Chapter 5 Exponential and Logarithmic Functions · 5 120 Intersection X = 11.5S2t-153,Y = 10 0 -1-1-1 15 53. x P.-. 1.4650 5 55. x —3.3219 —5 6 5 148 Chapter 5: Exponential and

Chapter 5

Exponential and Logarithmic Functions

SECTION 5.1 29. f (a) = f (b) 1. a —3a+2=-3b+2

3. (f o g)(x)= f(g(x)) —3a = —3b

jc( 1

—x+3 a = b

f is one-to-one 1

+3

) x

1

x + 3 31. f (a) = f(b)

5. 2a2 — 3 = 2b2 — 3

202 = 2b2

7. = ;W. il

4(1x3) = x b2 4

a = ±b 9. f (g(x)) = f (—x — 3) f is not one-to-one, since, for example,

—(—x —3)— 3=x+ 3-3 =x g( f (x))= g(—x — 3)

= —(—x —3)-3 =x+ 3-3 =x

11. f (g(x)) = fix)= 6(-1:x)= x 6 6

g( f (x)) = g(6x) = (6x) = x

13. f (g(x))= f (-1 x + 13) 3 3

f(3)= f(-3)= 15

33. f (a) = f(b)

—2a3 + 4 = —2b3 + 4

—2a3 = —2b3

= b3 a = b

f is one-to-one

35. f(x)=

2 y = --x

3 3 3

g( f (x))= g(-3x + 8)

1 13=x-43•=x 3 3 3 3

2 x = --y

2 15. f (g(x)) = f (VT--2)= (Vx — 2)3 +2 = x —2 + 2 = x

g( f (x)) = g(x3 + 2) = V(x3 + 2)— 2 =fJ= x

17. f(8.(x))= f (vx _ 3) 3)2 + 3 = x 3 + 3 = x

g(f (x))= g(x2 + 3) = V(x2 + 3) — 3 = .N[X = x

19. The function is one-to-one because there are no values of a and b in the chart such that f (a) = f(b) , where a b

21. The function is not one-to-one because there are values of a and b in the chart such that f (a) = f (b) , where a b . Here,

f (0) = f (2) = 9 .

23. Not one-to-one

25. One-to-one

27. Not one-to-one

f-1(x)= --3

x 2

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144 Chapter 5: Exponential and Logarithmic Functions

37. f(x)=-4x+ -1 5

y = -4x + -1 5

x = -4y + -1 5

5x = -20y + 1 5x-1=-20y 5x-1 -20 =

f (x) = -ix + —1

4 20

39. f (x) = - 6

y = x3 - 6

x= y3 - 6

x + 6 =

if

r(x) + 6

1 41. f (x) = - 4

1 y =-

2x-4

x = 1

-4 2

x+ 4 =-1

y 2

2x + 8 = y

f -1 (x)= 2x + 8

43. g(x) -x 2 + 8, x 0

y = -x2 + 8

x=-y2 +8,y?_ 0

x-8 = -y2

8 - x = y2

= y

e(x)

45. g(x) x 2 -5, x Lc. 0

y = x2 -5

x=y2 - 5,y0

x + 5 = y2

+.4x + 5 = if g-1(x)=

47. f (x) = -2x3 + 7

y=-2x3 +7

x = -2y3 +7

x -7 = -2y3

x-7 = Y3 -2

17- x _ - Y

2

= x

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to

49. f(x)= —4x5 + 9

y = —4x5 +9

x=-4y5 +9

x —9 = —4y5 x -9 5

= y

53. g(x) = (x —1)2 x

y = (x —1)2

x = (y —1)2 y ?_1

y —1

1+ = y

Chapter 5: Exponential and Logarithmic Functions 145

55. :.1T7T-3,x_?_ —3 (y 0)

y = ,/x + 3

x=-Vy+3,y-3(x?_0)

x2 =y+3

x2 — 3 = y

f-1(x)= x2 —3, x > 0

2x 57. f (x)=

x —1 2x

= x —1 2y

x= y-1

x(y — 1) = 2y xy — x = 2y

—x = 2y— xy

—x = y(2— x)

—x 2—x

= y P(x)— x x— 2

59. Domain off. [-3, 3]

Range off [0,4]

Domain of f-1 [0, 4]

Range of f-1 : —3,3]

4 (-1,2)3

(-3,0)

—2 / —3 • —4

= x


y y = x 6 (2, 5) /

(1,2) ,'(5,2) (-I, -I) 2

24 6

-3)/ / -6

C.

-3 -2 -I 0 2 3X


61. Domain off 1-3, 21

Range off [-5, 5]

Domain of f-1 : 1-5, 51

Range of f -1 : [-3, 2]

63. f-1(1)= x 1= f (x); For the inverse

function, its inputs are the outputs off. From the table, the output 1 corresponds to the input X = -2 . Thus, f -1(1) = -2 .

65. r(-2) = 2 so 1-1(f-1( 2)) f-1(2) 1

67. f -1(1) = 0

69. r1(-3) = -2 so f -1(f-1(_3)) _ f(2) _ 1.5

71. f (x)= 4x , where x is the number of gallons and

f (x) is the number of quarts in x gallons. The inverse,

1 . r(x)=-x , gives the number of gallons, 4

where x is the number of quarts.

