Exponential and Logarithmic Functions

Transcription

1 Chapter 4 Exponential and Logarithmic Functions 4. Exponential Functions. Use your HP48G. For e, punch, then e x, to get 3. e For the TI-85, press nd e x then ENTER. Similarly e 0.35, e , e , e 0, e.78, e.649, e d) 3 3 ) /7 8 9) / )3 ) ) 5/ [3 ) / ] 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 Bt) P + r ) kt k dollars and if interest is compounded continuously, the balance will be Bt) P e rt dollars. y 3 x and y 4 x. The x axis is a horizontal asymptote. 5. a) 7 /3 7 /3 ) 3 9 b) 8) 3/ c) 8 /3 +6 3/4 8 /3 ) +6 /4 ) a) If P, 000, r 0.07, t 0, and k, then B0), ) 0, ) 0 $, b) If P, 000, r 0.07, t 0, and k 4, then B0), ) 40 4 $,

2 6 CHAPTER 4. EXPONENTIAL AND LOGARITHMIC FUNCTIONS c) If P, 000, r 0.07, t 0, and k, then B0), ) 0 $, d) If P, 000, r 0.07, t 0, and interest is compounded continuously, then B0), 000e 0.7 $, B5) P ) 0 5, Thus 7. a) b) P Thus 5, 000 5, 000 $3, ) B5) P P ) 0 5, , 000 9, 000 $6, ) B5) 9, 000 e 0.07)5) $6, a) The population in t years will be P t) 50e 0.0t million. Thus P 0) 50 million. b) The population 30 years from now will be P 30) 50e 0.030) 50e million.. fx) e kx f) e k 0 f) e k e k ) 0) fx) A kx, f0) A 0, f) 0 k ) 40, so k. f8) 0 8k ) 0 k ) ) B5) P P ) + r ) 60 4 ) 0 B30) P ) + r 4 [ P ) + r ) 60 ] P )) 4P 4 The money has quadrupled, which is not surprising since 30 )5). 7. P t) P 0 e kt When t 0 P 0 5, 000, and when t 0 P 8, , 000 5, 000e 0k, so e 0k 8 5 and P 30) 5, 000e 0k ) 3 ) 8 3 5, 000 0, 480 bacteria 5 9. P t) P 0 e kt When t 0, P 0 5, 000 copies and when t P 0, 000 copies. 5, 000e k 0, 000, so e k 5) and P ) 5, 000e k 5, 000 4, copies. 3. The population density x miles from the center of the city is Dx) e 0.07x thousand people per square mile. a) At the center of the city, the density is D0) thousand people per square mile. b) Ten miles from the center, the density is D0) e 0.070) e that is 5,959 people per square mile. 33. Let Qt) denote the amount of the radioactive substance present after t years. Since the decay is exponential and 500 grams were present initially, Qt) 500e kt

3 4.. EXPONENTIAL FUNCTIONS 7 Moreover, since 400 grams are present 50 years later, 400 Q50) 500e 50k or e 50k 4 5. The amount present after 00 years will be Q00) 500e 00k 500e 50k ) 4 ) grams ft) e 0.t toasters will still be in use after t years. a) After t 3 years, f3) e b) The fraction of toasters still working after years is f) e , so the fraction of toasters failing is Similarly the fraction failing after 3 years will be f3) The fraction of toasters failing during the third year is ) 0.5. c) The fraction of toasters in use before the first year is e 0., so the fraction of toasters failing during the first year is.0 e Let k be the number of compounding periods per year. a) P ) + i) k P ) + r e ) from which b) In 950 t 0, so t in 950. N) 0e 0.7) 69 This prediction is very close to the actual count of 67. c) Writing Exercise Answers will vary. P i 43. M + i) n P 50, 000, i 0.09 M , n , ) $, ) a) If the loan of $5,000 is amortized over 3 years, the monthly payment would be ) In this case you would receive which is $, )36) $6, 83 6, 83, )60) 6, 83 6, 760 $63 less than the buyer s way. b) Writing Exercise Answers will vary 47. An) + n) n. r e + i) k b) P )e r P ) + r e ) from which r e e r 39. Let the period of investment be year and the principal $00. Bt) ) Bt) 00e Thus 8.% is the winner. 4. a) Nt) N 0 e 0.7t Let t 0 be 00 B.C. In 000 A.D. t.. N00) 500e 0.7.) A, 000) 0.00), , A, 000) ), , ) 50,000 A 50, 000) , 000 The values approach e.788. An) 5 ) n/3 n A0) 5 ) 0/ A00) 5 ) 00/ which suggests lim n + 5 n ) +

