Exponential and Logarithmic Functions · SECTION 4.1 ExponentialFunctions 303 27. g(x) = 2X —3....

Exponential and Logarithmic Functions

Exponential Functions

1. f(x) =4*;/(0.5) = 2,/(v/2) = 7.103, f {%) = 77.880, /(J) = 1.587

2. / (ar) = 3"+1; / (-1.5) - 0.577, / (\/3) = 20.115, / (e) = 59.439, /(-§) = 0.760

3. fl (x) = (f)""1; g(1.3) = 0.885, g(Vb) = 0.606, p(2tt) = 0.117, gr (-§) = 1.837

4- 5(*) =(|)2r; fl (0-7) =0.668, 5(^) =0.467, <? (A) =0.833, r; (§) =0.681

5. /(*) = 2* y 6. »(*) = 8"

a; 1/

-4l

16

-21

•i

0 1

2 4

4 16

7. /(*)-(i:

a: y

-2 9

-1 3

0 1

11

3

21

!)

9. / (x) = 3^

X y

-2 0.406

-1 1.104

0 3

0.5 4.946

1 8.155

1.5 13.445

2 22.167

X y

-2i

64

-11

8

0 1

1 8

2 64

8. frfart-ri.il

X y

-5 0.620921323

-1 0.90

0 1

5 1.61051

10 2.59374246

10. g(x) = 2e-

X y

-4 14.7761

-3 8.96338

-2 5.43656

-1 3.29744

0 2

1 1.21306

2 0.73576

3 0.44626

4 0.27067

301

302 CHAPTER4 Exponential and LogarithmicFunctions

11. /(x) = 2zandp(a:) = 2-x 12./(x)=3xand5(x) = (±)

13. / (x) = 41 and g (x) = V 14./(*) = (§)*and <,(*) = (!)=

15. From thegraph, / (2) = a2 = 9,soa = 3. Thus / (x) = 3Z.

16. From thegraph, / (—1) = a-1 = A, soa = 5. Thus / (x) = 5X.

17. From the graph, / (2) = a2 = ±, so a = \. Thus / (x) = (J)x.

18. From the graph, / (-3) = a~3 = 8,so a = ±. Thus / (x) = (|)x.

19. Ill 20. V 21.1 22. VI 23.11 24. IV

25. Thegraph of / (x) = -3* is obtained by reflecting the 26. Thegraph of/ (x) = 10"1 is obtained by reflecting the

graph ofy = 3X about thex-axis. Domain: (-oo, oo). graph ofy = 10*about they-axis. Domain: (-oo, oo).

Range: (-oo, 0). Asymptote: y = 0. Range: (0, oo). Asymptote: y = 0.

SECTION 4.1 Exponential Functions 303

27. g (x) = 2X —3. The graph ofg isobtained byshifting the 28. g (x) = 2X~3. The graph ofg isobtained by shifting the

graph of y = 2X downward3 units. Domain: (-oo, oo). graph of y = 2X to the right 3 units. Domain: (—oo, oo).

Range: (-3, oo). Asymptote: y = -3. Range: (0, oo). Asymptote: y = 0.

29. h(x) = 4+ (I)*. The graph ofh isobtained by shifting 30.h(x) = 6 - 3X. The graph ofh isobtained by reflecting

the graph ofy = (\)x upward 4units. Domain: tne graph ofy = 3X about the x-axis and shifting upward(-oo,oo). Range: (4, oo). Asymptote: y = 4. 6units- Domain: (-oo,oo). Range: (-oo,6).

Asymptote: y = 6.

31. / (x) = 10x+3. The graph of/ is obtained by shifting the 32. / (x) = - (\)x. Note that / (x) = - (i)1 = -5_x.graph ofy = 10x to the left3 units. Domain: (-oo, oo). thegraph of/ isobtained by reflecting thegraph of

Range: (0, oo). Asymptote: y = 0. y = 5X aboutthe y-axis and aboutthe x-axis. Domain:

(-00,00). Range: (-00,0). Asymptote: y = 0.

So

304 CHAPTER 4 Exponential and LogarithmicFunctions

33. y = -ex. Thegraph of y = -ex is obtained from the 34. / (x) = 1 - ex. Thegraph of / (x) = 1 - ex is obtained

graph ofy = ex by reflecting it about the x-axis. Domain: by reflectingthe graph ofy = ex about the x-axis and then

(—oo, oo). Range: (-oo, 0). Asymptote: y = 0. shifting upward 1 unit. Domain: (—oo, oo). Range:

yt (-oo, 1). Asymptote: y = 1.

35. y = e~x - 1.The graph ofy = e~x - 1 isobtained from 36. / (x) = -e"x. The graph of/ (x) = -e x isobtainedthe graphof y = ex by reflecting it about the y-axis then

shiftingdownward 1 unit. Domain: (-co, oo). Range:

(—l,oo). Asymptote: y = —1.

by reflecting the graphof y = ex about the y-axis and then

aboutthe x-axis. Domain: (—00,00). Range: (—00,0).

Asymptote: y = 0.

37. y= ex"2. The graph ofy = ex_2 is obtained from the 38. / (x) = ex~3 + 4. The graph of/ (x) = ex~3 + 4isgraph ofy = exbyshifting ittothe right 2 units. Domain: obtained byshifting the graph ofy = ex totheright(-00,00). Range: (0,00). Asymptote: y = 0. 3units, and then upward 4units. Domain: (-00,00).

Range: (4,00). Asymptote: y = 4.

39. Using the points (0,3) and (2,12), we have / (0) = Ca° = 3 <* C= 3. We also have / (2) = 3a2 = 12 <&a2 = 4 «• a = 2 (recall that for an exponential function / (x) = ax we require a > 0). Thus / (x) = 3 •2x.

SECTION 4.1 ExponentialFunctions 305

40. Using the points (-1,15) and (0,5) we have / (0) = Ca° = 5 & C = 5. Then / (-1) = 5a_1 = 15 &a"1 =3 <& a= i.Thus/(x)=5(i)x.

41. (a)

(b) Since g (x) = 3 (2X) = 3/ (x) and / (x) > 0, the

height of the graph of g (x) is always three times the

height of the graph of/ (x) = 2X, so the graph of g is

steeper than the graph of/.

42. (a)

(b) /(x) = 9X'2 = (32)I/2 = 32x/2 = 3X = g(x). Sof(x)=g (x), and the graphs are the same.

43. / (x) = 10x, so

44.

f(x + h)-f(x)h

,x+h10 -10a 10x10ft-10x 1rtX10h-l

: = 10 : ,

45.

X /(x) = x3 g(x)=3x

0 0 1

1 1 3

2 8 9

3 27 27

4 64 81

5 125 243

6 216 729

7 343 2187

8 512 6561

9 729 19,683

10 1000 59,049

15 3375 14,348,907

20 8000 3,486,784,401

46.

10

1,000,000

306 CHAPTER 4 Exponential and Logarithmic Functions

47. cosh(—x) =-* _i_ --(-*) „-*_!_ „* „* J,e~xe~- +e e--" + e" e" + e

= cosh x

>_a: _ C-(-x) o~x _ ^ _r._a: -1- o* a* _ a~x48. sinhi—x) = = = = = —smhx

v ' 2 2 2 2

49. (coshx)2 -(sinhx)2 ={^f1)2 ~{f^f1)2 =\ (e2* +2+e~2x) -J(e2x -2+e-2x) =\+\=1Qx _ e-x ey + e-y ex _j_ e-x &y _ &-y

50. sinh x cosh y + cosh x sinh y = • 1 •y 2 2 2 2

ex+v ^_ ex~y —ev~x —e~^x+y^ ex+v —ex~y + ey~x —e~^x+w^ ex+v —e~^x+y^= + = = sinh (x + y)

51. (a) From the graphs below,we see that the graph of/ ultimately increasesmuch more quickly than the graph ofg.

(i) [0,5] by [0,20] (ii) [0,25] by [0,107] (Hi) [0,50] by [0,108]10,000.000

8,000,000 /( 9

6,000,000

4,000,000

2,000,000

0J10 20

100.000,0001

80,000,000

60,000.000

40,000,000

20,000,000

/ 9

0 20 40

(b) From the graphs in parts (a)(i) and (a)(ii), we see that the approximate solutions are x « 1.2 and x « 22.4.

52. (a) (i) [-4,4] by [0,20] (ii) [0,10] by [0,5000] I 0,20] by [0,105]100,000-I

80,000

60.00091 //

40,000

20.000

010 20

(b) From the graphs inparts (i)and (ii), we see that the solutions of3X = x4are x « -0.80, x « 1.52and x w 7.17.

54.53.

The larger the value of c, the more rapidly the graph of

/ (x) = c2x increases. Also notice that the graphs are just

shifted horizontally 1 unit. This is because ofour choice

of c; each c inthis exercise isof theform 2k. So

/ (x) = 2k •2X = 2x+fe.

,0i= 4 c-1 c=l

8

6 .0= 0.5

4

2— c = 0.25

-2 0 2 4

The larger the value of c, the more rapidly the graph of

/ (x) = 2CX increases. In general, / (x) = 2CX = (2C)X ;

so, for example, / (x) = 22x = (22)x = 4X.

55.

)

2

1

010 » so 40

Note fromthe graph that y = [1 + (l/x)]x approaches e

as x get large.

57. (a)... • lOn

\\ \ 1 8\\\ 1 6

xv\ 4

a=1.5-> ,a=\

a=2, J J ,a=0.5

-4 -2 A 2 4

(b) As a increases the curve y= - fex/a + e x/a ]

flattens out and the y intercept increases.

59. y = — has vertical asymptote x = 0 and horizontalx

asymptote y = 0. As x —• —oo, y —> 0, and as x —• oo,

61. g (x) = ex + e~3x. The graph ofg (x) isshown inthe

viewing rectangle [—4,4] by [0,20]. Fromthe graph,we

see that there is a local minimum ofapproximately 1.75

when x « 0.27.


56.

Fromthe graph,we see that / (x) = [1 - (l/x)]x

approaches 1/e « 0.368 as x get large.

58. y = 21/x has vertical asymptote x = 0 and horizontalasymptote y = 1.