73. Solve for q: p = 100- 0.1q

p - 100 = -0.1q

p -100 _

q q = 1000 -10p

75. a. The smallest value for s is 2 so the smallest value for f (s) is 2 + 30 = 32. Similarly the

largest value for f (s) is 24+ 30 = 54.

Range off even integers in the interval [32, 54].

b.

f(s):58++330 y

s=y+ 30

6-30 = y

f -1(s)= s -30 This function converts a woman's dress size in France to American size.

77. a. x=-2

b. Since 2 is the input of the inverse function g, it is the output of the function f. Looking at y = 2 on the graph off we see that it

corresponds to the input I. Thus, g(2) =1 .

79. No. Linear functions that are horizontal lines are not one-to-one, so they do not have inverses.

81. Quadrant II

83. The function f (x) = x 5 + x 3 - x is an odd

function that is not one-to-one.

85. One possible way is to define f with the restriction x < 0 .

SECTION 5.2

1. 53 = 125

3. _2 1 1

_

5. 2(32 ) = 2(9) =18

7. 2.11/3 =1.2806

9. 41.6 = 9.1896

11. 3'ri = 4.7288

13. e3 = 20.0855

15. e-15 = 0.0821

17.

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4X

Y 16

12

-16-12-3 -4 -4

4 X

Y

-16-12-8 -4 -4

25.

4 8 12 16X

-

_

29.

27.


19. 31.

21. 33.

23. 35.

37. a. y-intercept: (0, —1)

b. Domain: (—co, co) ; Range: (—co, 0)

C. Horizontal Asymptote: y = 0

d. f (x) —> 0 as and x .— —co f (x) -- —oo as X —> CO

-8 -6 -4 -2

x -2

- 4

- 6

39. a. y-intercept: (0, 2)

b. Domain: (—co, oo) ; Range: (—co, 3)

c. Horizontal Asymptote: y = 3


5

120

Intersection X = 11.5S2t-153,Y = 10 0 -1-1-1 15

53. x P.-. 1.4650

5

55. x —3.3219

—5

6

5


—5

b. Domain: (—oo, oo) ; range: (-00, 0.36791

c. x-intercept: (0, 0) ; y-intercept: (0, 0)

d. f(x) —co as x —> —oo and f(x) —* 0

as X—' oo

61. A = 1500(1+ 1 oD6 (1)(5) )

= $2007.34

63. A =- 1500[1 + "6)'12'5) =$202328 12

65. A = 1500e°'(3) = $1795.83

67. A =1500e"'5155) = $1793.58

d. f(x) —* 3 as x —co and f(x) —> —oo

as x -4 co

41. a. y-intercept: (0,7)

b. Domain: (—co, co) ; Range: (0, oo)

c. Horizontal Asymptote: y = 0

d. f (x) —> 0 as x —4 —oo and f (x) —4 co as X — 00

11

5 5

—5

57. x 11.5525 10

k —8 —6 —4 —2 0 2 —3

43. a. y-intercept: (0, —1) 59. a.

b. Domain: (—co, co) ; Range: (-4, co)

c. Horizontal Asymptote: y = —4 5

d. f(x) -4 co as x —> —oo and f(x) —>

as

—4

X —* 00

45. This graph does not represent an exponential function since it does not include a horizontal asymptote.

47. This graph does not represent an exponential function since it includes a vertical asymptote.

49. d

51. a


69. S(t)= 10,000(1.05)i

Jears'at Work Annual Salary _a

0 $10,000.00

1 $10,500.00

2 $11,025.00

3 $11,576.25

4 $12,155.06

71. V(t)= 20,000(0.9)'

YeaN Since.

- :Purchase

Value

0 $20,000

1 $18,000

2 $16,200

3 $14,580

4 $13,122

73. V(t)= 18,000(01)1'

i. Years Since Purchase Value

1 $12,600.00

2 $8820.00

3 $6174.00

4 $4321.80

5 $3025.26

V 20,000

16,000

12,000

8000

4000

o0 2 4 6 8

0 0314"0) 75. A = 100011 + = 1348.35 .

4 The bond would be worth $1348.35. If the bonds continued paying interest, their value would have no upper bound. Thus it would be financially onerous for the government to guarantee this rate over a long period of time. For instance, after 80 years, a bond purchased for $1000 would be worth $10,924,90, nearly 11 times its purchase price.

V(0... 18,000(0.70)'


77. a. W(t)= 17.48(1.027)'

b.

Year Hourly Wage

2000 $17.48

2001 $17.95

2002 $18.44

2003 $18.93

2004 $19.45

2005 $19.97

2006 $20.51

2007 $21.06

C. 18.95 - 18.93

= 0.02

= 0.00106 18.95 18.95

The predicted value is within about 0.1% of the actual value.

79. a. The maximum height occurs where x=0:

h(0) = -34.38(e'(am eo.ot(o)) 693.76

= -34.38(1+1)+693.76

.--- 625 feet

b.

h(100)= -34.38(e-°' moo) + eo.o1000)) + 693.76

= 587.66 feet

c. x s-.1-243.06, 243.06

700

All&

Intersection X= 21-13.059-1Sa = 300

81. a. As x -> +co , = -1

0 so e'

f (x) = 2 + e-x-p2.

b. You could observe the graph to determine whether the function is asymptotic to the line y = 2 .

-300 300

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83. a. 11.