4 8 CHAPTER 4. EXPONENTIAL AND LOGARITHMIC FUNCTIONS 4. Logarithmic Functions. 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 ln 0 ln ln ln ln Now, for ln e n, enter n, then press the e x key, followed by the ln key HP48G) or enter LN followed by e n TI-85). Thus ln e and ln e. These results should not be surprising since the natural logarithm and the exponential functions are inverses. ln e n n. For ln 0 and ln ) 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? Answer: e x is impossible since e x > 0 for all real x. Remember that log means exponent.) 3. ln e 3 3 ln e 3 3 since ln u v v ln u and ln e. 5. Let s give the expression a name, say A, so that we can handle it. which can be written as A e ln 5 ln A ln 5. A solution the only solution) is A5. Thus e ln 5 5. Actually this result is immediate from the inverse relationship between e x and ln x. 7. Let s call the given expression A. since v ln u ln u v ). A 3 ln ln 5 e e ln 3 ln 5 A ln 8 ln 5 e e ln8/5) since ln u ln u ln v). v A 8 because of the inverse relationship 5 between e x and ln x. 9. e 0.06x. is equivalent to ln 0.06x, from which x ln using arithmetic), 3 + 5e 4x 5 e 4x 4x ln 5 ln 5 since ln u ln u ln v and ln 0), from v which x ln ln x t 50 + C or ln x t 50 C. Thus x e C t/50 e C )e t/50 ) because a r+s a r a s which) certainly applies when a e.

5 4.. LOGARITHMIC FUNCTIONS 9 5. ln x ln 6 + ln ) 3 3 ln 6 + ln ) 3 ln[6) )] 3 ln 6 ln 6 ) /3 ln. Thus ln x ln 4 which is valid when x x e is valid if the logarithms of both members are taken. ln 3 x ln e ln e x ln 3, or x ln a x+ b if ln a x+ ln b x + ) ln a ln b x ln b ln a. log x 5 5 x ln 5 5 ln ln x or ln x log 5 x) 7 5. ln 5 7 x ln 5 7 lnx) ln + ln x ln x 7 ln 5 ln ab 3 0 lnab 3 ) / lnab3 ) ln a + ln b3 ) ln a + 3 ln b) 7. Bt) P e rt. After a certain time the investment will have grown to Bt) P at the interest rate of Thus P P e 0.06t e 0.06t ln 0.06t and t ln.55 years The balance after t years is Bt) P e rt, where P is the initial investment and r is the interest rate compounded continuously. Since money doubles in 3 years, P B3) P e 3r e 3r ln 3r or r ln Thus the annual interest 3 rate is 5.33%. 3. Qt) Q 0 e kt Q 0 e,690k, 690k ln k ln, 690 Q, 690) Q 0 e,690k Since Q 0 50, 5 50e kt, e kt, kt ln ln 0, 0 0 ln 0, 690 ln 0 t 5, 64 years k ln 33. The number of bacteria is Qt) Q 0 e kt. + 9) 5.5.

6 0 CHAPTER 4. EXPONENTIAL AND LOGARITHMIC FUNCTIONS Since 6,000 bacteria were present initially, Q 0 6, 000 so that Qt) 6, 000e kt Q0) 9, 000 6, 000e 0k 0k ln 3 k and Qt) 6, 000e 0.003t. 35. Qt) 500 Ae kt and Thus A 00 and It follows that Q0) A. Qt) e kt, Q6) e 6k. e 6k 9 0 or k 0.33 Qt) e 0.33t. 37. From the half-life of 4 C R 0 R 0 e 5,730k ln 5, 730k or k ln 5, 730 Now Rt) R 0 e kt 0.8R 0 ln 0.8 ln 5, 730 t t 5, 730 ln 0.8 ln 0, 53years. The artifacts at the Debert site in Nova Scotia are about 0,500 years old. 39. From the half-life of 4 C R 0 R 0 e 5,730k ln 5, 730k or k ln 5, 730 Now Rt) R 0 e kt 0.997R 0 ln ln t or t years. 5, 730 The forged Rembrandt painting is only approximately 5 years old. In the year 000 t, 000, Let p be the percentage of 4 C left. Rt) pr 0 ln p 360 ln or p , 730 The original Rembrandt painting will contain approximately % of 4 C. 4. a) e ln /0.9)4) or 45% of the original 33 I should be detected. b) e ln /0.9)5) or 43.64% of 33 I remains in the thyroid % of 33 I still remains in the rest of the patient s body. T T a + T d T a )0.97) t so )0.97) t, ) t, ) 30 t ln 0.97 ln or t The murder occurred around :30 a.m. on Wednesday morning. Blohardt was in the slammer, so Scélerat must have done it. 45. a) R ln I ln ln I 8.3 ln I a e b) ln I b 7. ln I b e I a I b times more intense.

