22. 23. (a) k = ln(2) 20 ≈ 0.0346 (b) N (t) = 1000e0.0346t (c) t = ln(9) 0.0346 ≈ 63 minutes 3 ln 118 52 24. ln(6) 2.5 ≈ 0.4377 (a) k = 1 2 (b) ... |
3e−0.0138629t 6.5 Applications of Exponential and Logarithmic Functions 493 31. (a) T (t) = 75 + 105e−0.005005t (b) The roast would have cooled to 140◦F in about 95 minutes. 32. From the graph, it appears that as x → 0+, y → ∞. This is due to the presence of the ln(x) term in the function. This means that Fritzy will n... |
) The plot of the data and the curve is below. (c) y(140) ≈ 232889, so this model predicts 232,889 people in Lake County in 2010. (d) As x → ∞, y → 242526, so the limiting population of Lake County based on this model is 242,526 people. 494 Exponential and Logarithmic Functions Chapter 7 Hooked on Conics 7.1 Introducti... |
k)2 By squaring both sides of this equation, we get an equivalent equation (since r > 0) which gives us the standard equation of a circle. Equation 7.1. The Standard Equation of a Circle: The equation of a circle with center (h, k) and radius r > 0 is (x − h)2 + (y − k)2 = r2. Example 7.2.1. Write the standard equatio... |
x + 3y2 + 4y − 4 = 0. Solution. 3x2 − 6x + 3y2 + 4y − 4 = 0 3x2 − 6x + 3y2 + 4y = 4 3 x2 − 2x + 3 y2 + 4 3 y = 4 add 4 to both sides factor out leading coefficients 3 x2 − 2x + 1 + 3 y2 + (x − 1)2 + 3 y + (x − 1)(1) + 3 4 9 complete the square in x, y = = 25 3 25 9 factor divide both sides by 3 From Equation 7.1, we iden... |
)2 (2 − (−1))2 + (4 − 3)2 32 + 12 √ 10 2 Finally, since 2 √ 10 2 = 10 4, our answer becomes = 10 4 7.2 Circles 501 We close this section with the most important1 circle in all of mathematics: the Unit Circle. Definition 7.2. The Unit Circle is the circle centered at (0, 0) with a radius of 1. The standard equation of th... |
which satisfies the given criteria. 13. center (3, 5), passes through (−1, −2) 14. center (3, 6), passes through (−1, 4) 15. endpoints of a diameter: (3, 6) and (−1, 4) 16. endpoints of a diameter: 1 2, 4, 3 2, −1 17. The Giant Wheel at Cedar Point is a circle with diameter 128 feet which sits on an 8 foot tall platfor... |
π −e −e + π √ 2 − π e2 + 3√ 91 e2 e2 − 3√ 91 π − 3√ 91 π x π + 3√ 91 504 Hooked on Conics 7. (x − 2)2 + (y + 5)2 = 4 Center (2, −5), radius r = 2 9. (x + 4)2 + (y − 5)2 = 42 Center (−4, 5), radius r = √ 42 11. x2 + (y − 3)2 = 0 This is not a circle. 8. (x + 9)2 + y2 = 25 Center (−9, 0), radius r = 5 2 10. x + 5 2 Cent... |
to derive the equation of a parabola and, if all is right with the universe, we should get an expression much like those studied in Section 2.3. Let p denote the directed1 distance from the vertex to the focus, which by definition is the same as the distance from the vertex to the directrix. For simplicity, assume that... |
7.3. Equation 7.2. The Standard Equation of a Verticala Parabola: The equation of a (vertical) parabola with vertex (h, k) and focal length |p| is (x − h)2 = 4p(y − k) If p > 0, the parabola opens upwards; if p < 0, it opens downwards. aThat is, a parabola which opens either upwards or downwards. Notice that in the st... |