-1

f(x) = 2x

-2 0.5

-0.5 -1 ,5 2

0 0 1

0.5 1 ,5

1 2 2

1.5 3 2,5

2 4 4

2.5 5 4,5

3 6 8

3.5 7 8,5

4 8 16

b. (-oo, 1)U (2, oo)

c. (1,2)

d. When x increases by 1 unit, f (x) increases by 2 units.

e. When x increases by 1 unit, g(x) doubles.

f. For values above x = 2 , doubling produces a greater value than adding 2 units, so the value of g(x) increases much faster than

the value of f (x) .

SECTION 5.3

1. -4-j.= 31/2

3. -;/i.d=iolo

5. True

7. True

9. 8.45x106

Logarithmic

Statement

Fxponential

Statement

log, 1 = 0 3° = 1

log 10 = 1 101 =10

log, 1 = -1 1 5-1 .=__. _

5 5

logn x -= b, a > 0 te = x

13.

k

Exponential

: :Statement-

Logarithmic

Statement

34 = 81 log, 81 = 4

51/3 =. ,5 log, -;15- =- -1 3

6 -1 1

= - 1

log, - = -1 6 6

av=u,a> 0 loga u=v,a>0

15. 104 = 10,000 , we get log 10,000 = 4 . Since

1 17. Since 101/3 = VTO , we get log = - .

3

19. Since e2 =e2 , weget ln e2 = 2 .

1 21. Since e1/3 = e1/3 ,we get in e113 -=

3

23. log10''=x+ij

25. log 10k = k

1 27. log2 12- = log, 21/2 -= -2-

1 29. log3 = log3 I = log, 3-4 = -4 -8'1 34

31. logio 4 = -2

33. log4 4'2+1 = x2 +1

35. 2log 4 1.2041

37. 1n12-. 0.3466

39. log 1400 3.1461

41. 2log -1 -1.3979 5



43. log3 1.25 = In 1.25

0,2031 In 3

45. log, 0.5 = In 0.5

— 0.4307 In 5

47. log, 12 -= In 12

3.5850 in 2

49. log, 150 = In 150 2.5750 In 7

51. log2 x = 3

x=23 = 8

1 53. log, x =

x = 31/3 =

55. log, 216= 3

x3 = 216

x = =6

57. x> 0 Domain: (0, co) ; vertical asymptote:

x= 0 ; x-intercept: (1, 0) ; y-intercept: none

59. x> 0 = Domain; (0, cx)) ; vertical asymptote:

x= 0 ; x-intercept; (1, 0) ; y-intercept: none

61. x> 0 = Domain: (0, co) ; vertical asymptote:

x= 0 ; g(x)=0 = log x— 3 =

log x = 3

x = 103 = 1000

x-intercept: (1000, 0) ; y-intercept: none

63. x+1>0 x>-1;Domain: (-1, co) ;

vertical asymptote: x =- —1;

f(x)=0 log, (x + 1) =

x + 1 = 40

x = 0 x-intercept:

(0, 0) ; f(0) = log, (0 + 1) _= log, 1 = 0 = y-intercept:

(0, 0)

65. x + 4 > 0 —4 ; Domain: (-4, co) ;

vertical asymptote: x = —4; f(x)=0 = In(x+4)= 0

x+4=e° x= —3 x-intercept:

(-3, 0) ;

f(0) = In (0 + 4) = in 4 y-intercept:

(0, ln 4)



67. x —1> 0 x> 1; Domain: (1, co) ;

79. log 7 = t ; t 0.8451

vertical asymptote: x =1; 15 g(x) 0 = 2 log, (x — 1) =

log, (x —1) = 0

x —1= 10°

x = 2=x-intercept:

(2, 0) ; -5 g(0) = 2 log, (0 — 1) = 2 log, (-1) = not real

81. 4(10') = 20

y-intercept: none 10' = 5

log 5 = x; x 0.6990

30

Intersect X =

on = 20 2

83. I = 10,0004 = R(I) = log{10,0004)

= log(10,000) = 4

85. R(I)

7.8 = )

= 1078 = 63, 095, 734.45 To

—3 —2 —1

—2

73. log 7 8.5. The graph corresponds to The ratio is about 794.33 : 1.

f (x) = 10' , so log f (x) =- log 10' = x. In this

case, when f(x) -= 7, x 8.5. 89. pH = —log [H]= —log 10-4 = —(-4) = 4

75. c 91. a. Bubble sort: 100 =- 10,000 operations

77. b heap sort: 100 log 100 -= 100 2 = 200

operations

5

4 2 0

—2 4 6 8 x

-4 —6 -s

69. t > 0 =o t >0 ;Domain: (0, oo) ;

vertical asymptote: t = 0 ; t-intercept: (1, 0) ;

y-intercept: none

71. Ix( >0 for all real x except x 0

domain: (—oo, 0)U (0, co);

vertical asymptote: x = 0 ; f (x) = 0 = log lx1= 0

lx1= 10°

lx1= 1 x = ±1

x-intercepts: (-1, 0), (1, 0) ; y-intercept: none

= 63,095, 734.451,

87. R = 7.1 = 7.1 = log(—I and 4

= 107.1 =- 12,589,254.12 1;

I = 12,589,254.121,

R = 4.2 4.2 = lo{g L. Io

-170. = 104 2 = 15,848.93192

I = 15,848.931921,

12, 589, 254.12% 794.33

15,848.93192K

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II Operations, n2

5 25

10 100

15 225

20 400

b. 101. The graphs of the two functions are identical.


