Chapter 4 Exponential and Logarithmic Functions 4.1 Exponential Functions (d) 7. 1. Use your HP48G. For e2 , punch 2, then ex , to get (23 − 32 )11/7 = (8 − 9)11/7 = −1 (33 )(3−2 ) = 33−2 = 3 9. e2 = 7.389. 11. For the TI-85, press 2nd ex then 2 ENTER. Similarly e−2 = 0.135, e0.05 = 1.051, √ e−0.05 = 0.951, e0 = 1, e = 2.718, e = 1.649, 1 √ = 0.607. e (32 )5/2 52 1 = 52−3 = 53 5 2 1/2 5 = [(3 ) ] = 35 = 243 13. If P dollars is invested at an annual interest rate r and interest is compounded k times per year, the balance after t years will be B(t) = P 3. 1+ r kt k dollars and if interest is compounded continuously, the balance will be B(t) = P ert dollars. (a) If P = 1, 000, r = 0.07, t = 10, and k = 1, then 0.07 10 B(10) = 1, 000 1 + 1 = 1, 000(1.07)10 = $1, 967.15. y = 3x and y = 4x . The x−axis is a horizontal asymptote. 5. (a) (b) (c) 272/3 = (271/3 )2 = 32 = 9 (−128)3/5 = −18.379 82/3 +163/4 = (81/3 )2 +(161/4 )3 = 4+8 = 12 115 (b) If P = 1, 000, r = 0.07, t = 10, and k = 4, then 0.07 40 B(10) = 1, 000 1 + 4 = $2, 001.60. 116 CHAPTER 4. EXPONENTIAL AND LOGARITHMIC FUNCTIONS (c) If P = 1, 000, r = 0.07, t = 10, and k = 12, then 0.07 120 B(10) = 1, 000 1 + 12 = $2, 009.66. B(30) (d) If P = 1, 000, r = 0.07, t = 10, and interest is compounded continuously, then 15. = (P )(2)2 = 4P When t = 0 P0 = 5, 000, and when t = 10 P = 8, 000. 8, 000 = 5, 000e10k , so e10k = P = 5, 000 5, 000 = = $3, 534.12. (1.0175)20 1.4178 17. (a) B(5) = P 1+ 0.07 4 20 29. 9, 000 9, 000 P = = = $6, 361.42 20 (1.0175) 1.4148 B(5) = 9, 000 e−(0.07)(5) = $6, 342.19 19. (a) The population in t years will be P (t) = 50e0.02t million. Thus P (0) = 50 million. (b) The population 30 years from now will be 0.02(30) P (30) = 50e 0.6 = 50e = 91.11 million. f (x) = e f (1) = ek = 20 f (2) = e2k = (ek )2 = (20)2 = 400 f (x) = A2kx , f (0) = A = 20, f (2) = 20 (22k ) = 40, so 22k = 2. f (8) = 5, 000(e10k )3 3 8 = 5, 000 × = 20, 480 bacteria 5 P (t) = P0 e−kt When t = 0, P0 = 25, 000 copies and when t = 1 P = 10, 000 copies. 2 25, 000e−k = 10, 000, so e−k = and 5 2 2 P (2) = 25, 000e−2k = 25, 000 = 4, 000 5 copies. 31. The population density x miles from the center of the city is D(x) = 12e−0.07x thousand people per square mile. (a) At the center of the city, the density is D(0) = 12 thousand people per square mile. (b) Ten miles from the center, the density is kx 23. P (30) 8 and 5 = 5, 000. Thus 21. 2 P (t) = P0 ekt 27. Thus (b) r 120 = (P ) 1 + 4 r 60 = (P ) 1 + 4 The money has quadrupled, which is not surprising since 30 = (2)(15). 0.7 B(10) = 1, 000e = $2, 013.75. 0.07 20 = 5, 000. B(5) = P 1 + 4 r 60 B(15) = 2P = (P ) 1 + 4 25. = 20 (28k ) = 20(22k )4 = 20(24 ) = 320. D(10) = 12e−0.07(10) = 12e−0.7 = 5.959 that is 5,959 people per square mile. 33. Let Q(t) denote the amount of the radioactive substance present after t years. Since the decay is exponential and 500 grams were present initially, Q(t) = 500e−kt 4.1. EXPONENTIAL FUNCTIONS Moreover, since 400 grams are present 50 years 4 later, 400 = Q(50) = 500e−50k or e−50k = . 5 The amount present after 200 years will be Q(200) = 500e−200k = 500(e−50k )4 4 4 = 500 = 204.8 grams 5 35. f (t) = e−0.2t toasters will still be in use after t years. (a) After t = 3 years, f (3) = e−0.6 = 0.5488. (b) The fraction of toasters still working after 2 years is f (2) = e−0.4 = 0.6703, so the fraction of toasters failing is 1.0 − 0.6703. Similarly the fraction failing after 3 years will be 1 − f (3) = 1.0 − 0.5488. The fraction of toasters failing during the third year is 1.0 − 0.5488 − (1.0 − 0.6703) = 0.1215. (c) The fraction of toasters in use before the first year is e−0.1 , so the fraction of toasters failing during the first year is 1.0 − e−0.1 = 0.0952. 37. Let k be the number of compounding periods per year. (a) (P )(1 + i)k = (P )(1 + re ) from which re = (1 + i)k − 1 (b) (P )er = (P )(1 + re ) from which re = er − 1 39. Let the period of investment be 1 year and the principal $100. 0.082 4 = 108.46 B(t) = 100 1 + 4 B(t) = 100e0.081 = 108.44 Thus 8.2% is the winner. 41. (a) N (t) = N0 e−0.217t Let t = 0 be 200 B.C. In 2000 A.D. t = 2.2. N (2200) = 500e−0.217(2.2) = 310.2 117 (b) In 950 t = 0, so t = 1 in 1950. N (1) = 210e−0.217(1) = 169 This prediction is very close to the actual count of 167. (c) Writing Exercise — Answers will vary. Pi 43. M= 1 − (1 + i)−n 0.09 P = 150, 000, i = = 0.0075, n = 360. 12 M= 150, 000(0.0075) = $1, 206.93 1 − (1.0075)−360 45. (a) If the loan of $5,000 is amortized over 3 years, the monthly payment would be 5000 (1 + 0.12) −36 = 161.75 In this case you would receive $1, 000 + (161.75)(36) = $6, 823 which is 6, 823−1, 000−(36)(160) = 6, 823−6, 760 = $63 less than the buyer’s way. (b) Writing Exercise — Answers will vary 1 n 47. A(n) = 1 + . n A(−1, 000) = (1 − 0.001)−1,000 = 2.71964, A(−2, 000) = (1 − 0.0005)−2,000 = 2.71896, − 50,000 1 A(−50, 000) = 1 − = 2.71831. 50, 000 The values approach e ≈ 2.71828. 