7 4.3. DIFFERENTIATION OF LOGARITHMIC AND EXPONENTIAL FUNCTIONS 47. Let t denote the number of years after 960. If the population P measured in billions) is growing exponentially and was 3 billion in 960 when t 0), then P t) 3e kt. Since the population in 975 when t 5) was 4 billion, 4 P 5) 3e 5k or k 5 ln 4 3 For P t) 40 so 3e kt 40 e kt 40 or t 35 3 The population will reach 40 billion in the year a) 0.05 e 3k, ln0.05) 3k or k e 0.999x or x ln0.0) 4.6 m b) Writing exercise Answers will vary. 5. a) λ ln k. Therefore b) ln /λ)t Qt) Q 0 e Q 0 e ln /λ)t Q 0 0.5) kt k ln ln 0.5)λ ln ln λ λ 53. log a b log b b log b a log b a Therefore log a b log b a , 500e 0.3x e 3.5x + 57e.x Using a graphing utility and tracing, we find x lnx + 3) ln x 5 lnx 4) x + 3 ln xx 4) 5 0 x + 3 or xx. Using a graphing utility we 4) 5 find x k ln After 4 hours e 0.054) mg will be left. The time required for the isotope to decline to 5 mg is ln 0.5 t 9.4 hours Differentiation of Logarithmic and Exponential Functions x and log 0 x are reflections about y x.. fx) e 5x f x) e 5x d 5x) 5e5x dx 3. fx) e x +x f x) e x +x d dx x + x ) x + )e x +x.

8 CHAPTER 4. EXPONENTIAL AND LOGARITHMIC FUNCTIONS 5. fx) e 0.05x f x) 0 + 0e 0.05x d dx 0.05x) 0.5e 0.05x. 7. fx) x + 3x + 5)e 6x f x) x + 3x + 5) d dx e6x +e 6x d dx x + 3x + 5) 6x + 0x + 33)e 6x. 9. fx) 3e x ) f x) 3e x ) d dx 3ex ) 6e x 3e x ).. fx) xe x 3. f x) xe x + e x e x x) f 0) and f0) 0. An equation of the tangent line is fx) ex x y 0 )x 0) or y x f x) x e x ) e x x) x 4 x )ex x 3 f ) 0 and f) e. An equation of the tangent line is y e. fx) e 3x e 3x)/ f x) e 3x d dx 3/ x / ) [ ) ] e 3x 3 / x / 3. fx) ln x 3 3 ln x ) d f x) 3 x dx x 3 x. 5. fx) x ln x f x) x d dx ln x) + ln x d dx x ) x + ln x). 7. fx) 3 e x e x/3 ) f x) e x/3 3 e x x e 3x fx) x ln x x ln x f x + x ln x x) f ) and f) 0. An equation of the tangent line is y 0 ) x ) or x y 0 fx) x + )5 6 3x 5. ln fx) 5 lnx + ) ln3x 5). 6 f x) 5 fx) x x 5), f x) [ x + ) x 5 x + 3x 5) ] 9. ) x + fx) ln x f x) x ) d x + x + dx x x x + x ) x. 9. fx) x + ) 3 6 x) x + ) /3. ln fx) ln[x + ) 3 6 x) x + ) /3 ] lnx + ) 3 + ln6 x) + lnx + ) /3 3 lnx + ) + ln6 x) + lnx + ). 3

9 4.3. DIFFERENTIATION OF LOGARITHMIC AND EXPONENTIAL FUNCTIONS 3 Differentiating leads to f x) fx) 3 x + + ) 6 x + 3x + ), f x) x + ) 3 6 x) 3 x + [ 3 3. fx) x x + 6 x + 3x + ) ln fx) x ln f x) fx) x ln f x) x + x ln 33. Cx) e 0.x a) C x) 0.e 0.x b) Ax) e0.x x A x) e0.x 0.x ) x c) Rx) xe 3x d) R x) C x) when e) or x or x 5. R x) e 3x 3x) e 3x 3x) 0.e 0.x 0.e 0.x e0.x x 35. Cx) x + a) C x) x b) Ax) x + x A x) x ]. c) x lnx + 3) Rx) x + 3 R x) [x + 3) x + 3) x lnx + 3)] x + 3 lnx + 3) x + 3) d) R x) C x) when e) x + 3 lnx + 3) x + 3) x ) x + lnx + 3) x + 3 or x obtained with a graphing utility.) or x. x x + x 37. a) The population t years from now will be P t) 50e 0.0t million. Hence the rate of change of the population t years from now will be P t) 50e 0.0t 0.0) e 0.0t and the rate of change 0 years from now will be P t) e 0.. million per year. b) The percentage rate of change t years from now will be [ P ] ) t) e 0.0t P t) 50e 0.0t % per year, which is a constant, independent of time. 39. a) The value of the machine after t years is Qt) 0, 000e 0.4t dollars. Hence the rate of depreciation after t years is Q t) 0, 000e 0.4t 0.4) 8, 000e 0.4t