The corresponding increase would be 300 operations.

C.

Ii Operations, n log n

5 3,49

10 10

15 17.64

20 26.02

e. n2 grows faster than flog n

100

20

93. Because 102 = 100 and 103 =- 1000 , the value of x for which 10' = 400 is between 2 and 3, as is the value of log 400.

SECTION 5.4

1. V7x--=x1 / 5

3.

5. True

7. log 35 ---- log (5 x 7)

= log 5 +log 7 = 0.6990 + 0.8451 =1.5441

2 9. log -

5 = log 2 - log 7

-= 0.3010 -0,8451 = -0.3980

13. log 125 = log 53

3log 5 = 3(0.6990) = 2.097

15. log(xy3) -= log x + log y3

-= log x + 3 log y

17. log V-4-27) = log .6+ log

= log x113 + log y114

1 -1

1og x +-log y 3 4

The corresponding increase would be 16 operations.

d. The heap sort is more efficient because n2 11. log if= log 2' /2 grows faster than flog n.

-1

log 2 =- -1

(0.3010) = 0.1505 2 2

95. log 1000 = x where f(x)=10 and 19. log -71- =- log (y:V"-x)

f(x)=1000 , so x=3 .

97. log 0.5 = x where f(x) = 10' and f (x)= 0.5,

so x= -0.3010.

99. (10,3) = 3 = alog 10 = 3 = a;

f(x)= 3log x

= log y + log

= log y + log x1/4

1 log y + -log x

4

21. log x2H--.1-- = log x2y5 - log 10

= log x2 + log y5 -1

2log x+ 5log y-1

4 3 2 1

-1 -2 -3 -4


y2

43. ln (x2 —1) — ln (x —1) = ln x-1

= in (x +1)(x —1)

x-1 =-1n(x+1)


23. ln ---y- ln — ln

= in e3 -2 2 =-1n x-2 3

V x2 + y 25. log„

a3 =log V X2 + y — log, a3

-= log, (x2 + y)1/2 — 3 1 = —2

log, (x2 + y) — 3

27. log, x6 = loga y3z5 Vy3z5

= log, logn Vy3z5

= log, x3 — log„ y3/22/2

= 3 loge x —(logn y312 + log„ 212 )

= 3loga x— —3 log, y --5 log„ z 2 2

1 xy3 29. log ,3,1_33_ = _3 log —

2 1

= —3

(log xy3 — log z5 )

1(log x + 3log y — 5log z) 3 1 5

= —3

log x + log y --log z 3

31. log 6.3 —log 3 = log-6,3 = log 2.1 3

33. log 3 + log x + log = log 3x +log Ty

= log 3x.,[ti

35. 3log x +-1log y — log z log x3 + log — log z 2

= log x3 — log z

=log

37. log 8 +1= log 8 + log 10= log (8 x 10) = log 80

39. 21n y +3 = ln y2 +ln e3 = ln y2 e3

41. —1 log, 81y5 + log, y3 = log, 4F3T.7 + log, y3 4

= log, 3y2 + log, y3

= log, 3y5

45. 1 [log (x2 — 9)— log (x — 3)1— log x 3 1x2 -9 = —log log x 3 x-3

1 log (x + 3)(x — 3) = log x

= —1

log (x + 3)— log x 3

-= log Vx + 3 — log x

= logVx 3

47. log 10k = log 10 + log k = 1+ b

49. log k3 -= 3log k = 3b

1 51. log —k = log 1— log k = 0 — b -= —b

53. log 10'fi = Vflog 10 = -= 4-f

55. ln e' f5 = Nid ln e = .4 • 1= -4

57. 101°8(55) = 5x

59. 1010(35+1) = 3x +1

61. 1og2 8 =log2 2.3 = 3log,2 =3.1=- 3

63. loga loga a215 = —21oga a —2 • 1 = —2 5 5 5

65. f (x) = log 10x = log 10 + log x = 1+ g(x)

5

67. f (x) = ln e2x = ln e2 + ln x = 2 + g(x)-

-5

—5

5 10

10



69. Solution A: pH= —log [H+]

5= —log [F1+ ]

—5 = log [Fr ]

[H+]=10-5

10-5 1 The ratio is — = —

10-9 1 10,000: 1.

71. pH = —log [H+1

= —log (6<10-°)

= —(log 6 log10-8) = —(log 6 — 8log 10) = —(log 6 — 8) -= —(0.778 — 8) = 7.2

73. pH = —log [H+ ]

3.4 = —log [H+]

= log [H+ ]

[H]= 10-3 4

Pi low 75. D 10 log — .= 10 log .= 10 log 20 = 13.01 dB P2 0.5W

77. a. f (x)-= 2'

b. Domain off. (—co, co) ; range off (0, oo)

C.

d. Domain of f-1 : (0, oo) ; range of f-1 :

(—oo, oo)

79. a. Domain off (0, co) ; domain of g:

(—oo, oo)

b. The functions agree for all real values of x > 0 .