5 n/3 49. A(n) = 2− 2n 5 10/3 A(10) = 2− = 6.4584 20 100/3 5 A(100) = 2− = 7.12 × 109 200 which suggests lim (2 − n→+∞ 5 ) = +∞ 2n 118 CHAPTER 4. EXPONENTIAL AND LOGARITHMIC FUNCTIONS 4.2 Logarithmic Functions 7. Let’s call the given expression A. A = e3 ln 2−2 ln 5 3 2 = eln 2 −ln 5 1. Use your calculator. For the HP48G, first enter the number a, then press the ln key. For the TI-85, press LN and the number. This gives (since v ln u = ln uv ). ln 1 = 0 ln 2 = 0.6931472 ln 5 = 1.6094379 1 ln = ln 0.2 = −1.6094379 5 Now, for ln en , enter n, then press the ex key, followed by the ln key (HP48G) or enter LN followed by en (TI-85). Thus ln e1 = 1 and ln e2 = 2. A = eln 8−ln 25 = eln(8/25) u (since ln = ln u − ln v). v 8 A= because of the inverse relationship 25 between ex and ln x. is equivalent to These results should not be surprising since the natural logarithm and the exponential functions are inverses. ln en = n. ln 2 = 0.06x, from which x = For ln 0 and ln(−2) the calculator displays an error signal (such as the letter E or a flashing light on the HP48G or a complex number – a pair of numbers in parentheses – on the TI-85.) This is due to the domain of the logarithmic function which is x > 0. (Ask yourself “what exponent of e will give −2 ?” Answer: ex = −2 is impossible since ex > 0 for all real x. Remember that “log” means “exponent”.) 3. ln e3 = 3 ln e = 3 × 1 = 3 since ln uv = v ln u and ln e = 1. 5. Let’s give the expression a name, say A, so that we can handle it. A = eln 5 which can be written as 2 = e0.06x 9. 11. ln 2 = 11.55. 0.06 3 = 2 + 5e−4x 1 = e−4x 5 (using arithmetic), −4x = ln 1 = − ln 5 5 u = ln u − ln v and ln 1 = 0), from v ln 5 which x = = 0.402. 4 t − ln x = +C 13. 50 or t ln x = − − C. 50 Thus (since ln x = e−C−t/50 = (e−C )(e−t/50 ) ln A = ln 5. A solution (the only solution) is A=5. Thus eln 5 = 5. Actually this result is immediate from the inverse relationship between ex and ln x. because ar+s = ar as (which) certainly applies when a = e. 4.2. LOGARITHMIC FUNCTIONS 15. 1 (ln 16 + 2 ln 2) 3 1 (ln 16 + ln 22 ) 3 1 ln[(16)(22 )] 3 1 ln 26 = ln(26 )1/3 3 ln 22 . ln x = = = = = 119 After a certain time the investment will have grown to B(t) = 2P at the interest rate of 0.06. Thus 2P = P e0.06t 2 = e0.06t ln 2 = 0.06t Thus ln x = ln 4 which is valid when x = 4. 3x = e2 17. is valid if the logarithms of both members are taken. and t = 21. B(t) = P ert , 23. 25. where P is the initial investment and r is the interest rate compounded continuously. Since money doubles in 13 years, 2P = B(13) = P e13r 2 = e13r ln 2 = 13r ax+1 = b if ln ax+1 = ln b (x + 1) ln a = ln b ln b x = −1 ln a log2 x 25 ln 25 or ln x log5 (2x) 57 ln 57 ln x ln √ 1 ab3 = 5 = x = 5 ln 2 = ln x = 3.4657 = 7 = 2x = ln(2x) = ln 2 + ln x = 7 ln 5 − ln 2 = 10.5729 = 3 1/2 0 − ln(ab ) 1 = − ln(ab3 ) 2 1 = − (ln a + ln b3 ) 2 1 = − (ln a + 3 ln b) 2 1 = − (2 + 9) = −5.5. 2 ln 2 = 11.55 years. 0.06 29. The balance after t years is ln 3x = ln e2 = 2 ln e = 2 x ln 3 = 2, or 2 x = = 1.82 ln 3 19. B(t) = P ert . 27. ln 2 = 0.0533. Thus the annual interest 13 rate is 5.33%. or r = 31. Q(t) = Q0 e−kt Q0 = Q(1, 690) = Q0 e−1,690k 2 1 e−1,690k = 2 1 −1, 690k = ln 2 ln 2 k = 1, 690 Since Q0 = 50, 5 = 50e−kt , 1 1 e−kt = , −kt = ln = − ln 10, 10 10 ln 10 1, 690 ln 10 t= = ≈ 5, 614 years k ln 2 33. The number of bacteria is Q(t) = Q0 ekt . 120 CHAPTER 4. EXPONENTIAL AND LOGARITHMIC FUNCTIONS Now R(t) = R0 e−kt = 0.997R0 Since 6,000 bacteria were present initially, Q0 = 6, 000 so that ln 0.997 = − kt Q(t) = 6, 000e Q(20) = 9, 000 = 6, 000e20k 3 20k = ln 2 The forged Rembrandt painting is only approximately 25 years old. In the year 2000 t = 2, 000 − 1, 640 = 360. Let p be the percentage of 14 C left. R(t) = pR0 360 ln 2 or p = 0.9574 ln p = − 5, 730 k ≈ 0.0203 and Q(t) = 6, 000e0.0203t . Q(t) = 500 − Ae−kt 35. ln 2 t or t = 24.8372 years. 5, 730 and The original Rembrandt painting will contain approximately 95.74 % of 14 C. Q(0) = 300 = 500 − A. Thus A = 200 and Q(t) = 500 − 200e−kt , or 45% of the original 133 I should be detected. e−(ln 2/20.9)(25) = 0.4364 (b) Q(6) = 410 = 500 − 200e−6k . e−6k = 9 or k =≈ 0.1331 20 or 43.64% of 133 I remains in the thyroid. 43.64 − 41.3 = 2.34% of 133 I still remains in the rest of the patient’s body. It follows that Q(t) = 500 − 200e−0.1331t . 37. ¿From the half-life of 1 R0 2 1 ln 2 14 T = Ta + (Td − Ta )(0.97)t so 40 = 10 + (98.6 − 10)(0.97)t , 30 = 88.6(0.97)t , 30 t ln 0.97 = ln or t = 35.55 88.6 = R0 e−5,730k ln 2 5, 730 = R0 e−kt = 0.28R0 ln 2 ln 0.28 = − t 5, 730 5, 730 ln 0.28 t = − = 10, 523years. ln 2 Now R(t) The artifacts at the Debert site in Nova Scotia are about 10,500 years old. 39. ¿From the half-life of 1 R0 2 1 ln 2 43. C = −5, 730k or k = 14 C = R0 e−5,730k = −5, 730k or k = ln 2 5, 730 e−(ln 2/20.9)(24) = 0.45 41. (a) The murder occurred around 1:30 a.m. on Wednesday morning. Blohardt was in the slammer, so Scélerat must have done it. 45. (a) R ln I Ia (b) ln Ib Ib Ia Ib ln I = 8.3 ln 10 = 8.3 ln 10 = 19.1115 = e19.1115 = 1.995 × 108 = = 7.1 ln 10 = 16.3484 = e16.3484 = 1.2589 × 107 = 15.8472 times more intense. 4.3. DIFFERENTIATION OF LOGARITHMIC AND EXPONENTIAL FUNCTIONS 47. Let t denote the number of years after 1960. If the population P (measured in billions) is growing exponentially and was 3 billion in 1960 (when t = 0), then P (t) = 3ekt . 