10 4 CHAPTER 4. EXPONENTIAL AND LOGARITHMIC FUNCTIONS and the rate after 5 years is Q 5) 8, 000e $, per year. b) The percentage rate of change t years from now will be [ Q ] ) t) 8, 000e 0.4t Qt) 0, 000e 0.4t % per year, which is a constant, independent of time. a) C t) 0.e t/ + 0.t ) e t/ 0.e t/ t ) 0 at t. C) 0.47 or 4.7%. b) We want t such that Ct) ) Using a graphing utility, we find t 6.9 hours. 4. a) The first year sales of the text will be fx) 0 5e 0.x thousand copies when x thousand complementary copies are distributed. If the number of complimentary copies distributed is increased from 0,000, that is when x 0, by,000, that is, x, the approximate change in sales is f f 0) x. Since f x) 3e 0.x and x, f f 0) 3e thousand or 406 copies. b) The actual change in sales is f f) f0) 0 5e. ) 0 5e ) or 368 copies. 43. Qt) Q 0 e 0.005t a) b) e 0.005t or t 46 Half of the ozone will be depleted in 46 years. 0. e 0.005t ln 0., t, % of the ozone will be depleted in,073 years. 45. Ct) 0.te t/ 47. fx) x x f x) xx ) ln x x x ln x ) x 49. fx) x log 0 x x ln x ln 0 f x) + ln x) ln 0 5. By definition, the percentage rate of change of f with respect to x is 00 f x) fx) Since d dx ln fx) f x), it follows that the fx) percentage rate of change can be written as 00 d ln fx). dx 53. The population of the town x years from now will be P x) 5, 000 x + 4x + 9 From problem 5, the percentage rate of change x years from now will be 00 d dx ln 5, 000 x + 4x d dx [ln 5, lnx + 4x + 9)] 00x + ) x + 4x + 9 Hence the percentage rate of change 3 years from now will be ) 3.5 % per year + + 9

11 4.4. ADDITIONAL EXPONENTIAL MODELS fx) 3.7x x + )e 3x+ f.7) 48, y gt) 5 3e t. The line y 5 is a horizontal asymptote. 4.4 Additional Exponential Models. y ft) + e t The line y is a horizontal asymptote. 9. y fx) + 3e t. The lines y 0 and y are horizontal asymptotes. 3. y gx) 3e x. The line y is a horizontal asymptote.. fx) xe x f x) e x x + ) 0 when x. f x) e x + x + ) e x x + ) 0 5. y fx) 3 x ). The line y 3 is a horizontal asymptote. when x., 0.37) is a minimum,, 0.7) is a point of inflection.

12 6 CHAPTER 4. EXPONENTIAL AND LOGARITHMIC FUNCTIONS Note that fx) xe x < 0 if x < 0 and decreasing if x < ). 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 fx) f x) 0 + f x) fx) x e x. f x) e x [ x + x] xe x x) 0 when x 0,. f x) e x [ x + x + x)] e x x 4x + ) 0 3. when x. fx) xe x f x) e x x) 0 f x) e x [ + x) )] e x x ) 0 when x.,.7) is a maximum,, ) is a point of inflection. Note that fx) xe x > 0 if x > 0 and decreasing for x >. 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 fx).7 0 f x) + 0 f x) 0 + when x 4 ± 6 8, so x 0.59 or x , 0) is a minimum,, 0.54) is a maximum, 0.59, 0.9) and 3.4, 0.38) are points of inflection. Note that fx) x 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. x fx) f x) f x)

13 4.4. ADDITIONAL EXPONENTIAL MODELS 7 fx) 6 + e x 6 + e x ). f x) 6 + e x ) e x ) 6 e x + e x ) > 0. f x) 6 e x + e x ) 4 [0 e x )) + e x ) e x ) + e x ) e x ] 6 e x ) e x + e x ) 3 0 when e x or x 0. 0, 3) is a point of inflection. Note that if x + e x and y 0, so the x axis is a horizontal asymptote. Similarly if x + + e x and y 6, so the line y 6 is a horizontal asymptote also. When x < 0, fx) is concave up, and when x > 0, fx) is concave down. The y axis is a vertical asymptote. When x <, fx) is decreasing. When x >, fx) is increasing. When x < e, fx) is concave up. When x > e, fx) is concave down.. a) The reliability function is ft) e 0.03t. As t increases without bound, e 0.03t approaches 0 and so ft) approaches. Furthermore, f0) 0. The graph is like that of a learning curve. 9. when x. fx) ln x) ) ln x f x) 0 x f x) ln x) 0 x when x e., 0) is a minimum, e, ) is a point of inflection. Note that the curve seems to level off to the right, but this is an optical illusion. fx) will keep increasing beyond all bounds. For example f, 000) 47.7 and f0 98 ) 5, 964. b) The fraction of tankers that sink in fewer than 0 days is f0) e 0.3. The fraction of tankers that remain afloat for at least 0 days is therefore f0) e c) The fraction of tankers that can be expected to sink between the 5 th and 0 th days is f0) f5) e 0.6 ) e 0.45 ) e e The temperature of the drink t minutes after leaving the refrigerator is ft) 30 Ae kt.