81. No, because there is no real value of y for which aY is equal to a negative number, in this case x = —3 .

SECTION 5.5

1. True

3. False

5. 5' = 125

5' _53 x = 3

7. 10' =1000

10' = 103 x = 3

1 9. 4 = 176 4X 1

42

4' = 4-2 x = —2

11. 4e' = 36 ex 9

ln ex = In 9 xln e = ln 9

x -=ln 9 2.197

13. 2' = 5

in 2' = ln 5 xln 2 = ln 5

ln 5 X = tz-2, 2.322

ln 2

Solution B: pH= —log [14+1

9= —log [H1

—9 = log [H1

[H] =10-9

o' = 104 =10,000 , or

Os

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Intersection X= .521-11-17981 Y = 2

5

-5

5

5 5

-5


15. 3(1.3') = 5

1.3' = —5 3

ln 1.3' =in —5 3

xln 1.3 -= In 5 — In 3

x —ln 5 —ln 3

R: 1.947 In 1.3

17. bc =

log 10' = log 2-'4 xlog 10 = (—x + 4)log 2

x —xlog 2 + 4log 2 x xlog 2 = 4 log 2

x(1 +log 2) = log 16 log 16

x R0.926 1+ log 2

I

19. 3-2x-1 =__ 2x

In 3-2" =in 2' (-2x-1)1n 3 = xln 2

—2x1n 3 — In 3 = x In 2 —2x1n 3 — xln 2 = ln 3 —x(21n 3 +ln 2)= In 3

—In 3 0.380

2In 3 +ln 2

21. 1000eu4x = 2000 e0.04x = 2

In eu4x = ln 2 0.04x ln e = ln 2

0.04x =ln 2 ln 2 ,„

x = — 1/ ..1L7 0.04

23. 5ex + 7 = 32 Sex =25

ex = 5

In ex =ln 5 xln e = in 5

x = ln 5 R:1.609

25. 2(0.8x) — 3 = 8

2(0.8') = 11

0.8' =11

ln 0.8' =1 2

xln 0.8 = in 11—in 2

x = In 11—in 2

7.640 in 0,8

27. ex2+1 — 2 = 3

e'2+1 = 5

ln e' 2+1 = ln 5

x2 +1 = ln 5

=ln 5 —1

x = +.41n 5 —1 R: +0.781

29. 9 — e' 2-1 = 2

—es = —7

= 7

in e'2-1 = ln 7

x2 —1 =In 7

x2 =In7 +1

x = 41n 7 +1 R: +1.716

35. log x = 0

10in =i0°

x = 1

37. ln(x —1)= 2

eln(x -1) = e2

x —I.= e2

x = e2 +1, - 8.389



49. ln(2x)=1+1n(x+ 3) ln (2x)— ln (x + 3) = 1

2x in 3 = 1

in 2x e x+3 = el

2x = e x + 3

2x = e(x + 3) 2x = ex +3e

2x — ex =3e x(2 — e) = 3e

3e

x —11.353 2 — e

For this equation, x can't be negative so the answer is: No solution.

51. log (3x +1) — log (x2 +1) = 0

log 3x +1

=0 x2 +1 log 3,x4 1

10 " =i0°

3x +1=1 x2 +1 3x +1 -= x2 + 1

=x2 -3x 0 = x(x— 3)

x =0, x =3

53. log (2x + 5) + log (x +1) = 1

log [(2x + 5)(x + 1)1= 1

log (2x2 + 7x + 5) = 1

101°g (2r2+7'+5) = 101

2.X2 +7x+5=10

2x2 + 7x — 5 = 0

—7 + V(7)2 —4(2)(-5) x

2(2)

—7 ±../0 --= 0.608, _7-4 AtIg. 4

x = —7 +-A-9-

4

55. log2 (x + 5) = log2 (x)+ log2 (x —3)

log, (x + 5) = log2 (x2 — 3x)

x+ 5 =x2 —3x

0= x2 — 4x —5 0 = (x + 1)(x —5)

, x =5

39. log(x+ 2)=1

10108( `+2) =101 x+2=10

x = 8

41. log, (x + 4) = 2 31083 (x+4) = 32

4 = 9 x = 5

43. log (x +1) + log (x —1) = 0

log [(x +1)(x —1)] = 0

log (x2 —1) = 0

10105 °2-11 = 100

x2 —1 = 1

x2 =2

x = 74'

45. log x + log (x+ 3)=1

log [x(x + 3)] = 1

log (x2 + 3x) = 1

101°g( `2+31) = 101

+ 3x = 10

+ 3x — 10 = 0 (x+ 5)(x— 2)= 0

, x = 2

47. log, x = 2 — log, (x — 3) log2 x + log, (x — 3) = 2

log, (x2 — 3x) = 2 21082(12 -3A) _ 22

— 3x =4

— 3x— 4 = (x + 1)(x — 4) = 0

, x = 4

0 2010 Cengage Learning, All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.

57. x 3.186

5

5 5

-5

59. x 5.105, x -3.105

5

5 5

-5


61. A = Pert

2000 =1000emf r =In 1.5

= 0.05068; r -= 5.07%

2 = e°.065 8

1n2 = In eu6t 1n2 = 0.06t

t In 2

t=-11.55 years 0.06

63. A = Per' In -20

= In e'r 17

In -20= 5r 8000 = 4000ea°575t

17 2 = ea.0575t In -20

In 2 = In e5.057" r = 17 = 0.03250; r = 3.25%

1n2 = 0.0575t

5 In 2

12.05 years 73. P = piek? = 0.0575

t

2 =-- (1)ek(12) 65. A = Pert 2= el2k

5400 = 2700ea75t ln 2 =12k1n e 2= In 2 -= 12k

1n2 ln eu75' k = -

ln 2 0.0578

1n2 0.075t 12 To reach 8 million:

ln 2 To reach 4 million:

075 924 years t - P = ea05781 p = e0.0578t

0. 4 = eci."" 8 = e005"t

In 4 = 0.0578t in e In 8 = 0.0578t ln 4 = 0.0578t 1n8

- 36 hr ln 4

t = - 24 hr

0.0578 t =

0.0578

75. a. s(0) = 2297.1e°331610) = 2297.1 In 1971, there were about 2297 transistors per chip.