57. e−3.5x 1 + 257e−1.1x Using a graphing utility and tracing, we find x = −5.06. 3, 500e0.31x = 59. ln(x + 3) − ln x = 5 ln(x2 − 4) x+3 ln = 0 x(x2 − 4)5 Since the population in 1975 (when t = 15) was 4 billion, 4 = P (15) = 3e15k or k = For P (t) = ekt = x+3 = 1. Using a graphing utility we x(x2 − 4)5 find x = 2.28. 1 4 ln 15 3 40 so 3ekt = 40 40 or t ≈ 135 3 or 61. k = ln 2 = 0.015. After 24 hours 46.5 100e−0.015(24) = 69.77 mg The population will reach 40 billion in the year 1960 + 135 = 2095. will be left. The time required for the isotope to decline to 25 mg is ln 0.25 t=− = 92.4 hours. 0.015 49. (a) 0.05 = e−3k , ln(0.05) = −3k or k = 0.999 0.01 = e−0.999x ln(0.01) = 4.61 m. 0.999 (b) Writing exercise — Answers will vary. or x = − 51. (a) λ = ln 2 . Therefore k Q(t) = Q0 e−(ln 2/λ)t (b) Q0 e−(ln 2/λ)t = Q0 (0.5)kt − ln 2 − ln 2 1 = = (ln 0.5)λ = ln 2λ λ logb b 1 loga b = = logb a logb a k= 53. 4.3 Therefore loga b logb a = 1. 55. 10x and log10 x are reflections about y = x. 121 1. Differentiation of Logarithmic and Exponential Functions f (x) = e5x f 0 (x) = e5x 3. 2 f (x) = ex d (5x) = 5e5x dx +2x−1 d (x2 + 2x − 1) dx 2 = (2x + 2)ex +2x−1 . 2 f 0 (x) = ex +2x−1 122 5. 7. CHAPTER 4. EXPONENTIAL AND LOGARITHMIC FUNCTIONS 30 + 10e−0.05x d f 0 (x) = 0 + 10e−0.05x (−0.05x) dx = −0.5e−0.05x . f (x) = f (x) = 2 21. f 0 (0) = 1 and f (0) = 0. An equation of the tangent line is 6x (x + 3x + 5)e d f 0 (x) = (x2 + 3x + 5) e6x dx d +e6x (x2 + 3x + 5) dx = (6x2 + 20x + 33)e6x . f (x) = xe−x f 0 (x) = −xe−x + e−x = e−x (1 − x) y − 0 = (1)(x − 0) or y = x 23. f 0 (x) 9. = x 2 (1 − 3e ) f (x) = f 0 (x) d (1 − 3ex ) dx = −6ex (1 − 3ex ). f 0 (1) = 0 and f (1) = e2 . An equation of the tangent line is y = e2 2(1 − 3ex ) = √ e2x x2 x2 (2e2x ) − e2x (2x) 2(x − 1)e2x = x4 x3 f (x) = 1/2 √ x2 ln x 25. = e(3x) f (x) = x2 ln x = 2 √ d x + 2x ln x f 0 (x) = e 3x (31/2 x1/2 ) f 0 (x) = dx 2 √ √ 1 3 = e 3x 31/2 x−1/2 = √ e 3x . 1 2 2 3x f 0 (1) = and f (1) = 0. An equation of the 2 tangent line is 13. f (x) = ln x3 = 3 ln x 1 d 3 y − 0 = ( 12 (x − 1) or x − 2y − 1 = 0 f 0 (x) = 3 x= . x dx x 11. f (x) 15. 17. 19. = e 3x = x2 ln x d d f 0 (x) = x2 (ln x) + ln x (x2 ) dx dx = x(1 + 2 ln x). f (x) f (x) = f 0 (x) = f (x) = f 0 (x) = = = √ 3 e2x 27. ln f (x) f 0 (x) f (x) 2x/3 =e 2 2√ 3 e2x/3 = e2x 3 3 x+1 ln x−1 x−1 d x+1 x + 1 dx x − 1 x − 1 −2 x + 1 (x − 1)2 −2 . x2 − 1 f (x) = 1 ln(3x − 5). 6 5 3 − , x + 2 6(3x − 5) 5 (x + 2)5 1 √ − 6 3x − 5 x + 2 2(3x − 5) = 5 ln(x + 2) − = f 0 (x) = 29. (x + 2)5 √ . 6 3x − 5 f (x) = (x + 1)3 (6 − x)2 (2x + 1)1/3 . ln f (x) = ln[(x + 1)3 (6 − x)2 (2x + 1)1/3 ] = ln(x + 1)3 + ln(6 − x)2 + ln(2x + 1)1/3 = 3 ln(x + 1) + 2 ln(6 − x) 1 + ln(2x + 1). 3 4.3. DIFFERENTIATION OF LOGARITHMIC AND EXPONENTIAL FUNCTIONS Differentiating leads to f 0 (x) f (x) f 0 (x) (c) 3 x+1 2 2(−1) + , + 6−x 3(2x + 1) √ = (x + 1)3 (6 − x)2 3 2x + 1 3 2 2 − + . x + 1 6 − x 3(2x + 1) = 2 31. f (x) = 2x ln f (x) = x2 ln 2 f 0 (x) = 2x ln 2 f (x) f 0 (x) 2 2x = +1 x ln 2 C(x) = e0.2x 33. C 0 (x) = 0.2e0.2x (a) (d) R0 (x) = C 0 (x) when x + 3 ln(x + 3) = 2x (x + 3)2 or x = 0.1805 (obtained with a graphing utility.) 2 (e) 2x = x + x √ or x = 2. 37. (a) The population t years from now will be (c) e = x 0.2x A0 (x) e = (0.2x − 1) x2 = xe−3x = e−3x (1 − 3x) R(x) R0 (x) (d) R0 (x) = C 0 (x) when or x = 0.2049 (e) 0.2e0.2x = e0.2x x or x = 5. C(x) = x2 + 2 C 0 (x) = 2x A(x) A0 (x) million. Hence the rate of change of the population t years from now will be P 0 (t) = 50e0.02t (0.02) = e0.02t and the rate of change 10 years from now will be P 0 (t) = e0.2 = 1.22 million per year. e−3x (1 − 3x) = 0.2e0.2x (b) = 0.2x A(x) (a) x ln(x + 3) x+3 1 x 0 [(x + 3) + ln(x + 3) R (x) = (x + 3)2 x+3 −x ln(x + 3)] x + 3 ln(x + 3) = (x + 3)2 R(x) P (t) = 50e0.02t (b) 35. 123 = x + 2x−1 = 1 − 2x−2 (b) The percentage rate of change t years from now will be 0 0.02t P (t) e 100 = 100 P (t) 50e0.02t 100 = =2 50 % per year, which is a constant, independent of time. 39. (a) The value of the machine after t years is Q(t) = 20, 000e−0.4t dollars. Hence the rate of depreciation after t years is Q0 (t) = 20, 000e−0.4t (−0.4) = −8, 000e−0.4t 124 CHAPTER 4. EXPONENTIAL AND LOGARITHMIC FUNCTIONS and the rate after 5 years is (a) Q0 (5) = −8, 000e−2 = −$1, 082.68 per year. (b) The percentage rate of change t years from now will be 0 Q (t) −8, 000e−0.4t 100 = 100 = −40 Q(t) 20, 000e−0.4t % per year, which is a constant, independent of time. 41. (a) The first year sales of the text will be 47. f (x) = 20 − 15e−0.2x thousand copies when x thousand complementary copies are distributed. If the number of complimentary copies distributed is increased from 10,000, that is when x = 10, by 1,000, that is , ∆x = 1, the approximate change in sales is 49. (b) The actual change in sales is ∆f = f (11) − f (10) = (20 − 15e−2.2 ) − (20 − 15e−2 ) = 0.368 or 368 copies. Q(t) = Q0 e−0.0015t 43. 1 (a) = e−0.0015t or t = 462 2 Half of the ozone will be depleted in 462 years. (b) 0.2 = e−0.0015t , t = − ln 0.22 = 1, 073 0.0015 80% of the ozone will be depleted in 1,073 years. 45. C(t) = 0.2te−t/2 −t/2 at t = 2. C(2) = 0.147 or 14.7%. (b) We want t such that C(t) = 0.3(0.147) = 0.0441 Using a graphing utility, we find t = 6.9 hours. 