-I .2

60. p (x) = xx. Notice that g (x) is only defined for x > 0.

The graphofg (x) is shownin the viewingrectangle

[0,1.5] by [0,1.5]. Fromthe graph, we see that there is a

local minimum ofabout 0.69 when x « 0.37.

62. y = 10*

(a) From the graph, we see that the function is increasing

on (-oo, 0.50] and decreasingon [0.50,oo).

(b) From the graph, we see that the range is

approximately (0,1.78].


63. y = xe" (a) From thegraph, we see that the function / (x) = xex is increasing on

(-co, 1]anddecreasing on [1,oo).

(b) From thegraph, we see thatthe rangeis approximately (-oo, 0.37).

-I

64. D (t) = 50e_0-2t. So when t = 3we have D (3) = 50e_0-2(3) » 27.4 milligrams.

65. m(i) = 13e-°015t

(a) m (0) = 13 kg.

(b) m (45) = i3e-0015(45) _ i3e-°-675 _ 6 619 kg ^^ the mass ofme radioactive substance after 45days isabout6.6 kg.

66. (a) m (0) = 6e-°087<°> « 6 grams

(b) m (20) = 6e-0"087(2°) « 6 (0.1755) = 1.053. Thus approximately 1gram ofradioactive iodine remains after 20days.

67. v(i)=80(l-e-°-2t)

(a) v(0) = 80(1 -e°) =80(1-1)=0.

(b) v(5) =80 (l- e-02(5)) «80 (0.632) =50.57 ft/s. So thevelocity after 5 s is about 50.6 ft/s.

v(10) =80 (1 - e-°-2(10)) »80 (0.865) =69.2 ft/s. So thevelocity after 10 s is about 69.2 ft/s.

(d) The terminal velocity is 80 ft/s.

68. (a) Q(5) =15 (l- e-°04(5)) «15(0.1813) =2.7345. Thusapproximately 2.7 lb ofsalt are in the barrel after 5 minutes.

(b) Q(10) =15 (l- e-004(10)) *15(0.3297) =4.946.Thusapproximately 4.9 lb ofsalt are in the barrel after 10 minutes.

(d) The amount of salt approaches 15 lb. This is to be expected,since

50 gal x 0.3 lb/gal = 15 lb.

69. P(t) =1200

1 + lle-°-2t

1200(a)P(0) =

12001 + lle-°-2(°) "1 + 11

1200

= 100.

(c)

(c)

1200 1200<b> P(10) = i +iie-o^o) - 482. P(20) = , + lle_0,2(20) « 999. P(30) =TT_.2(30)

-0.2t(c) As t —> oo we have e 0, so P {t)1200

1 + 0= 1200. The graph shown confirms this.

1168.


70. n(t) =5600

0.5 + 27.5e-°044t(b)

(a)n(0) = S|29 =200

(c) From the graph, we see that n (t) approaches about 11,200 as t gets

large.

"• °W- 1+2.9e-o °u •S° D<20> - 1+2.!e-omM "160° *72. (a) Substituting no = 50 and t = 12, we have

„ (m = 300K' 0.05 + (^2 _ 0.05) e-°-M<12>

(c) The population approaches 6000 rabbits.

5164.

(b) 12,000 •]

10,000 \nB =12,0008000

"°=800° 6000

4000

2000//"o =50

n0 =2000-/ 0 2 4/ 6 t8 10 12 14

n0 = 500-/

73. Using the formula A(t) = P (1 + i)fc with P = 5000, 74. Using the formula A(t) = P (1 + i)fc with P = 5000,

i = 4% per year = -pr- permonth, and k = 12•number

ofyears, we fill in the table:

rate per year , , . , n _i = £r—^— per month, and k = 12 • 5 = 60 months,

we fill in the table:

75. P =

(a)

(b)

(c)

76. P =

(a)

(b)

(c)

77. P =

(a)

(b)

(c)

(d)

Time (years) Amount

1 $5203.71

2 $5415.71

3 $5636.36

4 $5865.99

5 $6104.98

6 $6353.71

Rate per year Amount

1% $5256.25

2% $5525.39

3% $5808.08

4% $6104.98

5% $6416.79

6% $6744.25

10,000, r = 0.10, and n= 2. So A(t) = 10,000 (l + ^)2t = 10,000 •1.052t.

A (5) = 10,000 •1.0510 « 16,288.95, andso thevalue of the investment is$16,288.95.

A (10) = 10,000 •1.0520 w 26,532.98, and sothe value of the investment is$26,532.98.

A (15) = 10,000 •1.0530 « 43,219.42, and sothe value of the investment is$43,219.42.

=4,000, r = 0.16, and n = 4. So A(t) = 4000 (1 + ^)4t = 4000 •1.044t.

A(4) = 4000 •(1.04)16 « 7491.92, and sothe amount due is$7,491.92.

A(6) = 4000 •(1.04)24 « 10,253.22, and so the amount due is$10,253.22.

A (8) = 4000 • (1.04)32 « 14,032.23, and sothe amount due is$14,032.23.

/ 0 09 \rat / 0 09=3000 and r= 0.09. Then we have A(t) =3000 f1+ ^-^ J ,and so A(5) = 3000 I1+^^Ifn = \,A (5) = 3000 (l + ^-f = 3000 •1.095 « $4,615.87.

Ifn = 2, A(5) = 3000 (l + P^2)10 = 3000 •1.04510 « $4,658.91.

Ifn = 12, A(5) = 3000 (l + ^)60 = 3000 •1.007560 * $4,697.04.

Ifn = 52, A(5) = 3000 (l + ^f)26° « $4,703.11.


(e) Ifn= 365, A(5) = 3000 (l + |f)1825 « $4,704.68.

(f) Ifn= 24 •365 = 8760, A(5) = 3000 (l + §^)438°° « $4,704.93 .(g) If interest iscompounded continuously, A (5) = 3000 •e045 « $4,704.94.

(7» \ 4(5) / 7. \;1+-J ,and so A(5) =4000 (1 +^J(a) Ifr = 0.06, ,4 (5) = 4000 (l + ^P)20 = 4000 •(1.015)20 « $5,387.42.

(b) Ifr = 0.065, A(5) = 4000 (l + «)20 = 4000 •(1.01625)20 « $5,521.68.

(c) Ifr = 0.07, A(5) = 4000 (l + ^f7-)20 = 4000 •(1.0175)20 « $5,659.11.

(d) Ifr = 0.08, A(5) = 4000 (l + ^f8-)20 = 4000 •(1.02)20 « $5,943.79.

79. We find the effective rate with P=1and i=1. SoA= (l+-)(i) n= 2, r = 0.085; ,4 (2) = (l + ^)2 = (1.0425)2 « 1.0868.

(ii) n = 4, r = 0.0825; A(4) = (l + 2^p)4 = (1.020625)4 « 1.0851.(iii) Continuous compounding: r = 0.08; A (1) = e008 » 1.0833.

Since (i) is larger than the others, the best investment is the one at 8.5% compounded semiannually.

80. We find the effective rate for P = 1 and t = 1.

(i) Ifr= 0.0925 and n= 2, then A(2) = (l + S^)2 = (1.04625)2 « 1.0946.(ii) If r = 0.09 and interest iscompounded continuously, then A (1) = e009 « 1.0942.

Since the effective rate in (i) isgreater than the effective rate in(ii), we can see that the account paying 9^% per yearcompounded semiannually is the better investment.

81. (a) We must solve for P in the equation 10000 = P (1 + P^5)2(3) = P (1.045)6 «*> 10000 = 1.3023P ^P = 7678.96. Thus the present value is $7,678.96.

(b) We must solve for P in the equation 100000 = P (l + ^)12(5) = P(1.00667)6«• P = $67,121.04.

82. (a) A(t) =5000 (1+^ J =5000 (1.045)2f =5000 (1.092025)'. (b)(c) A (t) = 25000 when t w 18.28 years.

v60

20

100000 = 1.4898P

83. (a) In this case the payment is $1 million.

(b) In thiscase thetotal pay is 2 + 22 + 23 H h 230 > 230 cents = $10,737,418.24. Since thisis much more thanmethod (a), method (b) is more profitable.

84. Since / (40) = 240 = 1,099,511,627,776, itwould take a sheet ofpaper 4 inches by1,099,511,627,776 inches. Since thereare 12 inches in a foot and 5,280 feet in a mile, 1,099,511,627,776 inches « 1.74 million miles. So the dimensions of the

sheet of paper required are 4 inches by about 1.74 million miles.

SECTION 4.2 Logarithmic Functions 311

V-^IHHHj

Logarithmic Functions

Logarithmic Exponential1.

form

log88 = l

og8 64 = 2

iog84=|og8512 = 3

8 8 = _1

"8 64 = ~2

3. (a) 52 = 25

(b) 5° = 1

7. (a) ex = 5

(b) e5 = y

11. (a)log8l=-l

(b)log2(|)=-3

15. (a) log33 = 1

(b)log3l=log33° = 0

(c)log332 = 2

17. (a)log636 = log662=2

(b)log981 = log992=2(c)log7710 = 10

19. (a)log3(£)=log33-3 = -3(b) log10 v/10 = log10 101/2 = §(C)log50.2 = log5(i)=log55-1 = -1