C 2010 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.

71. A = Pert

10,000 = 8500er 5)

••0

= e 17

67. A = Pe'

2000 =1500er°) 4

e5,

In -4 = In e'r 3

in-4 =5r 3

In -4

r = = 0.05754; r = 5.75% 5

69. A -= Per'

6000 = 4000er 8)

1.5 -= ea'

in 1.5 = in ear

ln 1.5 = 8r

83. D = 101og

0.7 -= 10log 75W

0.07 = log 75W


85. a.

s(t)= 2297.1e°3316t

2(2297.1) = 2297.1eu316'

2 = e°336'

0.3316t =- In 2

t 1n2 = 0.3316

2.09 yr

77. a. v(0) = $43,173

b. It loses 20% of its value each year.

C.

v(t) = 43,173(0.8)'

22,227 = 43,173(0.8)'

22, 227 -(0.8)'

43,173 in 22,227 -

tin 0.8 43,173

In 22'227

t = 43'173 - In 0.8

Z975 or about 3 yr

79. f (t)= 10e"

5 = 10e236°'

1 _ _ e24,360k

2 in 0.5= 24,360kln e In 0.5 = 24,360k

In 0.5 k =

24,360

P.1 -0.0000285 = f (t) = 10e-° °°°°285`

f (t) --= 10e-°0°285'

2 =

1 _-5 e-00000285t

In 0.2 = -0.0000285t

In 0.2 t 56, 472 yr

-0.0000285

81. pH = -log [H+1

1.5 = -log [H-1

-1.5 ---- log [H+]

10-4'5 = 10411-+ I

[H4 ] = 10-15 mole per liter

10°' = 75W

P, =- 10° '7 (75 W) 88 W

A(t) = ikoe'

A(t) = 5e"

2.5 -= 5esk

0.5 = esk 8k = in 0.5

In 0.5 k -= 0.0866

8 A(t)=. 5c0.0871

b. B(t) -= Boekt

B(t)= 5e kt

2.5 -= 5e3k

0.5 = e31

3k= In 0.5

k

in 0.5 0.231 = B(t) 5e-0 231t

3

C. A(t) = 5e-u0" 3 = 5e-0.0871

0.6 =

-0.087t = In 0.6

In 0.6 t =

-0.087 5.87 days

B(t) = 5e- 0.231t

3 = 5e-um 0.6 = e-0.231t

-0.231t = In 0.6

in 0.6 t

-0.231 2.21 days

d. The solution produced with water that has a pH of 6.0 should be used because more malathion will remain after several days than will remain in the solution produced with water that has a pH of 7.0.

b.



30

e.

6

87. A little under two years

89. V(t)= Poe"

V (t)= 5000e"

5309.18 = 5000e'(1)

1.061836 = er

9. 5= 10e-t

0.5=e'

ln 0.5 = -t

t = -ln 0.5 0.6931

11. f(1) = 4e1 = 4e P.-, 10.8731

13. 8 = 4et

2 =et

ln 2= t

t = ln 2 0.6391

10 10 10 15. 1(0) =

1+ 2e° 1+2(1) = 3

r = In 1.061836 ce, 0.060 10 _ 10

V(t)= 5000e°°65 and r =6% 17. f(10) -

1+ 2e ° 3t10 - 1+ 2e' 9.0944

91. This equation has no solution because there is no value of x for which the left side of the equation will equal a nonpositive value.

93. The step assumes that the logarithm of both sides of the equation have been taken, but this can be done only if the logarithms have the same base, which is not the case in this equation.

95. e)O8 e

log x = 1

101°81 =101

x =10

97. ln I2x - 31= 1

I2x - 31= el

12x - 31= e

2x - 3 = e or

2x= 3+e

x = :2.859 2

2x - 3 = -e

2x = 3 - e

x = 3-e

0.141

19. f (e) 31n e - 4 = 3(1)- 4 = -1

21. 2 = 31nx- 4

6 = 31nx

2= In x

e2 = x

x=e2 7,3891

23. c

25. a

27. a. A(t)= Aoek'

1 = A

°ewmk

1 _ emu

2 In 0.5 = 5700k

k = in 0.5 5700 -0.0001216 A(0= Ace-0.00012161

b. A(t)= AOC0.°11°1216t

1 A _ _ A ,,-0.0001216t 3 -`

2 3 - e

x = 3 + e

0.141, 2.859 x = P--;

1 -0 0001216t - = e 3

In -1

= -0.0001216t 3

ln -1

t 3 SECTION

2

5.6

2

= r.-.19035 years -0.0001216 1. growth

3. oo

5. 0

7. f (0) = 10e° =10.1=10

© 2010 Cengage Learning, All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.