2x f (x) = x x x(2 ) ln 2 − 2x 2x (ln 2x − 1) = f 0 (x) = x2 x2 x ln x f (x) = x log10 x = ln 10 1 0 f (x) = (1 + ln x) ln 10 51. By definition, the percentage rate of change of f with respect to x is ∆f = f 0 (10)∆x. Since f 0 (x) = 3e−0.2x and ∆x = 1, ∆f = f 0 (10) = 3e−2 = 0.406 thousand or 406 copies. 1 C (t) = 0.2e + 0.2t − e−t/2 2 t −t/2 = 0.2e 1− =0 2 0 100 f 0 (x) f (x) d f 0 (x) ln f (x) = , it follows that the dx f (x) percentage rate of change can be written as d 100 ln f (x). dx 53. The population of the town x years from now will be p P (x) = 5, 000 x2 + 4x + 19 Since ¿From problem 51, the percentage rate of change x years from now will be p d 100 ln 5, 000 x2 + 4x + 19 dx d 1 = 100 [ln 5, 000 + ln(x2 + 4x + 19)] dx 2 100(x + 2) = x2 + 4x + 19 Hence the percentage rate of change 3 years from now will be 100(3 + 2) = 12.5 % per year 32 + 12 + 19 4.4. ADDITIONAL EXPONENTIAL MODELS 125 f (x) = (3.7x2 − 2x + 1)e−3x+2 55. f 0 (−2.17) = −428, 640 y = g(t) = 5 − 3e−t . 7. The line y = 5 is a horizontal asymptote. 4.4 Additional Exponential Models 1. y = f (t) = 2 + et The line y = 2 is a horizontal asymptote. 9. 3. 2 . 1 + 3e−2t The lines y = 0 and y = 2 are horizontal asymptotes. y = f (x) = y = g(x) = 2 − 3ex . The line y = 2 is a horizontal asymptote. 11. f (x) = xex f 0 (x) = ex (x + 1) = 0 when x = −1. f 00 (x) = ex (1 + x + 1) = ex (x + 2) = 0 5. y = f (x) = 3 − 2(2−x ). The line y = 3 is a horizontal asymptote. when x = −2. (−1, −0.37) is a minimum, (−2, −0.27) is a point of inflection. 126 CHAPTER 4. EXPONENTIAL AND LOGARITHMIC FUNCTIONS Note that f (x) = xex < 0 if x < 0 and decreasing (if x < −1). The line y = 0 is an upper bound (the curve is never above this line for x < 0.) This strongly suggests that the x axis is a horizontal asymptote. x −∞ −2 −1 ∞ f (x) 0 −0.27 −0.37 ∞ f 0 (x) − − 0 + f 00 (x) − 0 + + f (x) = x2 e−x . 15. f 0 (x) = e−x [−x2 + 2x] = xe−x (2 − x) = 0 when x = 0, 2. f 00 (x) 13. f (x) f 0 (x) = e−x [−2x + 2 − (−x2 + 2x)] = e−x (x2 − 4x + 2) = 0 √ 4 ± 16 − 8 , when x = 2 so x = 0.59 or x = 3.41. (0, 0) is a minimum, (2, 0.54) is a maximum, (0.59, 0.19) and (3.41, 0.38) are points of inflection. Note that f (x) = x2 e−x > 0 if x > 0 and decreasing. The line y = 0 is a lower bound (the curve is never below this line.) This strongly suggests that the x axis is a horizontal asymptote. = xe2−x = e2−x (1 − x) = 0 when x = 1. f 00 (x) = e2−x [−1 + (1 − x)(−1)] = e2−x (x − 2) = 0 when x = 2. (1, 2.7) is a maximum, (2, 2) is a point of inflection. Note that f (x) = xe2−x > 0 if x > 0 and decreasing for x > 1. The line y = 0 is a lower bound (the curve is never below this line if x > 0.) This strongly suggests that the x axis is a horizontal asymptote. x −∞ 1 2 ∞ f (x) −∞ 2.7 2 0 0 f (x) + 0 − − f 00 (x) − − 0 + x −∞ · · · 0 · · · 0.59 2 3.41 ∞ f (x) ∞ · · · 0 · · · 0.19 0.5 0.38 0 f 0 (x) − − 0 + + 0 − − f 00 (x) + + + + 0 − 0 + 17. 4.4. ADDITIONAL EXPONENTIAL MODELS f (x) = f 0 (x) = = f 00 (x) = = 6 = 6(1 + e−x )−1 . 1 + e−x −6(1 + e−x )−2 (−e−x ) 6 > 0. ex (1 + e−x )2 6 2x e (1 + e−x )4 [0 − (ex )(2)(1 + e−x )(−e−x ) −(1 + e−x )2 ex ] 6(1 − e−x ) − x =0 e (1 + e−x )3 when e−x = 1 or x = 0. (0, 3) is a point of inflection. Note that if x → −∞ 1 + e−x → ∞ and y → 0, so the x axis is a horizontal asymptote. Similarly if x → +∞ 1 + e−x → 1 and y → 6, so the line y = 6 is a horizontal asymptote also. When x < 0, f (x) is concave up, and when x > 0, f (x) is concave down. 19. f (x) f 0 (x) = (ln x)2 ln x = 2 =0 x when x = 1. f 00 (x) = 2 (1 − ln x) = 0 x2 when x = e. (1, 0) is a minimum, (e, 1) is a point of inflection. Note that the curve seems to level off to the right, but this is an optical illusion. f (x) will keep increasing beyond all bounds. For example f (1, 000) = 47.7 and f (1098 ) = 51, 964. 127 The y–axis is a vertical asymptote. When x < 1, f (x) is decreasing. When x > 1, f (x) is increasing. When x < e, f (x) is concave up. When x > e, f (x) is concave down. 21. (a) The reliability function is f (t) = 1 − e−0.03t . As t increases without bound, e−0.03t approaches 0 and so f (t) approaches 1. Furthermore, f (0) = 0. The graph is like that of a learning curve. (b) The fraction of tankers that sink in fewer than 10 days is f (10) = 1 − e−0.3 . The fraction of tankers that remain afloat for at least 10 days is therefore 1 − f (10) = e−0.3 = 0.7408. (c) The fraction of tankers that can be expected to sink between the 15th and 20th days is f (20) − f (15) = (1 − e−0.6 ) − (1 − e−0.45 ) = −e−0.6 + e−0.45 = −0.5488 + 0.6373 = 0.0888. 23. The temperature of the drink t minutes after leaving the refrigerator is f (t) = 30 − Ae−kt . 128 CHAPTER 4. EXPONENTIAL AND LOGARITHMIC FUNCTIONS Since the temperature of the drink when it left the refrigerator was 10 degrees Celsius, f (x) = 15 − 20e−0.3x 29. x = 9 thousand. f (9) = 15 − 20e−2.7 = 13.656 10 = f (0) = 30 − A or A = 20. Thus −kt f (t) = 30 − 20e . (a) f 0 (9) = 6e−2.7 = 0.403 thousand (b) f (10) = 15 − 20e−3 = 14.004 Since the temperature of the drink was 15 degrees Celsius 20 minutes later, 15 = f (20) = 30 − 20e−20k or e−20k = 3 . 