21. (a) 2log237 = 37

(b) 3,oga 8 = 8

(c) eln ** = v/5

23. (a)log80.25 = log88-2/3 = -|(b) lne4 =4

(c) In (- l =lne-1 =-1

form

81 =8

82=64

82/3 = 4

83 = 512o-l _ 18 - 8o-2 _ 1

8 -64

4. (a) 10"! =0.1

(b) 83 = 512

8. (a) e2 = x + 1

(b) e4 = x - 1

2.Logarithmic

form

log, 64 = 3

log42=±

log48=|

log, & = -2iog41 = -ilog, 3L =-|

Exponential

form

43 = 64

41'2 = 2

43/2 = 8

4~2 = i* 16

4-1/2 - I1 — 2

4-5/2 J.^ 32

5. (a) 81/3 = 2

(b) 2"3 = I

9. (a) log5 125 = 3

(b) log10 0.0001 = -4

12. (a) log, 0.125 = -§ 13. (a) In 2 = a;(b) log7 343 = 3 (b) In y = 3

16. (a) log554 =4

(b)log,64 = log,43 = 3

(c)log99 = l

18. (a) log2 32 = log2 25 =5

(b)log8817 = 17

(c)logGl = logc6° = 0

6. (a) 34 = 81

(b) 82/3 = 4

10. (a) log101000 = 3

(b) log81 9 = I

14. (a) ln0.5 = .x + l

(b) lni = 0.5x

20. (a) log,, 125 = log5 53 = 3

(b)log,97 = log,9491/2 = ±

(c) log9 s/3 =log9 31/2 =log9 (9I/2) V2 =log9 9

22. (a)e1,,7r = tt

(b) 10log5 =5

(c) 10log87 = 87

1/4

1/224. (a) log, s/2 =log, 21/2 =log, (41/2) "=log, 41'4_ i

4

(b) log, (1) =log, 2-1 =log, (41'2) ~* =log, 4"1'2_ _i

1

25. (a) log2 a: = 5 «=> x = 25=32

(b) x = log2 16 = log2 24 = 4

(C) log, 8=log, 23 =log, (41/2)" =log, 43/226. (a) log5 x = 4 & x = 54 = 625

(b).,: = log10(0.1) = log1010-1 = -l

312 CHAPTER4 Exponential and Logarithmic Functions

27. (a) x = log3 243 = log3 35 = 5

(b) log3 x = 3 «• x = 33 = 27

29. (a) log10 x = 2 <^ x = 102 = 100

(b) log5 x = 2 «* x = 52 = 25

28. (a)x = log42 = log441/2 = ±(b) log,x = 2 <s> x = 42 = 16

30. (a) log* 1000 = 3 «» x3 = 1000 «» x = 10

(b) logx 25= 2 «• x2 = 25 «» x = 5

31. (a) logx 16 = 4 <* x4 = 16 <*> x = 2 32. (a) logx 6 = \ <* x1/2 = 6 & x = 36(b)logx8=| «- x3/2 =8 <* x=82/3 =4 (b)logx3 =i «- x1'3 =3 «• x=27

33. (a) log2« 0.3010

(b) log 35.2 « 1.5465

(c) log (|) « -0.1761

35. (a) ln5« 1.6094

(b) In 25.3 « 3.2308

(c) In (1 + V3) « 1.0051

34. (a) log 50 « 1.6990

(b) logy^« 0.1505

(c) log (3\/2) « 0.6276

36. (a) In 27 * 3.2958

(b) In 7.39 « 2.0001

(c) In 54.6 « 4.0000

37. Since the point (5,1) isonthe graph, we have 1 = loga 5 <=>• a1 = 5. Thus the function isy = log5 x.

38. Since the point (5, -l)is on the graph, we have -1 = loga (5) «=>• a-1 = | «=> a = 2. Thus the function isy = log2 x.

39. Since the point (3, §) is on the graph, we have 5 = loga 3 & a1/2 =3 «• a = 9. Thus the function isy = log9 x.

40. Since the point (9,2) isonthe graph, we have 2 = loga 9 «=> a2 =9 •£> o = 3. Thus the function isy = log3 x.

41. II 42. V 43. Ill

47. The graph of y = log, x is obtained from the graph of

y = 4X by reflecting it about the line y = x.

44. IV 45. VI 46.1

48. The graph ofy = log3x is obtained from the graph of

y = 3X by reflecting it about the line y = x.


49. / (x) = log2 (x - 4). The graph off isobtained from the 50. / (x) = - log10 x. The graph off isobtained from thegraph of y = log2 x byshifting it to theright 4 units. graph ofy = log10 x byreflecting it about thex-axis.

Domain: (4,oo). Range: (-co, oo). Vertical asymptote: Domain: (0,oo). Range: (-oo, oo). Vertical asymptote:x = 4. x = 0.

y ♦ i y

51. g (x) = log5 (-*). The graph ofg isobtained from the 52. g (x) = In(x + 2). The graph ofg isobtained from the

graph of y = log5x by reflecting it about the y-axis.

Domain: (-oo,0). Range: (-00,00). Vertical asymptote:

x = 0.

graph of y = In x by shifting it to the left 2 units. Domain:

(—2,00). Range: (—00,00). Vertical asymptote: x = —2.

53. y = 2 + log3 x. The graph ofy = 2 4- log3 x isobtained 54. y = log3 (x - 1) - 2. The graph ofy = log3 (x - 1) - 2from thegraph of y = log3 x by shifting it upward 2 units. is obtained from the graph of y = log3 x by shifting it to

Domain: (0,00). Range: (-00,00). Vertical asymptote: theright 1 unit andthen downward 2 units. Domain:x = 0. (1,00). Range: (—00,00). Vertical asymptote: x = 1.


55. y = 1 - log10 x. The graph ofy = 1- log10 x isobtained 56. y = 1 + In(-x). The graph ofy = 1 + In(-x) is

from thegraph ofy = log10 x byreflecting it about the obtained from thegraph ofy = Inx byreflecting itaboutx-axis, and then shifting it upward 1 unit. Domain: the y-axis and then shifting it upward 1 unit. Domain:

(0,oo). Range: (-oo, oo). Vertical asymptote: x = 0. (-oo, 0). Range: (—oo, oo). Vertical asymptote: x = 0.

57. y = |lnx|. The graph of y = |lnx| is obtainedfromthe

graph of y = In x by reflectingthe part of the graph for

0 < x < 1 about the x-axis. Domain: (0, oo). Range:

[0,oo). Vertical asymptote: x = 0.

>'4

58. y = In |x|. Note that y =lnx ifx>0

ln(-x) ifx<0

graphof y = In |x| is obtainedby reflecting the graph of

y = lnx about the x-axis. Domain: (—oo, 0) U (0, oo).

Range: (—oo, oo). Vertical asymptote: x = 0.

y*

The

59. / (x) = log10 (x + 3). We require thatx + 3 > 0 <$• x > -3, so thedomain is (-3, oo).

60. / (x) = logs (8 - 2a0- Tnenwe musthave 8 - 2x > 0 &

61. g (x) = log3 (x2 - 1). We require that x2 - 1 > 0 <=>(-00,-l)u(l,oo).

62. g (x) = In (x - x2). Then we must have x —x2 > 0 &x (1 —x) > 0. Using the methods fromChapter 1 with the

endpoints 0 and 1, we get the table at right. Thus the domain

is (0,1).

8 > 2x ^ 4 > x, and so the domain is (—oo,4).

x2 > 1 ^ x < -1 or x > 1, so the domain is

Interval (-00,0) (o,D (l,oo)

Sign ofx

Sign of 1 - x +

+

+

+

Sign of x (1 —x) - + -

63. h (x) = In x + In (2 - x). Werequire that x>0and2-x>0 «=> x>0andx<2 •«• 0 < x < 2, so thedomain is (0,2).

64. h (x) = \/x-2 - log5(10 _ x) • Tnenwe mustnave £ - 2 > 0 and 10 - x > 0 <=> x > 2 and 10 > x «=>2 < x < 10. So the domain is [2,10).


65. y= log10 (1 - x2) has domain (-1,1), vertical 66. y= In (x2 - x) = In (x (x - 1)) has domainasymptotesx = -1 and x = 1, and local maximum y = 0 (-co,0) U (1, oo), vertical asymptotes x = 0 and x = 1,

at x = 0. and no local maximum or minimum.

67. y = x + In x has domain (0, oo), verticalasymptote

x = 0, and no local maximum or minimum.

10

-10

68. y = x (In x)2 has domain (0,oo), no vertical asymptote,local minimum y = 0 at x = 1, and local maximum

y « 0.54 at x« 0.14.

69. y = hasdomain (0,oo),vertical asymptote x = 0, 70. y = x log10 (x + 10) has domain (-10, oo), verticalx

. . . , f/^„ n , ,„„„, „,„„;„,..„, „,^,nQ7 asymptote x = -10, and local minimum y « -3.62 athorizontal asymptote y = 0, and local maximum y « 0..37 J v ' yatx« 2.72. xw-5.87.

71. The graphofg (x) = y/x growsfaster than the graphof 72. (a)

/ (x) = In x.

to y

•to V, ^r

•10

(b) From the graph, we see that the solution to the

equation y/x = 1 + In (1 + x) is x «s 13.50.


73. (a)

(b) Noticethat / (x) = log (ex) = logc -+- logx, so as c

increases, the graphof / (x) = log (ex) is shifted

upward log c units.

75. (a) / (x) = log2 (log10 x). Since the domain of log2 x is the positive real numbers, we have: log10 x > 0 «•x > 10° = 1.Thus the domain off (x) is (1, oo).

(b) y = log2 (log10 x) <s> 2y = log10 x «- 102" = x. Thus f~l (x) = 102*.

76. (a) / (x) = In(In(Inx)). We must have In(Inx) > 0 & Inx > 1 & x > e. Sothedomain off is (e,oo).

(b) y = In (In (lnx)) <& ev = In (lnx) •& ee = lnx & ee = x. Thus the inverse function is

/-1(x) = ce'V

77. (a)/(x) =2X

1 + 2*" " 1+2

y + y2x = 2x &

y = 2x-y2x = 2x(l-y) &

y2X =1-0

r>(,)=log2(^_).

y = <&

<& x = log2 (*)• Thus

74. (a)

(b) Asc increases, thegraphoff (x) = c log (x)

stretches vertically by a factor ofc.

(b)1-x

> 0. Solving this using the methods from

Chapter 1, we start with the endpoints, 0 and 1.

Interval

Sign of x

Sign of 1 —x

Sign of1-x

(-oo,0) (0,1) (l,oo)

+

+

+

+

+

78. Using J=0.7/o we have C=-2500 In (y ) =-

79. Using D=0.73D0 we have A= -8267 In (—j =-8267 In 0.73 « 2601 years.

80. Substituting N=1,000,000 we get t=3logW50) =3log 20,000 ^ ^ g6 ^q^log 2 log 2

81. When r = 6% we have t = —— « 11.6 years. Whenr = 7% we have i = —— « 9.9 years. And when r = 8% we0.06 0.07

L , ln2 „„have t = ——- « 8.7 years.0.08

82. Using k = 0.25 and substituting C = O.QCq we have

t=-0.25In (l - —J=-0.25In(1 - 0.9) =-0.25In0.1 «0.58hours.83. Using A=100 and W=5we find the ID to be ^A/W) = log (2-100/5) = log40 ^ g32 Usj ^ =10Q

log 2 log 2 log 2

Thus the domain off 1(x)is(0,1).