_1111=011111 11111


29. A(t) = Aoe-00001216t

0.70,40 = Aoe' '1216'

0.70 =

in 0.70 = —0.0001216t

in 0.70 t = 2933 yr

—0.0001216

31. a. 16.0 million in 2000 (0, 16.0); C =16.0

17,4 million in 2004 = (4, 17.4)

P(t) 16ekt

17.4 = 16ek(4)

17.4 = e41

16

in —17.4=

4k 16

In 17,4

k= 16 0.02097; 4

P(t) = 16e 2O97

b. 2010 t = 2010 — 2000 = 10 P(10) =16e°2°97110)

=16e°2097 c.-,19.7 million

33. (0, $123,000), (14, $220,000)

P(t) = 123,000ekt

220,000 = 123, 000e14k

220 = el4k

123

in 220

=14k 123

in 220

k = 123 0.04153; 14

P(t) = 123,000e°°4193t

35. a. Purchase price $23,024 C = 23,024

P(t) = 23,024ekt

17,160 = 23,024ek

2145 — ek

2878 2145 k = in — —0.29396; 2878

P(t)= 23,024C° 293961

b. 2009 t = 2009 — 2006 = 3

P(3) = 23,024C°29396(3) $9532

37. a. N(0)= 300 300 = 20 deer 1+14e-m10 - 1+14(1)

b. N(10) = — 32 deer 1 + 14e °°10>

300

39. a. f (x) = 2.8265(1.1975)x

22

a

a

b. 2008 x = 2008 —1990 =18

f(18) = 2.8265(1.1975)1g = $72.4784 billion

c. This is not a realistic model over the long term because as x increases, the function produces values for f (x) that represent a

greater revenue than the global population can provide.

41. a. A logistic function fits this data well because there is an upper limit to the speed that a land vehicle can attain under the laws of physics.

330 (x) = 1+ 2.23 e-"232x

250

76

c signifies the limiting speed of the car.

2008 x = 2008 —1931 = 77

330 f(77)= 1 +

240 mph 2.23C°0232(77)

P(t) = 16.76(1.079)t

60 a

a a

b. 2009 x = 2009 —1992 =17

P(17) = 16.76(1.079)17

$61.04 per barrel

15

b.

C.

d.

43. a.

15

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45. a. The annual cost increases as the arsenic concentration is lowered.

b. C(x)-- 774.6092(0.8786)x

700

24

c. a must be less than 1 because the cost is decreasing as the concentration is increasing.

d. C(12)=774.6092(0.8786)12 $163.9 million

e. C(2) = 774.6092(0.8786)2 598

This means that it would cost nearly $600 million a year to keep the arsenic concentration to 2 micrograms per liter.

47. If the quantity of strontium-90 is half its original amount after the time deemed its "half-life" has passes, then after another half-life, the remaining quantity will be further reduced by half. So it

A will take two half-lives to reach the value .

4 Because the half-life is 29 years, it would take 58 years for the initial amount A, to be reduced to

An 4

49. The function f (t) = e(112)' cannot model

exponential decay because as t —> , the function f(t) increases. To model decay,

as t —>00, the function f (t) must decrease. This

would occur if e were raised to a negative exponent, as opposed to a fractional positive exponent.

CHAPTER 5 REVIEW EXERCISES

f(g(x))— f [ x 27)=2i x 271+7= x-7+7 x _2x x

2 2

f(g(x))= 8r1 = 8(x)— 2 2 x

3 8 3 = x g( f(x)) = g(8x3)=- = 2x

2 2

5 --x = y 4

(x) =— —5 x 4

7. f(x) = —3x + 6

y=-3x+6

x=-3y+6

x —6=-3y

x — 6 = Y —3

6— x3

9. f (x) + 8

y=x3 +8

x=y3 +8

x-8=y3

8 =1' f(x)=Ix-8

11. g(x)=—x2 +8, x 0

y=—x2 +8, x>0

x=—y2 +8, y>0

x-8=—y2, y>0

8—x=y2, y?_0

+'J8—x=y, y>0

V8—x=y,x<8

g-1(x)= J8 — x, x <8

13. f(x)=—x-7 y=—x —7

—y —7 x+7=—y

—x-7=y3 r(x)= —x-7

1.

3.



15. f (x) = —x3 +1

= —x3 +1 x=—y3 +1

x —1= —y3

1 — x y3

-= y

f -1(x) =/1—x

17. y-intercept: (0, — 1) ; other points:

(1, — 4), (2, — 16)

domain: (—oo, co) ; range: (—co, 0) ; f(x)—> 0 as

—> —oo and f (x) —oo as x co

19. y-intercept: (0,1) ; other points: 11, —21 (-1,1 3 2

domain: (—oo, oo) ; range: (0, co) ; g(x) oo

as x —> —oo and g(x) 0 as x —> co

23. y-intercept: (0, 3) ; other pts:

[1,-2 +1 ,(-1,2e+1)

domain: (—oo, co) ; range: (1, oo) ; g(x) —> oo

as x —oo and g(x)—+1 as x —> oo

-2

2 4 6 8 X

25.