parabola. Graphically, we have the following. the latus rectum F V D It turns out3 that the length of the latus rectum, called the focal diameter of the parabola is |4p|, which, in light of Equation 7.2, is easy to find. In our last example, for instance, when graphing (x + 1)2 = −8(y − 3), we can use the fact that the... |
of a (horizontal) parabola with vertex (h, k) and focal length |p| is If p > 0, the parabola opens to the right; if p < 0, it opens to the left. (y − k)2 = 4p(x − h) 4plural of ‘directrix’ 7.3 Parabolas 509 Example 7.3.3. Graph (y − 2)2 = 12(x + 1). Find the vertex, focus, and directrix. Solution. We recognize this as... |
. Consider the equation y2 + 4y + 8x = 4. Put this equation into standard form and graph the parabola. Find the vertex, focus, and directrix. Solution. We need a perfect square (in this case, using y) on the left-hand side of the equation and factor out the coefficient of the non-squared variable (in this case, the x) on... |
abolas 511 Every cross section through the vertex of the paraboloid is a parabola with the same focus. To see why this is important, imagine the dashed lines below as electromagnetic waves heading towards a parabolic dish. It turns out that the waves reflect off the parabola and concentrate at the focus which then become... |
= 4(y + 3) 7. (y − 4)2 = 18(x − 2) 2. (y + 4)2 = 4x 6. (x + 2)2 = −20(y − 5) 8. y + 3 2 2 = −7 x + 9 2 In Exercises 9 - 14, put the equation into standard form and identify the vertex, focus and directrix. 9. y2 − 10y − 27x + 133 = 0 10. 25x2 + 20x + 5y − 1 = 0 11. x2 + 2x − 8y + 49 = 0 12. 2y2 + 4y + x − 8 = 0 13. x2... |
�parabolic cylinder.’ 7.3 Parabolas 7.3.2 Answers 513 1. (x − 3)2 = −16y Vertex (3, 0) Focus (3, −4) Directrix y = 4 Endpoints of latus rectum (−5, −4), (11, −4) y 4 3 2 1 −5−4−3−2−1 − 10 11 x −2 −3 −4 2 Vertex − 7 3, − 5 Focus − 7 3, −2 Directrix y = −3 Endpoints of latus rectum − 10 2 3, −2, − 4 3, −2 3. (y − 2)2 = −... |
11 9 7 5 3 1 −1 −3 −5 7.3 Parabolas 515 8. y + 3 2 2 2 = − Vertex − 9 Focus − 25 Directrix x = − 11 4 Endpoints of latus rectum − 25 2 2 4, 2, − 25 4, −5 9. (y − 5)2 = 27(x − 4) Vertex (4, 5) 4, 5 Focus 43 Directrix x = − 11 4 11. (x + 1)2 = 8(y − 6) Vertex (−1, 6) Focus (−1, 8) Directrix y = 4 13. (x − 5)2 = −12(y − ... |
point called the center and considered all of the points which were a fixed distance r from that one point. For our next conic section, the ellipse, we fix two distinct points and a distance d to use in our definition. Definition 7.4. Given two distinct points F1 and F2 in the plane and a fixed distance d, an ellipse is th... |
(−a, 0) and (a, 0). We will label the y-intercepts of the ellipse as (0, b) and (0, −b) (We assume a, b, and c are all positive numbers.) Schematically, y (0, b) (−a, 0) (−c, 0) (c, 0) (x, y) x (a, 0) (0, −b) Note that since (a, 0) is on the ellipse, it must satisfy the conditions of Definition 7.4. That is, the distan... |
2 = 2a (x − (−c))2 + (y − 0)2 + (x + c)2 + y2 + In order to make sense of this situation, we need to make good use of Intermediate Algebra. (x + c)2 + y2 + (x + c)2 + y22 (x − c)2 + y2 = 2a (x + c)2 + y2 = 2a − 2a − (x + c)2 + y2 = 4a2 − 4a (x − c)2 + y2 = 4a2 + (x − c)2 − (x + c)2 (x − c)2 + y2 = 4a2 − 4cx (x − c)2 + ... |