2500 In 0.7 = 891.69 moles/liter.

and W= 10 we find .he ID to be ^fA{W) = !»«(»• MO/10) _ log20 Sq sma||erlog 2 log 2 log 2

5.23

4.321.23 times harder.

icon is

SECTION 4.3 Laws of Logarithms 317

84. (a) Since 2 feet= 24 inches, the height ofthe graph is 224 = 1677216 inches. Now, since there are 12 inches per foot and5280 feet per mile, there are 12 (5280) = 63,360 inches per mile. So the height ofthe graph is '-g™6 « 264.8, orabout 265 miles.

(b) Since log2 (22'1) = 24, we must be about 224 inchest 265 miles to the right ofthe origin before the height ofthegraph of y = log2x reaches 24 inchesor 2 feet.

85. log (log 10100) = log 100 = 2log (log (log lO80080')) = log (log (googol)) = log (log 10100) = log (100) = 2

86. Notice that log„x is increasing for a > 1. So we have log4 17 > log,, 16 = log442 = 2. Also, we havelog5 24 < loS5 25 = loS5 52 = 2- Thus' log5 24 < 2 < log4 17.

87. The numbers between 1000 and 9999 (inclusive)each have 4 digits, while log 1000 = 3 and log 10,000 = 4. Since

[logx] = 3 for all integers x where 1000 < x < 10,000, the number ofdigits is [logx] + 1. Likewise, ifx is an integerwhere 10""1 < x < 10", then x hasn digits and [logx] = n - 1. Since [logx] = n - 1 •» n= [logxj + 1, thenumberof digits in x is [logx] + 1.

4.3 Laws of Logarithms1. log., s/Tt = log3 3:,/2 = f 2. log2 160 - log2 5= log2 ±f = log2 32 = log2 25 = 5

3. log 4+log 25 = log (4 •25) = log 100 =2 4. log ^^ = log 10_3/2 = - §5. log, 192 - log, 3 = log, ±f = log, 64 = log, 43 = 36. log12 9+ log12 16 - log12 (9 •16) = log12 144 = log12 122 = 27. logo 6- log2 15 + log2 20 = log2 ± + !og2 20 = log2 (| •20) = log2 8 = log2 23 = 3

8. log3 WO - log3 18 - log3 50 =log, (rg^) =loS3 (I) =lo§3 3~2 =-29. log, 16100 = log, (42)100 = log, 4200 =200 10. log2 833 = log2 (23)33 = log2 299 = 99

11. log (log 1010000) = log (10,000 log 10) = log (10,000-1) = log (10,000) = log 10" =4 log 10 = 4

12. In (ln<r°°) =In (e200lne) =In e200 =200 lne =20013. log2 2x = log2 2 + log2 x = 1 + log2 x 14. log3 (5y) = log3 5 + log3 y

15. log2 [x (x - 1)] =log2 x+log2 (x - 1) 16. log., (|) =log5 x- log5 217. Iog610 =101og6 18.ln(V5)=ln(zl/2) =±\nz19. log2 (AB2) = log2 A+ log2 B2 = log2 A+ 2log2 B 20. log6 ^17= \ logc 17

21. log3 (xy/y) = log, x+ log3 ^y = log3 x+ \ log3 y 22. log2 (xy)10 = 10 log2 (xy) = 10 (logo X+ log2 y)

23. log5 ^TT = I log5 (x2 + 1)

24. log,, (-£g J=log,, x2 - log,, (yz3) =2log„ x- (log„ y+3log,, 2)25. In \/ab = \ In ab = § (In a + lnb)

26. In \/3r^s =JIn (3r2s) = \ [in 3+ln r2 +ln s] = |(ln3+2In r+In s)

27. log( ?-%- ) =log(x3y4) -log26 = 31ogx +41ogy-61og2

318 CHAPTER 4 Exponential andLogarithmic Functions

2

28. log -^-j=c =log a2 - log (64 VE) =2log a- (4 log6+±log c)

29. log2 ( =^J=log2 x+log2 (x2 +1) -Jlog2 (x2 -1)30> l0g5 Vfri =2l0g5 (xTl) =*[loS5 (* -1) -log5 (x +1)]31. \n(xJij =lnx+±ln(^) =lnx+ \ (lny - lnz)32. ln

3x2

(* + l)1033. log Vx^+F = \ log (x2 + y2)

x34. log 3/-_— = logx - log v"l-x = logx - \ log (1 -

x2+4

= ln (3x2) - ln(x+ 1)10 = ln3+ 2lnx - 10 ln(x+ 1)

35. logx' + 4

= ±log(x2 + 1)(x3 - 7)2 2*"6 (x2 + 1) (x3 - 7)2

= 1 [log (x2 + 4) - log (x2 +1) - 2log (x3 - 7)]

=1[log (x2 +4) - log (x2 +1) (x3 - 7)2]

36•log yjxy/yjz =1log (xy/yJP) =i (logx +log \/y^) =\ [logx +1log {yy/z)]= 1 [logx + 1 (logy + 1loga)] = 1logx + J logy + §logz

37. ln^^^ =ln(x3V^rl) -ln(3x +4) =31nx+iln(x-l)-ln(3x +4)3x + 4

10a

38' 1<>g x(x2 +l)(x4+2) =l0g 1QI ~l0g I* ^ +X) (a;4 +2)] =*- N* +lQg (** +1) +lQg (*4 +2)]39. log3 5+ 5log3 2 = log3 5+ log3 25 = log3 (5•25) = log3 160

40. log 12 +1log7- log 2=log (12V7) - log 2=log ^p =log (6V7)41. log2 A+log2 B- 21og2 C=log2 (XB) - log2 (C2) =log2 (^\42. log5 (x2 - 1) - log5 (x - 1) = log. x'-l

X - 1 "°° X - 1

43. 4logx - i log (x2 + l) + 2log(x- 1) = logx4 - log ^Tl + log(x - l)2

= logE(x-l)(x+l)

= log5(x + l)

MwTih^'-^M^^-)44. ln(a + 6) + ln(a-6)-21nc = ln[(a+ 6)(a-6)]-ln(c2) = lna2-62

45. In5 +21nx +31n (x2 +5) =ln (5x2) +ln (x2 +5)3 =ln [5x2 (x2 +5)3].2\246. 2[log5 x+2log5 y-3log5 z] =2log5 3L =log5 0^\ =log

47. |log(2x+l) + ±[log(x-4)-log(x4-x2-l)] = log V2x~+\ + \ log

xV5 ~6

x — 4

x4 - x2 - 1

-'-(v^-'J^tt)LJC

48. logab + clogad-rlogas = loga (bdc) - logasr = loga —

SECTION4.3 Lawsof Logarithms 319

49. log2 5=J^? « 2.321928log 2

log 16

log 3

log 2.61

50. log5 2= r^| « 0.4306775 log5

52. log6 92 =^? « 2.52365851. log316 =

53. log7 2.61 =log 7

2.523719

0.493008 54. log6 532 =

log 6

log 532log 6

3.503061

55. log4 125 =128125 ^ 3482892 56. log12 2.5 = t^? « 0.368743

58. Note that log,•"&)•• x (by the change of base formula). So

the graphof y = logc x is obtainedfrom the graphof y = ln x by

either shrinking or stretching vertically by a factor of -— depending

on whetherln c> 1 or ln c < 1. All of the graphspass through(1,0)

because logc 1 = 0 for all c.

lne 1

log4 °"~" log12

57. iog3 x=^ =_ =_ lnl. The graph of, =^ In* isshown in the viewing rectangle [—1,4] by [-3,2].

>

2-a = e

a = 5

0

— a = 10

/rf/ 2 4

-2-

59. loge =ln 10 ln 10

60. (log25)(log57) =log 5 log 7 _ log 7log 2 log 5 log 2

= log2 7

61. -h(.-^m) -,(7Z^=) -.(-^.=±£3) -(5££fJ)= ln (x + y/x2 - 1)

Po62. From Example 5(a), P =

(t + lYstudent should get a score of 30.

63. (a) logP =logc -klog W & logP =logc - log Wfc «*• logP =log(^) <^ P=^-.

(b) Using fc = 2.1 and c = 8000, when W = 2 we have P = -rjy « 1866 and when W = 10wehave P = —^

64. (a) log 5 = logc + A; log A & log 5 = log c+ log Ak & \ogS = log (cAk) & S = cAk.

(b) IfA = 2A0 when A; = 3 we get S = c(2A0f = c•23 •A% = 8•c.4o- Thus doubling the area increases the specieseightfold.

65. (a) M = -2.51og(B/B0) = -2.5log B + 2.51ogB0.

(b) Suppose Bi and B2 are the brightness of two stars such that B\ < Bi and let M\ and M2 betheir respective magnitudes. Since log is an increasing function, we have log B\ < logB2. ThenlogBi<logB2 <*> log Bi - logBo < logB2 - logBo ** log (Bi/B0) < log (B2/B0) «»-2.5 log(B1/B0) > -2.5 log (B2/B0) <& M\> M2. Thusthe brighter star has lessmagnitudes.

(c) Let Bi be the brightness of the star Albiero. Then 100Bi is the brightness of Betelgeuse, and its magnitude is

M = -2.51og(IOOB1/B0) = -2.5 [log 100+ log(Bi/B0)] = -2.5 [2 + log(Bi/B0)] = -5 - 2.5log(B1/B0)

= —5+ magnitude of Albiero

SubstitutingP0 = 80, f. = 24, and c = 0.3 we have P =80

(24 + 1)0.330.5. So the

64.


66. (a) False; log(x/y) = logx - logy ^ (logx) / (logy).

(b) False; log2 x - log2 y = log2 (x/y) ^ log2 (x - y).

(c) True; the equation is an identity: log5 (a/b2) = log5 a - log5 b2 = log5 a - 2log5 6.(d) True; theequation is an identity: log 2~ = z log2.