4 The temperature of the drink after 40 minutes is therefore f (40) 25. (a) = 30 − 20e−40k = 30 − 20(e−20k )2 2 3 = 30 − 20 = 18.75 degrees. 4 f (t) = The actual increase was 14.004 − 13.656 = 0.348 or 348 books. The estimate of 403 books was not too bad an approximation. Q(t) = 80(4 + 76e−1.2t )−1 31. Q0 (t) = 80(−1)(4 + 76e−1.2t )−2 (76)e−1.2t (−1.2) e−1.2t = 80 × 76 × 1.2 (4 + 76e−1.2t )2 After 2 weeks (at the end of the second week) 2 . 1 + 3e−0.8t Q(2) = Q00 (t) (b) f (0) = 0.5 thousand people (500 people). 2 (c) f (3) = = 1.572, so 1,572 1 + 3 × 0.0907 people have caught the disease. (d) The highest number of people who can 2 contract the disease is = 2 or 2,000 1+0 people. Note that the graph is shown only for t ≥ 0. 27. Q(t) = 40 − Ae−kt Q(0) = 20, 20 = 40 − A so that A = 20. 1 Q(1) = 30, 30 = 40 − 20e−k and e−k = . 2 Q(3) = 40 − 20e−3k = 37.5 units per day. = e−2.4 (4 + 76e−2.4 )2 5.576 or 5,576 people = 80 × 76 × 1.2 7296(−1.2)(4 + 76e−1.2t − 152e−1.2t ) e1.2t (4 + 76e−1.2t )3 which equals 0 if 76e−1.2t = 4 or t = 2.45. The disease is spreading most rapidly approximately 2 12 weeks after the outbreak. Cekt 33. P (t) = 1 + Cekt (a) P (0) = P0 = or C (b) = C 1+C P0 1 − P0 Cekt t→∞ 1 + Cekt C = lim −kt = 1 = 100% t→∞ e +C lim 35. (a) The profit per VCR is $x − 125. The number of units sold is 1, 000e−0.02x . The weekly profit is P (x) = 1, 000(x − 125)e−0.02x 4.4. ADDITIONAL EXPONENTIAL MODELS 129 200 200(6 + t) 1 = 6+t which will be equal to the prevailing interest 1 = 0.08 or rate of 8% when 6+t 1 t= − 6 = 6.5. 0.08 1 Moreover, > 0.08 when 0 < t < 6.5 6+t 1 < 0.08 when 6.5 < t. and 6+t Hence the percentage rate of growth of the value of the collection is greater than the prevailing interest rate when 0 < t < 6.5 and less than the prevailing interest rate when t > 6.5. Thus the collection should be sold in 6.5 years. c y= (e−at − e−bt ) 41. b−a c (a) y0 = (−ae−at + be−bt ) = 0 b−a a when ae−at = be−bt or = e(a−b)t b a 1 Thus t = ln . b a−b In the long run both exponential terms approach zero, so y → 0. = (b) P 0 (x) = 1, 000[−0.02(x − 125)e−0.02x +e−0.02x ] = 0 when 0.02(x − 125) = 1 or x = 175. 37. The percentage rate of change of the market price √ V (t) = 8, 000e t of the land (expressed in decimal form) is V 0 (t) V (t) √ = 8, 000e t 1 1 √ √ = √ t 2 t 2 t 8, 000e which will be equal to the prevailing interest rate of 6 % when 2 1 1 √ = 0.06 or t = = 69.44. 0.12 2 t 1 Moreover, √ > 0.06 when 0 < t < 69.44 and 2 t 1 √ < 0.06 when 69.44 < t. 2 t Hence the percentage rate of growth of the value of the land is greater than the prevailing interest rate when 0 < t < 69.44 and less than the prevailing interest rate when 69.44 < t. Thus the land should be sold in 69.44 years. 39. Since the stamp collection is currently worth $1,200 and its value is increases linearly at the rate of $200 per year, its value t years from now is V (t) = 1, 200 + 200t. The percentage rate of change of the value (expressed in decimal form) is V 0 (t) V (t) = 200 1, 200 + 200t (b) N (t) = 43. (a) 500(0.03)(0.4) t 0 N (0) = 500(0.03)(0.4) = 15 N (5) = 500(0.03)(0.4) ≈ 482 employees. 5 t 500(0.03)(0.4) ln 0.6 if (0.4)t = = 0.145677 ln 0.03 ln 0.145677 t = = 2.10 years. ln 0.4 0 lim N (t) = 500(0.03) = 500 employees. 300 t→∞ = 130 CHAPTER 4. EXPONENTIAL AND LOGARITHMIC FUNCTIONS (b) F (t) = 500(0.03)−(0.4) −t f (x) = x(e−x + e−2x ) 49. N (t) is bounded between 0 and 500 while 500 is the lower bound for F (t) and there is no upper bound. lim f (x) = 0 x→∞ Note: The portion of the graph in the third quadrant is likely to be hidden from view on your graphing utility unless you specificaly request a negative domain. The high point occurs at (0.76, 0.52). 45. Let’s assume continuous growth, so Q(t) = Q0 e0.06t Let t = 0 be 1947. Q0 = 1, 139. (a) 0.06(7) Q(7) = 1, 139e Q(53) = 27, 389.3 √ = 1, 733.5 (b) We want 2, 000 = 1, 139e0.06t or 1.756 = e0.06t . This leads to t = 9.38. 2, 278 = 1, 139e0.06t or 2 = e0.06t . This leads to t = 11.55. (c) Writing Exercise — Answers will vary. P (x) = λ2 xe−λx , 0 < λ < e 47. (a) (b) P 0 (x) = λ2 e−λx + λ2 x(−λe−λx ) = λ2 e−λx (1 − xλ) = 0 1 1 λ at x = . P = λe−1 = λ λ e P (t) = 20, 000te 51. 0.4t−0.07t A graphing utility indicates a maximum present value of $150, 543 at t = 44.38 years. 4.5 Review Problems Review Problems 1. (a) If f (x) = 5e−x , then f (x) approaches 0 as x increases without bound, and f (x) increases without bound as x decreases without bound. The y–intercept is f (0) = 5. 4.5. REVIEW PROBLEMS 131 If f (x) = 5 − 2e−x , (b) As x decreases without bound, e2x approaches 0 and thus f (x) approaches 2. 5 3+2 The y−intercept is f (0) = = . 1+1 2 Applying the quotient rule to determine f 0 (x) yields then f (x) approaches 5 as x increases without bound, and f (x) decreases without bound as x decreases without bound. The y–intercept is f (0) = 3. f 0 (x) = (c) If f (x) = 1 − 6 , 2 + e−3x the y–intercept is 6 6 f (0) = 1 − =1− = −1. 0 2+e 2+1 As x increases without bound, e−3x approaches 0, and so lim f (x) = 1 − x→∞ 2e2x . (e2x + 1)2 Since 2e2x > 0, there are no critical points and f (x) increases for all x. To find f 00 (x), differentiate f 0 (x) using the quotient rule obtaining 6 = −2. 