14 8 CHAPTER 4. EXPONENTIAL AND LOGARITHMIC FUNCTIONS Since the temperature of the drink when it left the refrigerator was 0 degrees Celsius, Thus 0 f0) 30 A or A 0. ft) 30 0e kt. Since the temperature of the drink was 5 degrees Celsius 0 minutes later, 5 f0) 30 0e 0k or e 0k 3 4. The temperature of the drink after 40 minutes is therefore f40) 30 0e 40k 30 0e 0k ) ) degrees a) ft) + 3e 0.8t. 9. fx) 5 0e 0.3x x 9 thousand. a) f9) 5 0e f 9) 6e thousand b) f0) 5 0e The actual increase was or 348 books. The estimate of 403 books was not too bad an approximation. 3. Qt) e.t ) Q t) 80 )4 + 76e.t ) 76)e.t.) e.t e.t ) After weeks at the end of the second week) e.4 Q) e.4 ) or 5,576 people Q t) 796.)4 + 76e.t 5e.t ) e.t e.t ) 3 b) f0) 0.5 thousand people 500 people). c) f3).57, so, people have caught the disease. d) The highest number of people who can contract the disease is or, people. Note that the graph is shown only for t Qt) 40 Ae kt Q0) 0, 0 40 A so that A 0. Q) 30, e k and e k. Q3) 40 0e 3k 37.5 units per day. which equals 0 if 76e.t 4 or t.45. The disease is spreading most rapidly approximately weeks after the outbreak. 33. P t) Cekt + Ce kt a) b) P 0) P 0 C + C or C lim t Ce kt P 0 P 0 + Ce kt C lim t e kt + C 00% 35. a) The profit per VCR is $x 5. The number of units sold is, 000e 0.0x. The weekly profit is P x), 000x 5)e 0.0x

15 4.4. ADDITIONAL EXPONENTIAL MODELS 9 b) P x), 000[ 0.0x 5)e 0.0x +e 0.0x ] 0 when 0.0x 5) or x The percentage rate of change of the market price V t) 8, 000e t of the land expressed in decimal form) is V t) V t) 8, 000e t 8, 000e t t t which will be equal to the prevailing interest rate of 6 % when ) 0.06 or t t 0. Moreover, > 0.06 when 0 < t < and t < 0.06 when < t. t Hence the percentage rate of growth of the value of the land is greater than the prevailing interest rate when 0 < t < and less than the prevailing interest rate when < t. Thus the land should be sold in years. 39. Since the stamp collection is currently worth $,00 and its value is increases linearly at the rate of $00 per year, its value t years from now is V t), t. The percentage rate of change of the value expressed in decimal form) is V t) V t) 00, t t) 6 + t which will be equal to the prevailing interest rate of 8% when 0.08 or 6 + t t Moreover, > 0.08 when 0 < t < t and < 0.08 when 6.5 < t. 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. 4. y c b a e at e bt ) a) y c b a ae at + be bt ) 0 when ae at be bt or a b ea b)t Thus t a ) a b ln. b In the long run both exponential terms approach zero, so y 0. b) 43. a) Nt) ) 0.4)t N0) ) 0.4)0 5 N5) ) 0.4)5 48 employees ) 0.4)t if 0.4) t ln 0.6 ln t ln years. ln 0.4 lim t )0 500 employees.

16 30 CHAPTER 4. EXPONENTIAL AND LOGARITHMIC FUNCTIONS b) F t) ) 0.4) t 49. fx) xe x + e x ) Nt) is bounded between 0 and 500 while 500 is the lower bound for F t) and there is no upper bound. 45. Let s assume continuous growth, so Qt) Q 0 e 0.06t Let t 0 be 947. Q 0, 39. a) Q7), 39e 0.067), Q53) 7, b) We want, 000, 39e 0.06t or.756 e 0.06t. This leads to t 9.38., 78, 39e 0.06t or e 0.06t. This leads to t.55. c) Writing Exercise Answers will vary. 47. P x) λ xe λx, 0 < λ < e a) P x) λ e λx + λ x λe λx ) λ e λx xλ) 0 ) lim fx) 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, 0.5). 5. P t) 0, 000te 0.4t 0.07t A graphing utility indicates a maximum present value of $50, 543 at t years. 4.5 Review Problems Review Problems. a) If fx) 5e x, then fx) approaches 0 as x increases without bound, and fx) increases without bound as x decreases without bound. The y intercept is f0) 5. at x λ. P λ λe λ e b)