(e) False; logP + logQ = log(PQ) ^ (logP) (logQ).

(f) False; loga - logb = log(a/6) ^ (loga) / (log/;).

(g) False; x log2 7 = log2 T # (log2 7)J:.

(h) True; theequation isan identity. log„ a" = a log„ a = a •1 = a.

(i) False; log(x-y) ^ (logx)/(logy). For example, 0 = log (3 - 2) ^ (log 3)/(log 2).

0) True; the equation is an identity: - ln (1/A) = - lnA-1 = -1 (- lnA) = lnA.67. Theerror ison the first line: log0.1 < 0, so21og0.1 < logO.l.

68. Let / (x) = x2. Then / (2x) = (2x)2 = 4x2 = 4/ (x). Now the graph of/ (2x) is the same as the graph of/ shrunkhorizontally by a factor of|, whereas the graph of4/ (x) is the same as the graph of/ (x) stretched vertically by a factorof4.

Let g (x) = ex. Then g (x+ 2) = ex+2 = e2ex = e2o (x). This shows that a horizontal shift of2 units to the right is thesameas a vertical stretch by a factor of e2.

Let /; (.;:) = lnx. Then h (2x) = ln2x = ln2 + lnx = ln2 + h (x). This shows that a horizontal shrinking by a factor of\ is the same asa vertical shift upward by ln2.

4.4 Exponential and Logarithmic Equations1. 10J' = 25 & loglOJ" = log25 & x log10 = log25 ^ x « 1.398

2. lO"* = 4 & logl0-a! = log4 <^ -x = log4 <=> x = - log4 « -0.6021

3. e~2x = 7 & ln e~2x = ln 7 & -2xlne = ln 7 <$ -2x = In 7 & x = -{ ln7 « -0.9730

4. e3x = 12 ^ lne3:r = ln 12 <* 3x = ln 12 <* x = — w 0.8283

5. 21-- = 3 «. k^1"* = log3 «• (l-x)log2 = log3 «• 1-x = j^ ^log 2

x=J_ |2I3 ^ _Q 585Qlog 2

6. 32-r"1=5 ^ logS2'-1 =log5 <=> (2x-l)log3 =log5 ^ 2x - 1= ^ «• 2x =1+^log 3 log 3

2 V~ ' log3,7. 3c1 = 10 «• ex = f «• x = ln(f) « 1.2040

8. 2e12* = 17 ^ e12a! = ¥ <=> 12x = ln(f) <* x= £ [hi(f)] »0.1783

9. <' •'•' 2 o I !./• In 2 •=• 4a: I I-In 2 <> .*• -' " — =0.07674

10. 4(l + 105x) = 9 <=> 1 + 105x = f «• 105x = f <*> 5x = log (|) <*X=\ [log5-log4] « 0.0194

11. 4+35x =8 <* 35r =4 «• log35x = log4 <£> 5xlog3 = log4 *> 5x = }^log 3

x=J°ll« 0.25245 log 3

x=i(l + 52lf J^i.2325

<3>

SECTION 4.4 Exponential and Logarithmic Equations 321

12. 23x =34 <£> log23x =log 34 «• 3xlog2 =log 34 <=> x= j^—^ ~16958

13. 804x =5 «*• log8°-4x =log5 & 0.4xlog8 =log5 <S> 0.4x =|^| e> x=Q°^ g»1.934914. 3I/14=0.1 <S> log3x/14 =log0.1 ^ (^)log3 =log0.1 <S> x=

15. 5"x/100 =2 «- Iog5-x/100 =log2 ^> --^-log5 =log2 <*> z=-10|01(?2 ^-43.0677100 log 5

16. e3_5x = 16 <=> 3- 5x =ln 16 <=> -5x =ln 16 - 3 <S> x=-^ (ln 16 - 3) « 0.0455o

17. e2x+1=200 «» 2x + l = ln200 <£• 2x = -l + ln200 <S> x= ~ + n «2.1492

18. (I)1 =75 «• 4_x =75 <* log4"x =log75 <*> (-x) (log4) =log75 <S> -x=-j^-j- ^

a; =-^«-3.1144log 4

19. 5X = 4X+1 <=> log5x =log4x+1 <£> xlog5 = (x + l)log4 = xlog4 + log4 «=> xlog5-xlog4 = log4

* x(log5-log4) =log4 ^ X=_M1_« 6.212620. 101_X=6X <S> loglO1-1 = log6x <=> l-x = x(log6) «*• l=x(log6) + x ^ l=x(log6 + l)

<* x=, * »0.5624log 6 +1

21. 23x+1 = 3X"2 & log23x+1 = log3x_2 <*> (3x + l)log2 = (x-2) log3 & 3xlog2 +log2= xlog3-2log3 <^ 3xlog2-xlog3 = -log2-21og3 <=> x(31og2 - log3) = - (log2+ 21og3)

log2 + 2log331og2-log3

22. 7x/2 =51_x <*> log7x/2 =log51"x <s> (|) log 7=(1 - x)log5 44> (|) log 7=log5 - xlog5*> (|)log7 +xlog5 =log5 & x(ilog7 +log5)=log5 <* x=y^^-g «0.6232

23. —^—=4 <=> 50 = 4 + 4e-x & 46 = 4e_x & 11.5 = e_x <& lnll.5 = -x &1 + e-1

x = -lnll.5 « -2.4423

24. ——— =2 & 10 = 2 + 2e_x «*- 8 = 2e_x <S> 4 = e_x <=> ln4 = -x «•l + e"x

x = -ln4«-1.3863

25. 100(1.04)2t = 300 & 1.042f = 3 <=> logl.042t = log3 <S> 2tlog 1.04 = log 3 <s>

t=>Ei3„ 14.00652 log 1.04

26. (1.00625)12' = 2 <S> log 1.0062512' = log2 «. 12r.log 1.00625 = log2 <S>

t = 1^ « 9.270812 log 1.00625

27. x22x - 2X = 0 «» 2X (x2 - 1) = 0=• 2X = 0(never) or x2 - 1= 0. Ifx2 -1=0, then x2 = 1=$• x = ±1. So theonly solutions are x = ±1.

28. x210x - xl0x = 2(10x) «• x210x - xl0x - 2(10x) =0 & 10x (x2 - x - 2) = 0 =» 10x = O(never) orx2 - x - 2 = 0. If x2 - x - 2 = 0, then (x - 2) (x + 1) = 0 =• x = 2, -1. Sotheonlysolutions arex = 2, -1.

29. 4x3e_3x - 3x4e"3x =0 «*• x3e_3x (4 - 3x) = 0 => x = 0 ore_3x = 0 (never) or4 - 3x = 0. If4 - 3x = 0, then3x = 4 <f> x = 4. So the solutions are x = 0 and x = 4.

14 log 0.1

log 3-29.3426


30. xV + xex - e1 = 0 «• ex (x2 + x - l) = 0=• ex = 0(impossible) or x2 + x - 1= 0. Ifx2 + x - 1= 0, then

x = . So the solutions are x = .2 2

31. e2x - 3ex + 2 = 0 <* (ex - 1)(ex - 2) = 0 => ex - 1 = 0ore1 - 2 = 0. Ifex - 1 = 0, then ex = 1 &x = ln 1 = 0. If ex - 2 = 0, then ex = 2 <* x = ln 2 « 0.6931. So the solutions are x = 0 andx « 0.6931.

32. e21 - ex - 6 = 0 <* (ex - 3)(ex + 2) = 0 =• ex + 2 = 0(impossible) orex - 3 = 0. Ifex - 3 = 0,then ex = 3& x = ln 3 « 1.0986. So the only solution is x « 1.0986.

33. e4x + 4e2x -21 = 0 & (e2x + 7) (e2x - 3) = 0=• e2x = -7 or e2x = 3. Now e2x = -7 has no solution, sincee2x > 0 for all x. But we can solve e2x = 3 <=> 2x = ln3 <=>• x = ±ln3 « 0.5493. So the only solution isx w 0.5493.

34. ex - 12e"x -1 = 0 <* ex - 1- 12e~x =0 «» ex (e1 - 1- 12e_x) = 0•ex <* e2x - ex - 12 = 0<=*> (ex - 4) (ex + 3) = 0 =*• ex + 3 = 0 (impossible) or ex - 4 = 0. If ex - 4 = 0, then ex = 4 ^x = ln4 « 1.3863. So the only solution is x « 1.3863.

35. lnx = 10 «» x = e10w 22026

36. ln (2 + x) = 1 <* 2 + x = e1 <s> x = e - 2 « 0.7183

37. logx = -2 <* x = 10_2=0.01

38. log(x - 4) = 3 <=> x - 4 = 103 = 1000 & x = 1004

39. log(3x + 5) = 2 <* 3x + 5= 102 = 100 <& 3x = 95 «- x = f «31.666740. log3 (2 - x) = 3 & 2 - x = 33 = 27 & -x = 25 *> x = -25

41. 2-ln(3-x)=0 «» 2 = ln(3-x) <s> e2 = 3-x <=> x = 3 - e2 * -4.3891

42. log2 (x2-x-2)=2 <* x2-x-2 = 22=4 «- x2-x-6 = 0 «• (x-3)(x + 2) = 0 «»x = 3 or x = —2. Thus the solutions are x = 3 and x = -2.

43. log2 3 + log2 x = log2 5 + log2 (x-2) «• log2 (3x) = log2 (5x - 10) & 3x = 5x - 10 & 2x = 10«*• x = 5

44. 2logx = log 2+ log (3x- 4) <=> log(x2) = log(6x - 8) «• x2=6x-8 <=> x2-6x + 8 = 0 <=>(x —4) (x - 2) = 0 «=• x = 4 or x = 2. Thus the solutions are x = 4 and x = 2.

45. logx + log(x - 1) = log(4x) 4» log[x(x - 1)] = log(4x) & x2 - x = 4x <* x2 - 5x = 0 <*x (x —5) = 0 => x = 0 or x = 5. Sothe possible solutions are x = 0 and x = 5. However, when x = 0, logx isundefined. Thus the only solution is x = 5.

46. logs x + loS5 (ar + 1) = log5 20 & log5 (x2 + x) = log5 20 & x2 + x = 20 & x2 + x - 20 = 0 «•(x + 5) (x - 4) = 0 «=> x = -5 orx = 4. Since log5 (-5) isundefined, theonly solution isx = 4.