− h)2 a2 + (y − k)2 b2 = 1 Some remarks about Equation 7.4 are in order. First note that the values a and b determine how far in the x and y directions, respectively, one counts from the center to arrive at points on the ellipse. Also take note that if a > b, then we have an ellipse whose major axis is horizontal, In ... |
, the endpoints of the minor axis, and the foci. 9 + (y−2)2 25 = 1. Find the center, the lines which contain the major Solution. We see that this equation is in the standard form of Equation 7.4. Here x − h is x + 1 so h = −1, and y − k is y − 2 so k = 2. Hence, our ellipse is centered at (−1, 2). We see that a2 = 9 so... |
, meaning a > b. Since the center is the midpoint of the foci, we know it is (3, 1). Since one vertex is (0, 1) we have that a = 3, so a2 = 9. All that remains is to find b2. Since the foci are 1 unit away from the center, we know c = 1. Since 9 − b2, so b2 = 8. Substituting all of our findings into the a > b, we have c ... |
)2 + 4(y + 3)2 = 4 (x − 1)2 + 4(y + 3)2 4 4 4 = (x − 1)2 4 (x − 1)2 4 + (y + 3)2 = 1 + (y + 3)2 1 = 1 Now that this equation is in the standard form of Equation 7.4, we see that x − h is x − 1 so h = 1, and y − k is y + 3 so k = −3. Hence, our ellipse is centered at (1, −3). We see that a2 = 4 so a = 2, and b2 = 1 so b... |
ipse, denoted e, is the following ratio: e = distance from the center to a focus distance from the center to a vertex In an ellipse, the foci are closer to the center than the vertices, so 0 < e < 1. The ellipse above on the left has eccentricity e ≈ 0.98; for the ellipse above on the right, e ≈ 0.66. In general, the c... |
other focus will hear the first person as if they were standing right next to them. We explore the Whispering Galleries in our last example. Example 7.4.5. Jamie and Jason want to exchange secrets (terrible secrets) from across a crowded whispering gallery. Recall that a whispering gallery is a room which, in cross sec... |
, the vertices, the endpoints of the minor axis, the foci and the eccentricity. 9. 9x2 + 25y2 − 54x − 50y − 119 = 0 10. 12x2 + 3y2 − 30y + 39 = 0 11. 5x2 + 18y2 − 30x + 72y + 27 = 0 12. x2 − 2x + 2y2 − 12y + 3 = 0 13. 9x2 + 4y2 − 4y − 8 = 0 14. 6x2 + 5y2 − 24x + 20y + 14 = 0 In Exercises 15 - 20, find the standard form ... |
distance in AU between the sun and aphelion and the distance in AU between the sun and perihelion. 24. The graph of an ellipse clearly fails the Vertical Line Test, Theorem 1.1, so the equation of an ellipse does not define y as a function of x. However, much like with circles and horizontal parabolas, we can split an ... |
e = 3 5) √ √ √ 4. (x + 5)2 16 + (y − 4)2 1 = 1 Center (−5, 4) Major axis along y = 4 Minor axis along x = −5 Vertices (−9, 4), (−1, 4) Endpoints of Minor Axis (−5, 3), (−5, 5) 15, 4), (−5 − Foci (−5 + √ e = 15, 4) √ √ 15 4 527 x 13 −13 y 1 5 4 3 2 1 −1 −1 −2 −3 −4 −5 y 5 4 3 2 1 −3−2−1 −1 1 2 3 x −2 −3 −4 −5 y 1 2 3 4... |
−2 −1 1 2 3 4 x x −1 −2 −3 −4 −5 y 10 6 −5 −4 −3 −2 −1 1 2 x 7.4 Ellipses 529 8. (x − 4)2 8 + (y − 2)2 18 = 1 √ Center (4, 2) Major axis along x = 4 Minor axis along y = 2 Vertices (4, 2 + 3 Endpoints of the Minor Axis (4 − 2 2, 2), (4 + 2 √ Foci (4, 2 + √ 5 e = 3 2, 2) 10), (4, 2 − √ √ √ 2), (4, 2 − 3 10) √ 21 −2 −. ... |
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