Logarithmic

•4„..., .$tatement

Exponential

SMternent.:

log, 9 = 2 32 = 9

log 0.1 = —1 10-1 = 0.1

log, —1 =

25 —2 5-2 _

1 _ 25

27. log, 625 = 4

29. log9 81 = 2

31. log 41.5 = log 101/2 =1

1 33. ln = ln el" = —

3

35. log 10 2 = x + 2

37. 4 log 2 1.2041

39. in 4-§ 1.0397

41. log, 4.3 = in 4.3

1.3277 in 3

-4-3-2-1 2 3 4X

43. log6 0.75 = in 0,75

— 0.1606 21. y-intercept: (0, 4) ; other points: (1, 4e), (-1, —4) in 6

domain: (—co, oo) ; range: (0, oo) ; f(x) --> 0 45. x> 0 domain: (0, co) ; y-intercept: none as x —4 — 00 and f (x) --> co as x —> co 0=logx-6 = 6=logx = x=106

x-intercept: (106, 0) ; vertical asymptote: x -= 0

8 Y- 2-

6

j0

4 0

2468 X

-2-

-4-

-8 -6 -4 -2 0 2 -8-



47. x> 0 = domain: (0, oo) ; y-intercept: none

0 = 3log4 x 0 = log, x x 4° = 1

x-intercept: (1, 0) ; vertical asymptote: x = 0

1 61. —tlog (x2 — 1)— log (x +1)] + 3log x

4

—1 log

x2-1 =

, + 3tog x 4 x + 1

6 (x +1)(x —1)

= —1

log + x 3 log 4 x + 1 4

2 = —

1log (x —1) + 3 log x 0 —2 4 —4 = log — 1 + log x3

=log x.34/x — 1

49. log 21 = log (7 x 3)

= log 7+ log 3 = 0.8451 + 0.4771 = 1.3222

51. log = log 5 — log 3 3

= 0.6990 — 0.4771 = 0.2219

53. log V-XV37 = log V-X- + log

= log X1/4 + log yi" 1 1

= —4

log x +-3

log y

21og3 x2 —log, ,Fc = log, (x2)2 — log, Vi) 3

= log, x 4 —log, (x1 / 2)"

= log, x 4 —log, x116

, x4 = —

x" = log, x23/6

65. 5' = 625 55 =54

x = 4

63. 1/3

\I X6 X3 67. 7'1 =

55. log, 5 5 = log, 3/2 5/2 49

,y z y z

=-- log, x3 — log, y3nz"2. 7' = 1 72

= log, x3 —(log, y3/ 2 + log, 2/ 2) 7' -= 7-2 3 5 x = —2

= 3log, x — •log, y — -log„ z

57. In 3lxy3 = x" Z5 Z5 / 3

= ln xli'y —ln z513

= in xi" + in y — in Z5/3 1 -= —in x + in y — —51n z 3 3

59. ln (x2 — 3x) — ln (x — 3) = ln x2 — 3x x — 3

= x(x — 3) ln

x — 3 = in x

69. 25ea°4x = 100 e0O4x = 4

0.04x = in 4

x = —ln 4

RI 34.6574 0.04

71. 421'1.3 = 16

421+3 = 42 2x+ 3= 2

2x = —1 1

x =

73. e2:c+1 = 4

2x+1=ln 4 2x=ln 4-1

x =

in 4-1

0.1931 2

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85. f(0)= 4e-2.5M = 4 d. f(3) = 4e-2'5(3) ' 0.0022

87. f(0) = 201n (0 + 2) +1 = 201n 2 +1 14.8629 f (3) = 201n (3 + 2) +1 = 201n5 +1 F.-, 33.1888

89. j.(0) 100 _ 100 _ 20 1 + 4e-02(0) — 1 + 4(1) —

f(3)..= 100 31

'2965

1 + 4e-°'213)


75. ln (2x —1) = 0 eln (2x-1)

2x — 1 = 1

2x = 2

x

77. in X1/ 2 = 2

—1

1n x = 2 2

in x = 4

eIns e4

x = e4 54.5982

79. log x + log (2x —1) = 1

log [x(2x —1)1= 1

log (2x2 — x) =1

2x2 — x = 10'

2x2 — x —10 = 0

(2x — 5)(x + 2) = 0

x =

81. log (3x +1) — log (x2 +1) = 0

log 3x +1

=0 X2 +1 3x+1

= 100 x2 + 1 3x+1=x2 +1

0= x2 — 3x 0 = x(x — 3)

x =0, x =3

83. log, x +log, (x + 3) = 1

log, [x(x + 3)] = 1

log, (x2 + 3x) = 1 41001 (x2 +3x) = 41

x2 +3x= 4

x2 +3x — 4 = 0 (x + 4)(x —1) = 0

x =1

0 014(6) 91. A = 150011 + =$2021.03

4

93. A = 1500e°°" = $2065.69

95.

Number of Years

i After Purchase \ ,11

1 $12,750.00

2 $9562.50

3 $7171.88

4 $5378.91

5 $4034.18

V(t) =17,000(0.75)1

16,000

12,000

8000

4000

oo 2468 't

97. R(101.0) log 0

—101_

log 10 = 1

450

16

About 14.3 months

450

Intersection X = 1L1.255551-1_Y = 16

99. a. N(0) = 450 450 =

45 trout 1 + 9C°30 — 1 + 9(1)

b. N(15) = 1 +

191;5_00 305) 409 trout

c. As t increases, the number of trout in the pond increases, but at an increasingly slower rate. The The graph is asymptotic to N(t)= 450.


Documents

Chapter 5 Exponential and Logarithmic Functions · 5 120 Intersection X = 11.5S2t-153,Y = 10 0 -1-1-1 15 53. x P.-. 1.4650 5 55. x —3.3219 —5 6 5 148 Chapter 5: Exponential and