2+0 f 00 (x) = 4e2x (1 − e2x ) (e2x + 1)3 There is one inflection point 5 (0, f (0)) = (0, ) since f 00 (0) = 0. The 2 function f (x) is concave down for x > 0 and concave up for x < 0. As x decreases without bound, e−3x increases without bound, and so lim f (x) = 1 − 0 = 1. x→−∞ 2. (a) If f (x) = Ae−kx and f (0) = 10, then 10 = Ae0 . Hence f (x) = 10e−kx . Since f (1) = 25, 25 = 10e−k or e−k = (d) If f (x) = 3 + 2e−2x , 1 + e−2x then f (x) approaches 3 as x increases without bound, since e−2x approaches 0. Rewriting the function, by multiplying numerator and denominator by e2x yields f (x) = 3e2x + 2 . e2x + 1 5 . 2 Then f (4) = 10e−4k = 10(e−k )4 4 5 = 10 = 390.625. 2 (b) If f (x) = Aekx and f (1) = 3 as well as f (2) = 10, then 3 = Aek and 10 = Ae2k , two equations in two unknowns. 3 Aek Division eliminates A, so = , 10 Ae2k k 3 e 10 = 2k , or ek = . 10 e 3 132 CHAPTER 4. EXPONENTIAL AND LOGARITHMIC FUNCTIONS 10 10 Since e = ,3=A 3 3 9 and so A = . Thus 10 9 k 3 100 9 3k f (3) = e = (e ) = 10 10 3 k (c) If and f (0) = 50, then 50 = 30 + Ae0 or A = 20. Hence, f (x) = 30 + 20e−kx . Since f (3) = 40, 40 = 30 + 20e−3k , 1 10 = 20e−3k , or e−3k = . 2 (d) V (10) = 50, 000e2 ln(2/5) = 50, 000eln(4/25) 4 = $8, 000. = 50, 000 25 4. The sales function is f (x) = 30 + Ae−kx Thus f (9) Hence V (t) = 50, 000e[1/5 ln(2/5)]t and so = 30 + 20e−9k = 30 + 20(e−3k )3 3 1 = 30 + 20 = 32.5. 2 6 1 + Ae−kx 6 and f (0) = 3 then 3 = , or A = 1. 1 + Ae0 6 Hence, f (x) = . 1 + e−kx 6 Since f (5) = 2, 2 = , 1 + e−5k −5k −5k 2 + 2e = 6, or e = 2. Then, Q(x) = 50 − 40e−0.1x units, where x is the amount (in thousands) spent on advertising. (a) As x increases without bound, Q(x) approaches 50. The vertical axis intercept is Q(0) = 10. The graph is like that of a learning curve. If f (x) = f (10) = = 6 6 = −10k 1+e 1 + (e−5k )2 6 6 = . 2 1 + (2) 5 3. Let V (t) denote the value of the machine after t years. Since the value decreases exponentially and was originally $50,000, it follows that V (t) = 50, 000e−kt . Since the value after 5 years is $20,000, 20, 000 −5k e or k −5k = V (5) = 50, 000e 2 = , 5 1 2 = − ln . 5 5 (b) If no money is spent on advertising, sales will be Q(0) = 10 thousand units. (c) If $8,000 is spent on advertising, sales will be Q(8) = 50 − 40e−0.8 = 32.027 thousand or 32,027 units. (d) Sales will be 35 thousand if Q(x) = 35, that is, if 50 − 40e−0.1x = 35, 3 e−0.1x = , 8 ln(3/8) x = − = 9.808 0.1 thousand or $9,808. , (e) Since Q(x) approaches 50 as x increases without bound, the most optimistic sales projection is 50,000 units. 4.5. REVIEW PROBLEMS 133 5. The output function is Q(t) = 120 − Ae−kt . Since Q(0) = 30, 30 = 120 − A or A = 90. Since Q(8) = 80, 80 = 120 − 90e−8k , 4 −40 = −90e−8k or e−8k = . Hence, 9 Q(4) = 120 − 90e−4k = 120 − 90(e−8k )1/2 1/2 4 = 120 − 90 = 60 units. 9 6. The population t years from now will be P (t) = 30 . 1 + 2e−0.05t (a) The vertical axis intercept is 30 P (0) = = 10 million. 1+2 As t increases without bound, e−0.05t approaches 0. Hence, lim P (t) t→∞ = lim t→∞ 30 = 30 1 + 2e−0.05t As t decreases without bound, e−0.05t increases without bound. Hence, the denominator 1 + 2e−0.05t increases without bound and 30 lim P (t) = lim = 30. t→∞ t→∞ 1 + 2e−0.05t (b) The current population is P (0) = 10 million. (c) The population in 20 years will be P (20) = = 30 1 + 2e−0.05(20) 30 = 17.2835 1 + 2e−1 million or 17,283,500 people. (d) In the long run (as t increases without bound), e−0.05t approaches 0 and so the population P (t) approaches 30 million. 7. (a) ln e5 = 5 since n = ln en . (b) eln 2 = 2 since n = eln n . (c) (d) 8. (a) e3 ln 4−ln 2 3 = eln 4 −ln 2 64 ln = e 2 = eln 32 = 32. ln(9e2 ) + ln(3e−2 ) = ln[(9e2 )(3e−2 )] = ln 27 = 3 ln 3. 8 = 2e0.04x , e0.04x = 4, 0.04x = ln 4 x = 34.657. (b) = 1 + 4e−6x , 4e−6x = 4 −6x = ln 1 = 0, or x = 0. (c) 4 ln x = 8, ln x = 2, or x = e2 = 7.389. (d) 5x = e3 , ln 5x = ln e3 , x ln 5 = 3, or 3 x = = 1.864. ln 5 5 9. Let Q(t) denote the number of bacteria after t minutes. Since Q(t) grows exponentially and 5,000 bacteria were present initially, Q(t) = 5, 000ekt . Since 8,000 bacteria were present after 10 minutes, 8 8, 000 = Q(10) = 5, 000e10k , e10k = , or 5 1 8 k= ln . 10 5 134 CHAPTER 4. EXPONENTIAL AND LOGARITHMIC FUNCTIONS The bacteria will double when = t log3 t2 , 2 t ln t = ln 3 2 (1 + ln t). g 0 (t) = ln 3 (f ) g(t) Q(t) = 10, 000, that is, when 5, 000ekt kt or t 10. (a) (b) = = 10, 000 11. (a) Using the formula ln 2 ln 2 10 ln 2 r kt = = 1 + B(t) = P k ln(8/5) k = 14.75 or 14. min. and 45 seconds. with P = 2, 000, B = 5, 000, r = 0.08, and k = 4, f (x) = 2e3x+5 , 0.08 4t 5, 000 = 2, 000 1 + 0 3x+5 d f (x) = 2e (3x + 5) 4 dx 3x+5 5 = 6e . (1.02)4t = 2 = x2 e−x , d d f 0 (x) = e−x (x2 ) + x2 e−x dx dx = x(2 − x)e−x . f (x) B(t) = P ert (c) (d) (e) 1 ln(5/2) = 11.57 years. 4 ln 1.02 (b) Using the formula or t = p g(x) = ln x2 + 4x + 1 1 ln(x2 + 4x + 1), = 2 1 1 g 0 (x) = 2 x2 + 4x + 1 d (x2 + 4x + 1) dx x+2 = . 2 x + 4x + 1 = x ln x2 = 2x ln x, d d 0 h (x) = 2 x ln x + ln x x dx dx = 2(1 + ln x). h(x) f (t) = t , ln 2t (ln 2t)(1) − t 0 f (t) = = with P = 2, 000, B = 5, 000, and r = 0.08, 5, 000 = 2, 000e0.08t , e0.08t = or t = 5 , 2 ln(5/2) = 11.45. 