47. log5(x+l)-log5(x-l) =2 «. log5 f^l =2 ^ ^ =52 & x+1=25x - 25 ^\x —1/ x —1

24x = 26 <!=> x = if

48. logx + log(x - 3) = 1 <=• log[x(x - 3)] = 1 <=> x2 - 3x = 10 «• x2 - 3x - 10 = 0 «*•(x + 2) (x - 5) = 0 «=> x = -2 or x = 5. Since log(-2) is undefined, theonlysolution is x = 5.

49. log9 (x - 5) + log9 (x + 3) = 1 o- log9 [(x - 5) (x + 3)] = 1 «• (x - 5) (x + 3) = 91 «•a; - 2x —24 = 0 o- (x - 6) (x + 4) = 0 ^ x = 6 or—4. However, x = —4 is inadmissible, so x = 6 is the onlysolution.

50. ln(x - 1)+ ln(x + 2) = 1 & ln[(x- 1)(x+ 2)] = 1 <* x2 + x - 2 = e & x2 + x - (2+ e) = 0 =•x= -i±Vi+4(a+«) = -i±^FFC Sincex_1<0whenx = -i-^FFC^onlysolutionisx= -i+^Fpc w1>7290

51. log(x+3) = logx+ log3 «*• log(x+ 3) = log(3x) «» x + 3 = 3x «• 2x = 3 «*• x = f


52. (logx)3 = 31ogx & (logx)3-31ogx = 0 & (logx) ((logx)2 - 3) & (logx) = 0or(logx)2-3 = 0.Now logx = 0 <s> x = 1. Also (logx)2 -3 = 0 «*• (logx)2 =3 & logx = ±>/3 4» x = 10±V3,

_ m>/3sox = lO^5 « 53.9574 orx = 10-n/3 « 0.0185. Thus the solutions tothe equation arex = 1,x = lO^3 » 53.9574 and

x = 10_n/3 0.0185.

53. 22/'og«x = i <=> log222/'og5x = log2(^) <*log5x

= -4 <=> log5x = -1 <*

T _ c-i/2 - _i_ : 0.4472

54. log2 (logs x) = 4 «- log3 x = 24 = 16 <s> x = 316 = 43,046,721

55. lnx = 3-x &• lnx + x-3 = 0. Let

/ (x) = ln x + x —3. Weneed to solvethe equation

/ (x) = 0. Fromthe graphoff, we get x « 2.21.

56. logx = x2 - 2 <*• logx - x2 + 2 = 0. Let

/ (x) = log x - x2 + 2. We need tosolve the equation

/ (x) = 0. From the graph off, we get x « 0.01 or

x « 1.47.

57. x3 - x = log10 (x + 1) <=>

x3 - x - log10 (x + 1) = 0. Let

0

58. x = ln (4 - x2) «*• x - ln(4 - x2) = 0. Let/ (x) = x —ln (4- x2). We need tosolve the equation

/ (x) = x3 - x - log10 (x + 1). We need tosolve the / (x) = 0. From thegraph off, we getxx « -1.96 or

equation / (x) = 0. From the graph off, we get x = 0 or x « 1.06.

a: * 1.14.

-J

-4

59. ex = -x <=> ex + x = 0. Let / (x) = ex + x. We 60. 2"x = x - 1 & 2"x - x + 1 = 0. Let

need tosolve the equation / (x) = 0. From the graph of/, / (x) = 2~x - x + 1. We need tosolve the equationwe get x ss -0.57. / (x) = 0. Fromthe graph of /, we get x «s 1.38.

o


61. 4_x = ^x <=• 4-x - y/x = 0. Let

/ (x) = 4_x - y/x. We need to solvetheequation

/ (x) = 0. From thegraph off, weget x « 0.36.

62. ex -2 = x3-x «» ex -2-x3 + x = 0. Let

/ (x) = ex —2 - x3 + x. We need tosolve the equation

/ (x) = 0. Fromthe graphoff, we get x « -0.89 or

x « 0.71.

0

63. log(x-2) + log(9-x) < 1 & log[(x - 2) (9 - x)] < 1 ^ log (-x2 + llx - 18) < 1 =>-x2 + llx - 18 < 101 & 0<x2-llx + 28 & 0<(x- 7) (x - 4). Also, since the domain ofalogarithmis positive we must have 0 < -x2 + llx - 18 <s> 0 < (x- 2) (9 - x). Using the methods from Chapter 1with theendpoints 2,4, 7, 9 forthe intervals, wemake the following table:

Interval (~oo,2) (2,4) (4,7) (7,9) (9,oo)

Sign ofx —7

Sign ofx —4

Sign ofx - 2

Sign of9 - x +

+

+

+

+

+

+

+

+

+

1+++-

Signof(x-7)(x-4) + + - + +

Signof (x-2) (9 - x)- + + + -

Thus the solution is (2,4) U (7,9).

64. 3 < log2 x < 4 <& 23<x<24 <& 8 < x < 16.

65. 2 < 10x < 5 & log 2 < x < log 5 <s> 0.3010 < x < 0.6990. Hence the solution to the inequality isapproximately the interval (0.3010,0.6990).

66. x2ex - 2ex < 0 & ex (x2 - 2) < 0 & ex (x - y/2) (x + >/§) < 0. We use the methods ofChapter 1withthe endpoints - \/2 and y/2, noting that ex > 0 for all x. We make a table:

Interval (-oo,-v^) (-V2,y/2) (V5,oo)Sign ofex

Sign of(x- V2)Sign of (x+ y/2)

+ +

+

+

+

+

Sign ofex (x - y/2) (x+ y/2) + - +

Thus -y/2 < X < y/2.

(0 085\4(3)1+-^— J =5000 (1.0212512) =6435.09. Thus the amount after 3years is $6,435.09.

(b)(0 085 \ 4t1+ ^—J = 5000 (1.021254*) & 2= 1.021254t «*• log 2= 4t log 1.02125 «*>

t =log 2

71—,, --,-_ « 8.24 years. Thus the investment will double in about 8.24 years.4 log 1.02125 J


68. (a) A{2) = 6500e006(2) « $7328.73

(b) 8000 =6500eoo6t & g =e006t «. ln (If) =0.06t <* t=-L ln (if) « 3.46. So theinvestment doubles in about 3| years.

69.8000 =5000^1 +^^") =5000(1.018754t) «- 1.6 =1.018754* <s> log 1.6 =4t log 1.01875 ^t = ^—'• « 6.33 years. The investment will increase to $8000 in approximately 6years and 4months.

4 log 1.01875

70. 5000 =4000 (l +2^21^ & 1.25 = (1.04875)2' & log 1.25 =2tlog 1.04875 «•_ —logl .25 s . ^ b 2 save $5000

2log 1.04875 3'In 9

71. 2 = e0085t <& ln 2 = 0.085* 4* t = -—— « 8.15 years. Thus the investmentwill double in about 8.15 years.0.085

72. 1435.77 =1000 (l +02(4> «* 1.43577 =(l +£)* «> 1+^=^5^43577 «- ^=^43577-1<* r = 2(#1.43577 - l) « 0.0925. Thus the rate was about 9.25%.

73. reff =(l +-)" - 1. Herer =0.08 and n=12,soreff= (l + \ -1=(1.0066667)12 -1=8.30%.74. rApv = er - 1. Here r = 0.05.5so rApy = e0055 -1« 1.0565 - 1 = 0.565. So the annual percentage yieldis about

5.65%.

75. 15e-0087t =5 <*• e-0087t =± <=>• -0.087* =ln (±) =- ln 3 & t=-^ «12.6277. So only5 grams remain after approximately 13 days.

76. We want to solve for t in the equation 80(e_02t - l) = -70 (when motion is downwards, the velocity is negative).Then 80 (e"02' - l) = -70 & e"02' - 1 = -f & e~°-2t = £ <*> -0.2* = In(1) *>

t = —— « 10.4 seconds. Thus the velocity is 70 ft/sec after about 10 seconds.-0.2

77. (a) P (3) = ttzt*; = 7.337, so there are approximately 7337 fish after 3 years.» ' y ' x + 4e— ' '

(b) We solve fort. ^-Ji—=5 «> 1+4e-°8t =f =2 ^ 4e-°8t =1 ^ e"08'= 0.25 «.-0.8* = In 0.25 <& t= -^—:— = 1.73. So the population will reach 5000 fish in about 1year and 9months.

—0.8

78. (a) J = ioe-°008(30) = 10e-0-24 = 7.87. So at30ft the intensity is7.87 lumens.

(b) 5=lOe"0 008x «• e-°008x =I & -0.008x =ln (1) <* x=^^ «86.6. So the intensitydrops to 25 lumens at 86.6 ft.

P

Po79. (a) ln (—\ =-- «» ^- =e_,,/fc «*• P= Poe_,,/fc. Substituting k=7and P0 = 100 we get

\Po J k "

P = 100e_,,/7.

(b) When /i = 4 we have P = 100e_4/7 » 56.47 kPa.

O.iu «. T~20 =e-011t <* T - 20 = 200e-011t ^ T = 20 + 200e"0 nt-80- (3) lD (w) =- 200(b) When t = 20we have T = 20+ 200e-°n(20) = 20+ 200e-22 « 42.2° F.


81.<a)J =f§(l-e-13*/5) «> i/ =l-e-13t/8 «. e-13</5 =1- l§7 ^ -f* =In (l - flj) *

(b) Substituting 7 = 2, we have *= -A ln [l - 1§ (2)] « 0.218 seconds.

82. (a) P = M- Ce & Ce~kl =M-P &

•--V * -"-m —-4in

M-P

(b) P(t) = 20 - 14e-°02". Substituting M = 20,C = 14, fc = 0.024,

and /' = 12 into /. = -- ln ( ——— ), we have

1 , /20-12t = — ———- ln ~ 23.32. So it takes about 23 months.

0.024 \ 14

83. Since 91 = 9, 92 = 81, and 93 = 729, the solution of9X = 20 must be between 1and 2(because 20 is between 9and 81),whereas the solution to9* = 100must be between 2 and 3 (because 100 isbetween 81and 729).