17 4.5. REVIEW PROBLEMS 3 b) If fx) 5 e x, then fx) approaches 5 as x increases without bound, and fx) decreases without bound as x decreases without bound. The y intercept is f0) 3. As x decreases without bound, e x approaches 0 and thus fx) approaches. The y intercept is f0) Applying the quotient rule to determine f x) yields f x) e x e x + ). c) If fx) 6 + e 3x, the y intercept is f0) 6 + e As x increases without bound, e 3x approaches 0, and so lim fx) 6 x + 0. As x decreases without bound, e 3x increases without bound, and so lim fx) 0. x Since e x > 0, there are no critical points and fx) increases for all x. To find f x), differentiate f x) using the quotient rule obtaining f x) 4ex e x ) e x + ) 3 There is one inflection point 0, f0)) 0, 5 ) since f 0) 0. The function fx) is concave down for x > 0 and concave up for x < 0.. a) If fx) Ae kx and f0) 0, then 0 Ae 0. Hence fx) 0e kx. Since f) 5, 5 0e k or e k 5. d) If fx) 3 + e x + e x, then fx) approaches 3 as x increases without bound, since e x approaches 0. Rewriting the function, by multiplying numerator and denominator by e x yields fx) 3ex + e x +. Then f4) 0e 4k 0e k ) 4 ) b) If fx) Ae kx and f) 3 as well as f) 0, then 3 Ae k and 0 Ae k, two equations in two unknowns. Division eliminates A, so 3 0 Aek Ae k, 3 0 ek e k, or ek 0 3.

18 3 CHAPTER 4. EXPONENTIAL AND LOGARITHMIC FUNCTIONS c) If Since e k 0 3, 3 A 0 3 and so A 9 0. Thus f3) 9 0 e3k 9 0 ek ) ) fx) 30 + Ae kx and f0) 50, then Ae 0 or A 0. Hence, fx) e kx. Since f3) 40, e 3k, 0 0e 3k, or e 3k. Thus f9) e 9k e 3k ) 3 ) d) 6 If fx) + Ae kx 6 and f0) 3 then 3, or A. + Ae0 6 Hence, fx) + e kx. 6 Since f5), + e 5k, + e 5k 6, or e 5k. Then, f0) 6 + e 0k 6 + e 5k ) 6 + ) 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 $0,000, 0, 000 V 5) 50, 000e 5k, e 5k 5, or k 5 ln 5. Hence V t) 50, 000e [/5 ln/5)]t and so V 0) 50, 000e ln/5) 4. The sales function is 50, 000e ln4/5) ) 4 50, 000 $8, Qx) 50 40e 0.x units, where x is the amount in thousands) spent on advertising. a) As x increases without bound, Qx) approaches 50. The vertical axis intercept is Q0) 0. The graph is like that of a learning curve. b) If no money is spent on advertising, sales will be Q0) 0 thousand units. c) If $8,000 is spent on advertising, sales will be Q8) 50 40e thousand or 3,07 units. d) Sales will be 35 thousand if Qx) 35, that is, if 50 40e 0.x 35, e 0.x 3 8, thousand or $9,808. x ln3/8) e) Since Qx) approaches 50 as x increases without bound, the most optimistic sales projection is 50,000 units.

19 4.5. REVIEW PROBLEMS The output function is Qt) 0 Ae kt. Since Q0) 30, 30 0 A or A 90. Since Q8) 80, e 8k, 40 90e 8k or e 8k 4 9. Hence, Q4) 0 90e 4k 0 90e 8k ) / ) 4 / units The population t years from now will be P t) 30 + e 0.05t. a) The vertical axis intercept is P 0) 30 0 million. + As t increases without bound, e 0.05t approaches 0. Hence, 30 lim P t) lim 30 t t + e 0.05t As t decreases without bound, e 0.05t increases without bound. Hence, the denominator + e 0.05t increases without bound and 30 lim P t) lim 30. t t + e 0.05t 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 e 5 5 since n ln e n. b) e ln since n e ln n. c) e 3 ln 4 ln e ln 43 ln e ln 64 e ln 3 3. d) ln9e ) + ln3e ) ln[9e )3e )] 8. a) 8 e 0.04x, e 0.04x 4, 0.04x ln 4 ln 7 3 ln 3. x b) 5 + 4e 6x, 4e 6x 4 6x ln 0, or x 0. c) 4 ln x 8, ln x, or x e d) 5 x e 3, ln 5 x ln e 3, x ln 5 3, or x 3 ln b) The current population is P 0) 0 million. c) The population in 0 years will be P 0) 30 + e 0.050) e million or 7,83,500 people. 9. Let Qt) denote the number of bacteria after t minutes. Since Qt) grows exponentially and 5,000 bacteria were present initially, Qt) 5, 000e kt. Since 8,000 bacteria were present after 0 minutes, 8, 000 Q0) 5, 000e 0k, e 0k 8 5, or k 0 ln 8 5.