0.08 12. Compare the effective interest rates. The effective interest rate for 8.25 % compounded quarterly is rk 0.0825 4 1+ −1 = 1+ −1 k 4 = 0.0851 or 8.51 %. The effective interest rate for 8.20 compounded continuously is (ln 2t)2 ln 2t − 1 . (ln 2t)2 1 2t (2) er − 1 = e0.082 − 1 = 0.0855 or 8.55 %. 13. (a) Using the present value formula r −kt P =B 1+ k 4.5. REVIEW PROBLEMS 135 with B = 2, 000, t = 10, r = 0.0625, and k = 12, 0.0625 -120 P = 2, 000 1 + 12 2, 000 = = $1, 072.26. 1.8652182 (b) Using the present value formula P = Be−rt (a) The rate of change of the carbon monoxide level t years from now is Q0 (t) = 0.12e0.03t , and the rate two years from now is Q0 (2) = 0.12e0.06 = 0.13 parts per million per year. (b) The percentage rate of change of the carbon monoxide level t years from now is 0 Q (t) .12e.03t 100 = 100 = 3% per year Q(t) 4e.03t which is a constant, independent of time. with B = 2, 000, t = 10, and r = 0.0625, P = 2, 000e−0.0625(10) = $1, 070.52. 14. (a) 1+ 0.0625 2 2 × 10 8, 000 = (P ) P = 8, 000 = 4, 323.25 (1.03125)20 (b) 15. 18. Let F (p) denote the profit, where p is the price per camera. Then F (p) = (number of cameras sold) (profit per camera) = 800(p − 40)e−0.01p . 0 F (p) = 800[e−0.01p (1) +(p − 40)e−0.01p (−0.01)] = 8e−0.01p (140 − p) = 0 8, 000 = P e0.0625×10 P = 8, 000e−0.625 = 4, 282.09 B(t) = r = 2, 054.44 = 1, 000e12r ln 2.05444 = 0.06, or r = 6% 12 16. At 6% compounded annually, the effective interest rate is rk 0.06 1 1+ −1 = 1+ −1 k 1 = 0.06. At r % compounded continuously, the effective interest rate is er − 1. Setting the two effective rates equal to each other yields er − 1 = 0.06, when p = 140. Since F 0 (p) > 0 (and F is increasing) for 0 < p < 140, and F 0 (p) < 0 (and F is decreasing) for p > 140, it follows that F (p) has its absolute maximum at p = 140. Thus the cameras should be sold for $140 apiece to maximize the profit. 19. (a) f (x) = xe−2x , f 0 (x) = xe−2x (−2) + e−2x (1) = e−2x (1 − 2x) = 0 whenx = 1 f 2 = f 00 (x) = 1 . 2e 4e−2x (x − 1) = 0 er = 1.06, r = ln 1.06 = 0.0583 or 5.83 %. when x = 1. f (1) = 17. The average level of carbon monoxide in the air t years from now is Q(t) = 4e0.03t parts per million. 1 . e2 1 . 2 136 CHAPTER 4. EXPONENTIAL AND LOGARITHMIC FUNCTIONS 4e−x > 0. (1 + e−x )2 1 f 00 (x) = [(1 + e−x )2 (1 + e−x )4 (4e−x )(−1) − 4e−x (2) (1 + e−x )(e−x )(−1)] 4e−x (e−x − 1) = = 0 when x = 0 (1 + e−x )3 1 1 ∞ 2 1 1 f (x) −∞ 0 2e e2 0 f (x) ∞ + 0 − − − 0 ↑ ↓ ↓ ↓ f 00 (x) − − − 0 + ↓ ↓ ↓ ↑ 1 1 , is the absolute maximum 2 2e 1 while 1, 2 is a point of inflection. e x −∞ = −∞ 0 1 x f (x) f 0 (x) 0 2 + ↑ 0 + + ↑ ↑ f 00 (x) + − up ↓ (0, 2) is a point of inflection. lim x→−∞ (b) f (x) f 0 (x) f 00 (x) x ∞ 4 .0707 4 =0 1 + e−x because the denominator increases beyond all bounds. −x = e −e = ex + e−x > 0 = ex − e−x = 0 when x = 0 lim x→∞ 4 =4 1 + e−x because e−x → 0. So y = 0 and y = 4 are horizontal asymptotes. There are no extrema. x −∞ 0 ∞ f (x) −∞ 0 ∞ f 0 (x) ∞ + + + ↑ ↑ ↑ f 00 (x) − 0 + ↓ ↑ (0, 0) is a point of inflection. (d) f (x) = f 0 (x) = ln(x2 + 1) 2x = 0 when x = 0 x2 + 1 f (0) = 0. (c) f (x) = f 0 (x) = 4 = 4(1 + e−x )−1 , 1 + e−x d 4(−1)(1 + e−x )−2 (1 + e−x ) dx f 00 (x) = 2(1 − x)(1 + x) =0 (x2 + 1)2 4.5. REVIEW PROBLEMS 137 when x = ±1. f (±1) = ln 2. x −∞ −1 0 1 f (x) ∞ ln 2 0 ln 2 f 0 (x) 0 − − − 0 + + ↓ ↓ ↓ ↑ ↑ f 00 (x) 0 − 0 + 2 + 0 ↓ ↑ ↑ (0, 0) is the absolute minimum while (±1, ln 2) are points of inflection. aR0 = R0 e−kt , a = e−15,000k . ∞ ∞ + 0 ↑ − 0 ↓ 22. 1 Since the half-life is 5,730, = e−5,730k or 2 ln 2 k= and 5, 730 a = e−15,000 ln 2/5,730 = e−1.8145 = 0.1629. (a) R(t) = R0 e−(ln 2)t/5,730 R(1, 960) = R0 e−(ln 2)(1,960)/5,730 = R0 e−0.2371 = 0.7889R0 Thus about 78.89% of in the shroud. (b) Since 92.3% of √ t √ = 2, 000e t √ 2, 000e t 1 √ 2 t 1 = √ 2 t which will be equal to the prevailing interest 1 rate of 7 % when √ = 0.07 or 2 t 2 1 t= = 51.02. 0.14 1 Moreover, √ > 0.07 when 0 < t < 51.02 2 t 1 and √ < 0.07 when 51.02 < t. 2 t Hence the percentage rate of growth of the collection is greater than the prevailing interest rate when 0 < t < 51.02 and less than the prevailing interest rate when 51.02 < t. Thus the coin collection should be sold in 51.02 years. 21. Let a be the desired ratio. Then R(t) = R0 e−kt , 14 C 1988 − 662 = 1326 so Pierre d’Arcis’s suspicions were well founded. . Hence, the percentage rate of change of the value of the collection (expressed in decimal form) is V 0 (t) V (t) should be left is left in the forgery, (ln 2)t 0.923 = exp − 5, 730 (ln 0.923)(5, 730) t = − = 662 ln 2 20. The value of the coin collection in t years is V (t) = 2, 000e 14 C f (t) = 70 − Ae−kt 23. f (0) = 70 − A = 212 or A = −142. Thus f (t) = 70 + 142e−kt . Let T0 be the ideal temperature. Then T0 + 15 = 70 + 142e−k(5) and T0 = 70 + 142e−k(7) Solving (with the SOLVE utility of our calculator) we get k = 0.091 and T0 = 1450 . 