84. Notice that log (x1/lo*x) = r— logx = 1, so 1/loBX = 101 for all x >0. Sov / log X

x1/ logr = 5has no solution, and x1/ logx = khas asolution only when k = 10.This isverified by the graph of/ (x) = xx/ logx.

(c)

85. (a) (x - l)1"^-1) = 100 (x - 1) *> log ((x - l)to«<«-l>) = log (100 (x - 1)) <*[log (x-1)] log (x - 1) = log 100 +log (x-1) «• [log (x-1)]2 - log (x-1) -2 = 0 &[log (x-1) -2] [log (x- 1) + 1] =0. Thus either log (x- 1) = 2 <=> x = 101 or log (x - 1) =-1 <=>x= 11x io-

(b) log2 x + log4 x + log8 x = 11 4* log2 x + logo, y/x + log2 tyx = 11 «• log2 (xv^v^) = 11 <=>

log2 (x11/6) =11 & ^ log2 x=11 <=> log2 x=6 & x=26=64(c) 4r - 2'+1 =3 <=> (2X)2 - 2(2s) -3 = 0 & (2X - 3)(2X + 1) = 0 & either 2X = 3 <=>

s = r~^ or 2X = -1, which has no real solution. So x = —- is the only real solution.In 2 In 2 J

.5 Modeling with Exponential and

Logarithmic Functions1. (a) n (0) =500.

(b) The relative growth rate is 0.45 = 45%.

(c) n (3) = 500e°',5(3) « 1929.

In 20(d) 10,000 = 500e°-lr" «• 20 = e045' & 0.45* = In20 <=>• i = ^ « 6.66 hours, or6 hours0.45

40 minutes.

SECTION 4.5 Modeling with Exponential and Logarithmic Functions 327

2. (a) The relative growth rate is0.012 = 1.2%.

(b) n(5) = 12e0012(5) = 12e006 « 12.74 million fish.(C) 30= 12e0012t <=> 2.5 = e0012t <=> 0.012* = ln2.5

ln2.5* =

0.012

about 76 years

76.36. Thus the fish population reaches 30 million after

3. (a) r = 0.08 and n (0) = 18000. Thusthe population is given by the

formulan(*) = 18,000e008t.

(b) t = 2008 - 2000 = 8. Then we have

n(8) = 18000e008(8) = 18000e064 « 34,137. Thus there shouldbe 34,137 foxes in the region by the year 2008.

(d)

(c)

4. n (*) = n0ert;no = 110million, *= 2020 - 1995 = 25

(a) r = 0.03;n (25) = 110,000,000e003(25) = 110,000,000e°-75 « 232,870,000. Thus ata 3% growth rate, theprojected population will beapproximately 233 million people bytheyear 2020.

(b) r = 0.02;n(25) = 110,000,000e002(25) = 110,000,000e0-50 « 181,359,340. Thus ata 2% growth rate, theprojected population will beapproximately 181 million people by theyear2020.

5. (a) n(*) = 112,000eoo4t.

(b) t = 2000 - 1994 = 6and n(6) = 112,000e° 04{6) « 142380. The projected population is aboutl42,000.(C) 200,000 = 112,000e004t & ff = e004t <* 0.04* = ln (ff) <* t = 25ln (ff) « 14.5. Since

1994 + 14.5 = 2008.5, the population will reach200,000duringthe year 2008.

6. (a) n (*) = n0ert with n0 = 85 and r = 0.18. Thus n(t) = 85e018f.

(b) n(3) = 85e018(3) « 146 frogs.

(C) 600 = 85e018t <* W= e°18t «* °-18f = ln (W) *• l = oAs ln (W) * 10-86' So thepopulation will reach600 frogs in about 11 years.

7. (a) The deer population in 1996was 20,000.

(b) Using the model n (*) = 20,000ert and the point (4,31000), we have 31,000 = 20,000e4r «• 1.55 = e4r4r = ln 1.55 ^ r = \ ln 1.55 « 0.1096. Thus n(i) = 20,000eolO96t

(c) n (8) = 20,000eolo96(8) w48,218, so the projected deer population in 2004 is about 48,000.In5

(d) 100,000 = 20,000eulUi „ „ _ ^ „ „. .. . Q1Qg61996 + 14.63 = 2010.63, the deer populationwill reach 100,000 during the year 2010.

*• 5 = e^ <^ 0.1096* = ln5 <& t = 14.63. Since

8. (a) Since the population grows exponentially, the population is represented by n (*) = noer\ with no = 1500and n (30) = 3000. Solving for r, we have 3000 = 1500e30r <(=J> 2 = e30r ^ 30r = ln2 &r = (ln2)/30 w 0.023. Thus n(*) = 1500e0023t.

(b) Since 2hours is 120 minutes, the number ofbacteria in 2hours is n(120) = 1500e° 023(120) « 24,000.(c) We need to solve 4000 = 1500eoo23t for *. So 4000 = 1500e0023t & f = e0023t «• 0.023* = lnf <*

t = n^ ' ' cs 42.6. Thus the bacteria population will reach 4000 in about 43 minutes.0.023


9. (a) Using the formula n (*) = n0ert with n0 = 8600 and ra (1) = 10000, we solve for r,giving 10000 = ra (1) = 8600er& l=er «. r = ln(f§)« 0.1508. Thus n(*)=8600e01508t.

(b) n (2) = 8600e01508(2) « 11627. Thus the number ofbacteria after two hours is about 11,600.

(c) 17200 =8600e01508t <* 2=e01508t <* 0.1508* =ln 2 <* t=-^- « 4.596. Thus the number0.1508

of bacteria will double in about 4.6 hours.

10. (a) Using ra (*) = n0ert with n (2) = 400 and n (6) = 25,600, we have n0e2r = 400 and n0e6r = 25,600. Dividing

the second equation by the first gives n°eo = ' =64 <* e4r = 64 <*• 4r = ln64 <«•noe2r 400

r = 1 ln64« 1.04. Thus the relative rate ofgrowth isabout 104%.

(b) Since r = J ln 64 = 1ln 8, we have from part (a) n(t) = n0e( *ln 8K Since n(2) = 400, we have 400 = n0eln 8

^ n° = ~nr8 = ^ = 50. So the initial size ofthe culture was 50.

(c) Substituting ran = 50and r = 1.04, we have ra (*) = n0ert = 50e104'.

(d) n (4.5) = 50e104{4-5) = 50e468 « 5388.5, so the size after 4.5 hours is approximately 5400.

(e) n(*) = 50,000 = 50e104' <& eimt = 1000 <=> 1.04* = ln 1000 <* *= ^222 „ 6>64. Hence the1.04

population will reach 50,000 after roughly 6§ hours.

11. (a) 2n0 = n0e002t <& 2 = e002t «*• 0.02* = ln2 «. * = 50ln2 « 34.65. So we have*= 1995 + 34.65 = 2029.65, andhence at thecurrent growth rate thepopulation will double bytheyear2029.

(b) 3ra0 = n0e002t & 3 = e002t «• 0.02* = ln3 «- * = 501n3 ss 54.93. So we havet = 1995 + 54.93 = 2049.93, and hence at thecurrent growth rate thepopulation will triple bytheyear2050.

12. (a) Calculating dates relative to 1950 gives ra0 = 10,586,223 and n(30) = 23,668,562. Thenra(30) = 10,586,223e30r = 23,668,562 «- e30r = f§|||§|| « 2.2358 & 30r = ln 2.2358 <*r = i in2.2358 w 0.0268. Thus n(*) = 10,586,223e00268t.

(b) 2(10,586,223) = 10,586,223e00268t ^ 2= e00268t & ln2 = 0.0268* & t = -^- « 25.86. So0.0268

the populationdoubles in about 26 years.

(c) *= 2000 - 1950 = 50; ra (50) * 10586223e00268(50) « 40,429,246 and sothe population in the year 2000 will beapproximately 40,429,000.

13. n(t) = n0e2t. When n0 = 1, the critical level is ra(24) = e2(24) = e48.We solve the equatione48 = n0e2t, where ra0 = 10. This givese48 = 10e2t «» 48 = ln10 + 2* «• 2* = 48 - ln10 <=>*= 1 (48- ln 10)« 22.85 hours.

In O14. From theformula forradioactive decay, wehave m (*) = moe~rt, where r = —r~.

hIn 9

(a) We have m0 = 22 and h = 1600, so r = —— « 0.000433 and theamount after *years is given by

m(t) = 22e-°00O433t.

(b) m(4000) = 22e-0000433(4000) « 3.89, so the amount after 4000 years isabout 4mg.

(c) We have to solve for*intheequation 18 = 22e-°-°oo433t j^ gives 18 _ 22e-0000433ti / 9 \

«• -0.000433* = ln (^) & t= Jl"'* « 463.4, so it takes about 463 years.-0.000433

ii

SECTION4.5 Modeling with Exponential and LogarithmicFunctions 329

In 9 In 0

15. (a) Usingm(*) = m0e-rt with m0 = 10and h = 30, we have r = — = —- « 0.0231. Thusm(t) = lOe-00231'./I oU

(b) m(80) = i0e-00231(8°) « 1.6grams.

' <=. In (11 = _n.n9.3H _> /. =-0.0231

(c) 2=10e-00231t <^ 1=e-°-023U & ln (1) =-0.0231* <* t= ln|L « 70years.

16. (a) m (60) = 40e-°0277(60) « 7.59, sothe mass remaining after 60days isabout 8 g.

(b) 10 =40e-00277t & 0.25 =e-°0277t & ln0.25 =-0.0277* <& t=~^p «50.05, so it takesabout 50 days.

(c) We need to solve for *in the equation 20 = 40e-°0277t. We have 20 = 40e_0-0277t «• e"0277' = \ <»lnl

-0.0277* = ln 1 <=• * = 2„„ « 25.02. Thusthe half-life of thorium-234 is about25 days.2 -0.0277

In 0

17. By the formula in the text, m(*) = m0e~rt where r = —, so m(*) = 50e"[(ln2)/28]t. We need to solve

for *in the equation 32 =bOe-^2"2*1. This gives e^2"2*1 =§ ** -^* =ln (§§) <*28t = -— •ln (§§) « 18.03, so ittakes about 18 years.