20 34 CHAPTER 4. EXPONENTIAL AND LOGARITHMIC FUNCTIONS The bacteria will double when that is, when 5, 000e kt 0, 000 Qt) 0, 000, kt ln or t ln 0 ln k ln8/5) 4.75 or 4. min. and 45 seconds. 0. a) fx) e 3x+5, f x) e 3x+5 d 3x + 5) dx 6e 3x+5. b) fx) x e x, c) f x) e x d dx x ) + x d dx e x x x)e x. gx) ln x + 4x + lnx + 4x + ), g x) x + 4x + d dx x + 4x + ) x + x + 4x +. d) hx) x ln x x ln x, h x) x ddx ln x + ln x ddx ) x + ln x). f) gt) t log 3 t, ln 3 t ln t g t). a) Using the formula Bt) P + ln t). ln 3 + r ) kt k with P, 000, B 5, 000, r 0.08, and k 4, 5, 000, ) 4t 4.0) 4t 5 or t ln5/).57 years. 4 ln.0 b) Using the formula Bt) P e rt with P, 000, B 5, 000, and r 0.08, 5, 000, 000e 0.08t, or t ln5/) 0.08 e 0.08t 5,.45.. Compare the effective interest rates. The effective interest rate for 8.5 % compounded quarterly is + r ) k k ) or 8.5 %. e) ft) f t) t ln t, ) ln t)) t ) t ln t) ln t ln t). The effective interest rate for 8.0 compounded continuously is e r e or 8.55 %. 3. a) Using the present value formula P B + r ) kt k

21 4.5. REVIEW PROBLEMS a) 5. with B, 000, t 0, r 0.065, and k, P, ) -0, 000 $, b) Using the present value formula P Be rt with B, 000, t 0, and r 0.065, P, 000e ) $, , 000 P ) ) 0 P 8, 000 4, ) 0 b) 8, 000 P e P 8, 000e , 8.09 Bt), , 000e r r ln , or r 6% 6. At 6% compounded annually, the effective interest rate is + r ) k ) k At r % compounded continuously, the effective interest rate is e r. Setting the two effective rates equal to each other yields e r 0.06, e r.06, r ln or 5.83 %. 7. The average level of carbon monoxide in the air t years from now is Qt) 4e 0.03t parts per million. a) The rate of change of the carbon monoxide level t years from now is Q t) 0.e 0.03t, and the rate two years from now is Q ) 0.e parts per million per year. b) The percentage rate of change of the carbon monoxide level t years from now is [ Q ] ) t).e.03t % per year Qt) 4e.03t which is a constant, independent of time. 8. Let F p) denote the profit, where p is the price per camera. Then F p) number of cameras sold) profit per camera) 800p 40)e 0.0p. F p) 800[e 0.0p ) +p 40)e 0.0p 0.0)] 8e 0.0p 40 p) 0 when p 40. Since F p) > 0 and F is increasing) for 0 < p < 40, and F p) < 0 and F is decreasing) for p > 40, it follows that F p) has its absolute maximum at p 40. Thus the cameras should be sold for $40 apiece to maximize the profit. 9. a) fx) xe x, f f x) xe x ) + e x ) ) e x x) 0 whenx. e. f x) 4e x x ) 0 when x. f) e.

22 36 CHAPTER 4. EXPONENTIAL AND LOGARITHMIC FUNCTIONS x fx) e e 0 f x) f x) 0 +, ) is the absolute maximum e while, e ) is a point of inflection. b) fx) e x e x f x) e x + e x > 0 f x) e x e x 0 when x 0 There are no extrema. x 0 fx) 0 f x) f x) 0 + 0, 0) is a point of inflection. f x) 4e x + e x ) > 0. + e x ) 4 [ + e x ) 4e x ) ) 4e x ) + e x )e x ) )] 4e x e x ) + e x ) 3 0 when x 0 x 0 fx) 0 4 f x) f x) + 0 up 0, ) is a point of inflection. lim x 4 + e x 0 because the denominator increases beyond all bounds. lim x 4 + e x 4 because e x 0. So y 0 and y 4 are horizontal asymptotes. c) fx) 4 + e x 4 + e x ), f x) 4 ) + e x ) d dx + e x ) d) fx) lnx + ) f x) x x + 0 when x 0 f0) 0. f x) x) + x) x + ) 0

23 4.5. REVIEW PROBLEMS 37 when x ±. f±) ln. x 0 fx) ln 0 ln f x) f x) , 0) is the absolute minimum while ±, ln ) are points of inflection. ar 0 R 0 e kt, a e 5,000k. Since the half-life is 5,730, e 5,730k or k ln 5, 730 and a e 5,000 ln /5,730 e a) Rt) ln )t/5,730 R 0 e R, 960) ln ),960)/5,730 R 0 e R 0 e R 0 0. The value of the coin collection in t years is V t), 000e t. Hence, the percentage rate of change of the value of the collection expressed in decimal form) is V t) V t), ) 000e t, 000e t t t which will be equal to the prevailing interest rate of 7 % when 0.07 or ) t t Moreover, > 0.07 when 0 < t < 5.0 t and < 0.07 when 5.0 < t. t Hence the percentage rate of growth of the collection is greater than the prevailing interest rate when 0 < t < 5.0 and less than the prevailing interest rate when 5.0 < t. Thus the coin collection should be sold in 5.0 years. Thus about 78.89% of 4 C should be left in the shroud. b) Since 9.3% of 4 C is left in the forgery, [ ] ln )t 0.93 exp 5, 730 ln 0.93)5, 730) t 66 ln so Pierre d Arcis s suspicions were well founded. 3. ft) 70 Ae kt f0) 70 A or A 4. Thus ft) e kt. Let T 0 be the ideal temperature. Then T e k5) and T e k7) Solving with the SOLVE utility of our calculator) we get k 0.09 and T a) Dt) D )e 0.6t With D , D0) )e and D5) b) lim Dt) 0 t. Let a be the desired ratio. Then Rt) R 0 e kt,