24. (a) D(t) = (D0 − 0.00046)e−0.162t + 0.00046 With D0 = 0.008, D(10) = (0.008 − 0.00046)e−1.62 + 0.00046 = 0.00195 and D(25) = 0.00590. lim D(t) = 0 (b) t→∞ 138 CHAPTER 4. EXPONENTIAL AND LOGARITHMIC FUNCTIONS 25. (a) To compute doubling time in terms of the rate set 2P = P ert and solve for t. 2 = ert or t = ln 2 r P (5, 000) = e−(ln 2)5,000/5,730 = 54.62% ≈ 55% In the chart below find a comparison of the rules of 69, 70, 72 with the true doubling time. r 0.04 0.06 0.09 0.10 69 17.25 11.50 7.67 6.9 70 17.5 11.67 7.78 7.0 72 18 12 8 7.2 True 17.33 11.55 7.7 6.93 28. The Bronze age began about 5,000 years ago (around 3,000 B.C.). The maximum percentage is 29. (a) 0.12 5.75 5.83 6 5.78 69 The rule of 69 is closest because is 100 closest to ln 2. (b) A(x) = A0 (x) = ln x − 2 110 , for 10 ≤ x. x 3 − ln x 110 =0 x2 when x = e3 = 20.0855 years. A(10) = 3.3284 so a person’s aerobic capacity is maximized at about age 20. 27. 30. f 0 (x) = 4e−(ln x) √ for x > 0 πx −(ln x)2 = 4 e √ π x2 [−2 ln x − 1] = 0 1 when x = √ = 0.6065. e According to the graphing utility, the most common age is at (0.6065, 2.8977). = Cke−kt 100k = ekt − 1 T (t) = 35e−0.32t 27 = 35e−0.32t or t = 0.811 min. Rescuers have about 49 seconds before the girl looses consciousness. dT = −35(0.32)e−0.32t dt 27 35 dT 27 = (−35)(0.32) = −8.64 dt 35 At T = 27 or e−0.32t = 2 f (x) df df 100 f (c) Writing Exercise — Answers will vary. (b) Writing Exercise — Answers will vary. 26. f (t) = C(1 − e−kt ) = 0.008C 0.992 = e−2k or k = .00402 Thus the girl’s temperature is dropping at 8.64 degrees per minute. 31. (a) y y0 = 0.125(e4x + e−4x ) = 0.5(e4x − e−4x ) = 0 when e8x = 1 at x = 0. 4.5. REVIEW PROBLEMS 139 f (t) = 30 − Ae−kt 34. Solve for −Ae−kt = f (t) − 30. f 0 (t) = −Ae−kt (−k) = kAe−kt = k[30 − f (t)] where k is a constant of proportionality and f (t) is the temperature of the drink. P H = − log10 [H3 O+ ] 35. (b) Writing exercise — Answers will vary. 32. k1 k2 k1 k2 k1 ln k2 33. (a) For milk and lime, P Hm = 3P Hl . For lime and orange, P Hl = 0.5P Ho . E0 = A exp − RT1 E0 = A exp − RT2 E0 E0 = exp − RT2 RT1 E0 1 1 = − R T2 T1 2 t V (t) = V0 1 − L 3.2 = 1.6 2 [H3 O+ ]l = 10−1.6 = 0.0251 202.31 P (t) = 36. 1 + e3.938−0.314t Hint: Use the equation writer on the HP48G to enter the function. For the TI-85, press 2nd CALC EVALF(202.31/(1 + e(3.938−0.314x) ), x, 0), etc. P Hl (a) V0 = 875, L = 8, t = 5, 2 5 V (5) = 875 1 − = 207.64. 8 The refrigerator will be worth $207.64 after 5 years. 875 − 207.64 5 133.47 875 (b) 0 V (t) = V0 = $133.47 and = 0.15 or 15% 2 1− L t ln 2 1− L = The percentage rate of change is 2 t 2 V 1 − ln 1 − 0 V 0 (t) L L 100 = 100 V (t) 2 t V0 1 − L 2 = 100 ln 1 − L year 1790 1800 1830 1860 1880 1900 1920 1940 1960 1980 1990 2000 t 0 1 4 7 9 11 13 15 17 19 20 21 P (t) 3, 867, 087 5, 256, 550 12, 956, 719 30, 207, 500 50, 071, 364 77, 142, 427 108, 425, 601 138, 370, 607 162, 289, 822 178, 782, 499 184, 566, 652 189, 034, 385 (b) This model predicts that the population will be increasing most rapidly when t = 12.5 or in 1915. 140 CHAPTER 4. EXPONENTIAL AND LOGARITHMIC FUNCTIONS intersects (c) Writing exercise — Answers will vary. 37. x 1 2 x 1 = 3 x 1 = 5 y = 2−x = y = 3−x y = 5−x (0.5)−x = 2x x 1 The graphs of y = bx and y = are b reflections of each other in the y–axis (0 < b < 1). The larger b the steeper the curve. y 38. y y y √ y = 4 − ln x at (1.2373, 3.8935) according to the graphing utility. = 40. can be rewritten as 1 1 1 ln(x + 5) − ln x = ln(x2 + 2x)2 ln 5 ln 2 ln 10 Hint: Use the equation writer on the HP48G to enter the equation (and make a copy on the stack) and store in the EQ field (use ’EQ’ and STOre). According to the SOLVE capability of the graphing utility. Use GRAPH, y(x) =, and ZOOM/TRACE on the TI-85. √ = 3x = 3x/2 √ = 3−x = 3−x/2 = 3−x = 3−x x = 1.06566543483 is a solution of this equation. Plot f (x) = log5 (x + 5) − log2 x − 2 log10 (x2 + 2x) The graphs of y = 3bx and y = 3−bx are reflections of each other in the y–axis (0 < b). The larger b the steeper the curve. because the message “sign reversal” (on the HP48G) indicates another root. x < 0 leads to a complex solution (as it should since the argument of a logarithm must be positive). Plotting on 0 < x < 500, 000 did not reveal another root. There is no other root because x2 increases much more rapidly than any other argument, making f (x) monotonically decreasing. 41. 39. Hint: Use the equation writer on the HP48G to enter the function or y(x) = on the TI-85. y = 3x log5 (x + 5) − log2 x = 2 log10 (x2 + 2x) = f (x) = ln(1 + x2 ) and 1 y = g(x) = x intersect at (1.166, 0.858) according to the graphing utility. y 4.5. REVIEW PROBLEMS 141 even when n is large. 42. √ √ √ √ n ( n) n+1 ( n + 1) n 8 22.63 22.36 9 32.27 31.62 12 88.21 85.00 20 957.27 904.84 25 3, 665 3, 447 31 16, 528 15, 494 37 68, 159 63, 786 38 85, 679 80, 166 43 261, 578 244, 579 50 1, 165, 565 1, 089, 362 100 1.12 × 1010 1.05 × 1010 1, 000 2.87 × 1047 2.76 × 1047 √ Thus (n + 1) n ≤n √ n+1 Mathematically one could reason as follows: √ (x + 1) x √ lim x→∞ x x+1 √x + 1 x+1 ≤ lim x→∞ x √x + 1 x+1 = lim exp ln x→∞ x x+1 ln x = lim exp (x + 1)−1/2 x→∞ x(−1/x2 ) x→∞ (x + 1)(−1/2)(x + 1)−3/2 √ 2 x+1 = exp lim x→∞ x 2 = e0 = 1 = exp lim √ x→∞ 2 x + 1 = exp lim √ Thus (n + 1) n ≤n √ n+1

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