In218. From the formula for radioactive decay, we have m(*) = moe_rt, where r = -7-. Since h = 30, we have

1 o

r = « 0.0231 and m(t) = m0e~00231t. In this exercise we have to solve for t in the equation 0.05m0 = m0e~00231toU

<* e-o.o23it _ 0 05 ^ _o.0231t = ln0.05 «• *= " ' « 129.7. So itwill take about 130 s.

ln O

19. By the formula for radioactive decay, we have m(*) = m0e~rt, where r = —, in other words ra(*) = moe~^n2)/h]t.

In this exercise we have to solve for h in the equation 200 = 250e_[(ln2)/'11-48 <s> 0.8 = e-^n2)/h],4S <*

ln (0.8) = -48 <=>> h = • 48 ss 149.1 hours. So the half-life is approximately 149 hours.h In 0.8

In 920. From the formula for radioactive decay, we have m(t) = m0e~rt,wherer = —. In other words, m(t) = moe~^n2)/h]t.

(a) Using m(3) = 0.58m0, we have to solve for h in the equation 0.58mo = ra(3) = moe~[(,n2)/,l,3.

Then0.58m0 = moe""31"2^ <*> e-[<31"2>/hl = 0.58 «• -^ = ln0.58 ^h_ _ 31n2_ ^ 3g2 Thus the half|ife 0f Radon-222 is about 3.82 days.

In0.58

(b) Here we have to solve for *in the equation 0.2m0 = raoe~((ln2)/3-82]t. So we have 0.2m0 = moe_t(ln2)/382lt <*0.2 =e-[(ln2)/3.82]t ^ _^t =ln0<2 ^ t=

3.82

sample of Radon-222 to decay to 20% of its original mass.

21. By the formula in the text, m(*) =m0e-[(,n2)/hIt, so we have 0.65 =1•e-K,n2>/873°H & ln (0.65) =~^<^ t= _5730 In 0.65 ^ 3561 _hus ^ ^.^ js abQut 356Q yeafs o|d

ln2

22. From theformula forradioactive decay, wehave ra (i) = moe~rtwhere r = —r-. Sinceh = 5730,r = r-rr « 0.000121

and m (*) = moe"0000121'. We need tosolve for *in the equation 0.59mo = moe"00001214 & c-o.ooomt = Q59InO 59

*> -0.000121* = ln 0.59 **• * = ' » 4360.6. So it will take about 4360 years.~~ U • VJUU X £» _

0.2 =e-K1"2)/3-82)* & -^-t =ln0.2 <* *=_3-8^"0-2 ~8.87. So it takes roughly 9days for ao.oZ In Z


23. (a) T(0) = 65+ 145e-°05(0) = 65+ 145 = 210° F.

(b) T (10) = 65+ 145e"005(10) » 152.9. Thus the temperature after 10minutes isabout 153° F.

(C) 100 = 65 + 145e-°05' <* 35 = 145e-°05t & 0.2414 = e-005' <S> ln 0.2414 = -0.05*

28.4. Thus the temperature will be 100° F in about 28 minutes.* = —0.05

24. (a) We use Newton's Law ofCooling: T (*) = Ts + D0e~kt with k = 0.1947, T8 = 60,and D0 = 98.6- 60 = 38.6SoT (*) = 60+ 38.6e-°1947t.

12

38.6 **1

(b) Solve T(t) = 72. So 72 = 60 + 38.6e-°1947t

-0.1947t =ln(iy „ t._aiM7 In

38.6e-°1947t = 12 <^ e"0194'

6.00, and the time ofdeath was about 6 hours ago.(—)\38.6J

25. Using Newton's Uw ofCooling, T(t) = Ts+D0e-kt withTa = 75andD0 = 185-75 = 110. SoT(*) = 75+110e_fct.

(a) Since T(30) = 150, we have T (30) = 75 + llOe-30* = 150 ^ llOe"30* = 75 «- e_30fc = 1§ &-30k = In (1§) «• k = -£ ln(±§). Thus we have T(45) = 75 + n0e(45/3o)in(i5/22) ^ 136 9> ^ SQ ^temperature of the turkey after 45 minutes is about 137° F.

(b) The temperature will be 100°F when 75 + 110e(t/30) ln<15/22) = 10q «. e(t/30) ln<15/22> = 2*L = A ^

(|)ln(M)=ln(i) <S> * = 30Milln(l§) 116.1. So the temperature will be 100° F after 116 minutes.

26. We useNewton's Law ofCooling: T (*) = Ts + D0e~kt, with Ts = 20and

Do = 100 - 20= 80. So T (*) = 20+ 80e_fct. Since T (15) = 75, we have

20 + 80e-15fc = 75 <* 80e"15fc = 55 & e~15k = 11 «*•

-15fc = ln(li) «. fc = -^ln (11). Thus

T (25) = 20 + 80e(25/15)-,n(n/16) « 62.8, and sothe temperature after another

10min is63° C.The function T(t) = 20+ 80e(1/15)ln(11/16)t isshown inthe

viewing rectangle [0,30] by [50,100].

27. (a) pH = - log [H+] = - log (5.0 x 10~3) » 2.3

(b) pH = - log [H+] = - log (3.2 x 10~4) * 3.5

(c) pH = - log [H+] = - log (5.0 x 10"9) « 8.3

28. pH = - log [H+] = - log (3.1 x 10~8) « 7.5 and the substance is basic.

29. (a) pH = - log [H+] = 3.0 <s> [H+] = 10~3 M

(b) pH = - log [H+] = 6.5 «• [H+] = lO"6"5 « 3.2 x 10"7 M

30. (a) pH = - log [H+] = 4.6 <* [H+] = 10"4'6 M« 2.5 x 10-5 M

(b) pH = - log [H+] = 7.3 <*> [H+] = lO"73 M« 5.0 x 10"8 M

31. 4.0 x 10~7 < [H+] < 1.6 x 10"5 & log (4.0 x lO"7) < log [H+] < log (1.6 x 10-5) <&- log (4.0 x 10-7) > pH > - log (1.6 x 10-5) <* 6.4 > pH > 4.8. Therefore the range ofpH readings for cheeseis approximately 4.8 to 6.4.

32. 2.8 < pH < 3.8 «• -2.8 > -pH > -3.8 «- 10-28 > 10_pH > 10-38 &1.58 x 10~3 > [H+] > 1.58 x 10-4. The range of[H+] is 1.58 x 10-4 to 1.58 x 10-3.

SECTION 4.5 Modeling with Exponential and Logarithmic Functions 331

33. Let Iq be the intensity of the smaller earthquake and h the intensity of the larger earthquake. Then h = 20Iq. Notice

that Mo =log (^\ =log Jo - logSand Mi =log (^\ =log (^J=log 20 +log J0 - log 5. ThenMi - Mo = log 20 + log Jo - log S - log Jo + log S = log 20 « 1.3. Therefore the magnitude is 1.3 times larger.

34. Let the subscript S represent the San Francisco earthquake and J the Japan earthquake. Then we have

Ms =log (^\ =8.3 <=> Is =S•lO83 and Mj =log (^ J =4.9 <* Ij =S•1049. Sot -|p|8.3

~T = 1fl4 9 = in3 4* 2511.9, and so the San Francisco earthquake was 2500 times more intense than the Japanearthquake.

35. Let the subscript A represent the Alaska earthquake and S represent the San Francisco earthquake. Then

8.3 <=> Is = S • 1086.SoMA =log (^\ =8.6 <* Ia =S- 1086 ;also, Ms =log f^-\ =t Q • in®-®T" = r, ^^o o = 100'3 « 1.995, andhence theAlaskan earthquake was roughly twice as intense as the San FranciscoIs o • 108--5

earthquake.

36. Let the subscript N represent the Northridge, California earthquake and K the Kobe, Japan earthquake. Then

MN =log (lj\ =6.8 & IN =S•1068 and MK =log (^\ =7.2 & Ik = S- 10'"*. So

J 107'2I- = ma » = lft°4 ^ 2.51, and so the Kobe, Japan earthquake was 2.5 times more intense than theNorthridge,In 10do

California earthquake.

37. Let the subscript M represent the Mexico City earthquake, and T represent the Tangshan earthquake. We have

£- = 1.26 «- log 1.26 = log-^- = log-p^f = log^ - log%- = MT - MM. Therefore1m 1m Im/o o b

Mt = Mm + log 1.26 « 8.1 + 0.1 = 8.2. Thus the magnitude of the Tangshanearthquake was roughly 8.2.

38. 0=101og(£) =Mkgf^QxiO-") =101°g(2x 10?) =10 (log2+log 107) =10(log2 +7) «73.Therefore the intensity level was 73 dB.

39.98 = 101°S ( io3T2 j «* log(J1012) = 9.8 «- log J = 9.8 - loglO12 = -2.2 «-J = lO-22 « 6.3 x 10~3. Sotheintensity was 6.3 x 10~3 watts/m2.

40. Let the subscript Mrepresent the power mower and Cthe rock concert. Then 106 = 10 log ( -™l2 ) <*

log (IM •1012) =10.6 <* JM-1012 =1010-6.Alsol20 =101og^^[2-^ «- log (Jc •1012) =12.0 <*J in12Jc •1012 = lO120. So y- = —^ = 101,4 « 25.12, and so the ratio of intensity is roughly 25.

41. (a) fa =10 log (J±\ and ^=A =i0lOg f^-) =10 log (j-\ -21ogd1 =10 log (|-) _201ogc*i.Similarly, f32 = 10 log (—J- 201ogcfo. Substituting the expression for/3X gives

/32 =10log (—J- 201ogdi +201ogdi - 201ogd2 =&x +201ogdi - 201ogd2 =/^ +20log (j-\

(b) /3X =120,di =2,andd2 =10.Then/32=/31+201ogf^-j =120 +20log (^) =120 +20log0.2 «106, andso the intensity level at 10 m is approximately 106 dB.

Exponential and Logarithmic Functions · SECTION 4.1 ExponentialFunctions 303 27. g(x) = 2X —3....

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Transcript of Exponential and Logarithmic Functions · SECTION 4.1 ExponentialFunctions 303 27. g(x) = 2X —3....