24 38 CHAPTER 4. EXPONENTIAL AND LOGARITHMIC FUNCTIONS 5. a) To compute doubling time in terms of the rate set P P e rt and solve for t e rt or t ln r In the chart below find a comparison of the rules of 69, 70, 7 with the true doubling time. r True The rule of 69 is closest because is closest to ln. b) Writing Exercise Answers will vary. Ax) 0 ln x, for 0 x. x A x) 0 3 ln x x 0 when x e years. A0) so a person s aerobic capacity is maximized at about age 0. fx) f x) when x e e ln x) πx for x > 0 ln x) 4 e π x [ ln x ] 0 According to the graphing utility, the most common age is at ,.8977). 8. The Bronze age began about 5,000 years ago around 3,000 B.C.). The maximum percentage is P 5, 000) e ln )5,000/5, % 55% 9. a) ft) C e kt ) 0.008C 0.99 e k or k.0040 b) df Cke kt 00 df f c) Writing Exercise Answers will vary. 30. T t) 35e 0.3t 00k e kt 7 35e 0.3t or t 0.8 min. Rescuers have about 49 seconds before the girl looses consciousness. dt dt 350.3)e 0.3t At T 7 or e 0.3t 7 35 dt dt 35)0.3) 7 35 ) 8.64 Thus the girl s temperature is dropping at 8.64 degrees per minute. 3. a) y 0.5e 4x + e 4x ) y 0.5e 4x e 4x ) 0 when e 8x at x 0.

25 4.5. REVIEW PROBLEMS 39 b) Writing exercise Answers will vary. [ 3. k A exp E ] 0 RT [ k A exp E ] 0 RT [ k E0 exp E ] 0 k RT RT ln k E 0 ) k R T T 33. a) V t) V 0 ) t L b) V 0 875, L 8, t 5, V 5) ) The refrigerator will be worth $07.64 after 5 years $33.47 and 0.5 or 5% V t) V 0 L) t ln ) L The percentage rate of change is 00 V V 0 t ln t) L) ) V t) 00 L V 0 ) t L 00 ln ) L 34. ft) 30 Ae kt Solve for Ae kt ft) 30. f t) Ae kt k) kae kt k[30 ft)] where k is a constant of proportionality and ft) is the temperature of the drink. 35. P H log 0 [H 3 O + ] For milk and lime, P H m 3P H l. For lime and orange, P H l 0.5P H o. P H l 3..6 [H 3 O + ] l P t) + e t Hint: Use the equation writer on the HP48G to enter the function. For the TI-85, press nd CALC EVALF0.3/ + e x) ), x, 0), etc. a) year t P t) , 867, , 56, , 956, , 07, , 07, , 4, , 45, , 370, , 89, , 78, , 566, , 034, 385 b) This model predicts that the population will be increasing most rapidly when t.5 or in 95.

26 40 CHAPTER 4. EXPONENTIAL AND LOGARITHMIC FUNCTIONS 37. c) Writing exercise Answers will vary. y x y 3 x 3 y 5 x 5 ) x ) x ) x y 0.5) x x intersects y 4 ln x at.373, ) according to the graphing utility. ) x The graphs of y b x and y are b reflections of each other in the y axis 0 < b < ). The larger b the steeper the curve. 38. y 3 x 3 x/ y 3 x 3 x/ y 3 x 3 x The graphs of y 3 bx and y 3 bx are reflections of each other in the y axis 0 < b). The larger b the steeper the curve. 39. Hint: Use the equation writer on the HP48G to enter the function or yx) on the TI-85. y 3 x 40. log 5 x + 5) log x log 0 x + x) can be rewritten as lnx + 5) ln 5 ln ln x ln 0 lnx + x) 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, yx), and ZOOM/TRACE on the TI-85. x is a solution of this equation. Plot fx) log 5 x + 5) log x log 0 x + x) 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 x increases much more rapidly than any other argument, making fx) monotonically decreasing. 4. y fx) ln + x ) and y gx) x intersect at.66, 0.858) according to the graphing utility.

27 4.5. REVIEW PROBLEMS 4 even when n is large. 4. n n) n+ n + ) n , 665 3, , 58 5, , 59 63, , , , , , 65, 565, 089, , Thus n + ) n n n+ Mathematically one could reason as follows: x + ) x lim x x x+ x + lim x lim x exp lim x exp [ exp lim x [ exp lim [ exp x lim x x [ ln ) x + ) ] x + x + x ) x + ln x x + ) / x /x ] ) x + ) /)x + ) 3/ ] x + x ] e 0 x + Thus n + ) n n n+