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*
a;
y
1/
-4
-2
0
»(*) = 8"
y
X
l
-2
16
1
-1
•i
1
i
64
1
8
0
1
2
4
1
8
4
16
2
64
7. /(*)-(i:
8.
frfart-ri.il
a:
y
X
y
-2
9
-5
0.620921323
-1
3
-1
0.90
0
1
0
1
5
1.61051
1
2
9.
6.
1
3
1
10
!)
/ (x) = 3^
10.
X
-2
-1
y
2.59374246
g(x) = 2eX
y
0.406
-4
1.104
-3
14.7761
8.96338
0
3
-2
5.43656
0.5
4.946
-1
3.29744
1
8.155
0
2
1.5
13.445
1
1.21306
2
22.167
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, so a = 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. The graph 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 the y-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 4 units. 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. So
graph of y = 10x to the left3 units. Domain: (-oo, oo).
thegraph of/ is obtained 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.
304
CHAPTER 4 Exponential and Logarithmic Functions
33. y = -ex. The graph of y = -ex is obtained from the
34. / (x) = 1 - ex. The graph of / (x) = 1 - ex is obtained
graph of y = 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 isobtained
the graph of y = ex by reflecting it about the y-axis then
by reflecting the graph of y = ex about the y-axis and then
shiftingdownward 1 unit. Domain: (-co, oo). Range:
aboutthe x-axis. Domain: (—00,00). Range: (—00,0).
(—l,oo). Asymptote: y = —1.
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 + 4is
graph ofy = ex byshifting ittothe right 2 units. Domain:
obtained byshifting the graph ofy = ex totheright
(-00,00). Range: (0,00). Asymptote: y = 0.
3 units, 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
40. Using the points (-1,15) and (0,5) we have / (0) = Ca° = 5
a"1 =3
<&
&
C = 5. Then / (-1) = 5a_1 = 15 &
a= i.Thus/(x)=5(i)x.
42. (a)
41. (a)
(b) /(x) = 9X'2 = (32)I/2 = 32x/2 = 3X = g(x). So
(b) Since g (x) = 3 (2X) = 3/ (x) and / (x) > 0, the
height of the graph of g (x) is always three times the
f(x)=g (x), and the graphs are the same.
height of the graph of/ (x) = 2X, so the graph of g is
steeper than the graph of/.
43. / (x) = 10x, so
f(x + h)-f(x)
,x+h
10
-10a
10x10ft-10x
:
h
1rtX10h-l
= 10
:
44.
/(x) = x3
X
45.
305
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
10
1,000,000
46.
,
306
CHAPTER 4 Exponential and Logarithmic Functions
47. cosh(—x) =
-* +e
_i_ --(-*)
e~-
„-*_!_
„*
e--"
+ e"
>_a: _ C-(-x)
48. sinhi—x) =
v
'
2
„* +
J,e~x
e"
e
o~x _ ^
=
2
= cosh x
_r._a: -1- o*
=
a* _ a~x
=
2
= —smhx
2
49. (coshx)2 -(sinhx)2 ={^f1)2 ~{f^f1)2 =\(e2* +2+e~2x) -J(e2x -2 +e-2x) =\+\=1
Qx _ e-x
50. sinh x cosh y + cosh x sinh y =
y
=
2
ey + e-y
•
ex+v ^_ ex~y —ev~x —e~^x+y^
ex _j_ e-x
1
2
+
&y _ &-y
•
2
2
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 increases much more quickly than the graph of g.
(i) [0,5] by [0,20]
(ii) [0,25] by [0,107]
(Hi) [0,50] by [0,108]
100.000,0001
10,000.000
/(
8,000,000
9
80,000,000
6,000,000
60,000.000
4,000,000
40,000,000
2,000,000
20,000,000
0J
10
/
0
20
9
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]
0,20] by [0,105]
I
100,000-I
80,000
91
60.000
//
40,000
20.000
0
10
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.
=
,0i
4
c-1
c=l
8
6
. 0 = 0.5
4
— c = 0.25
2
-2
0
2
4
The larger the value of c, the more rapidly the graph of
The larger the value of c, the more rapidly the graph of
/ (x) = c2x increases. Also notice that the graphs are just
/ (x) = 2CX increases. In general, / (x) = 2CX = (2C)X ;
shifted horizontally 1 unit. This is because of our choice
so, for example, / (x) = 22x = (22)x = 4X.
of c; each c inthis exercise is of theform 2k. So
/ (x) = 2k •2X = 2x+fe.
SECTION 4.1 ExponentialFunctions
307
56.
55.
)
2
1
0
10
»
so
40
Note fromthe graph that y = [1 + (l/x)]x approaches e
Fromthe graph, we see that / (x) = [1 - (l/x)]x
as x get large.
approaches 1/e « 0.368 as x get large.
57. (a)
a=1.5->
...
•
,a=\
J J ,a=0.5
lOn a=2,
58. y = 21/x has vertical asymptote x = 0 and horizontal
asymptote y = 1.
\\ \ 1 8
\\\ 1 6
xv\ 4
-4
-2
A
2
4
-I
.2
(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 horizontal
60. p (x) = xx. Notice that g (x) is only defined for x > 0.
x
asymptote y = 0. As x —• —oo, y —> 0, and as x —• oo,
The graphofg (x) is shown in the viewingrectangle
[0,1.5] by [0,1.5]. From the graph, we see that there is a
local minimum of about 0.69 when x « 0.37.
61. g (x) = ex + e~3x. The graph ofg (x) isshown inthe
62. y = 10*
viewing rectangle [—4,4] by [0,20]. Fromthe graph, we
see that there is a local minimum of approximately 1.75
when x « 0.27.
(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].
308
CHAPTER 4 Exponential and Logarithmic Functions
63. y = xe"
(a) From the graph, we see that the function / (x) = xex is increasing on
(-co, 1] and decreasing on [1,oo).
(b) From thegraph, we see thatthe rangeis approximately (-oo, 0.37).
-I
64. D (t) = 50e_0-2t. So when t = 3 we 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 isabout
6.6 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)
(c)
(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 the
velocity after 5 s is about 50.6 ft/s.
v(10) =80 (1 - e-°-2(10)) »80 (0.865) =69.2 ft/s. So the
velocity 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. Thus
(c)
approximately 2.7 lb of salt are in the barrel after 5 minutes.
(b) Q(10) =15 (l- e-004(10)) * 15(0.3297) =4.946.Thus
approximately 4.9 lb of salt 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
(a)P(0) =
1200
1200
1 + lle-°-2(°) "1 + 11
= 100.
1200
1200
1200
<b> P(10) = i +iie-o^o) - 482. P(20) = , +lle_0,2(20) « 999. P(30) = TT_.2(30)
(c) As t —> oo we have e -0.2t
0, so P {t)
1200
1 + 0
= 1200. The graph shown confirms this.
1168.
SECTION 4.1 ExponentialFunctions
70. n(t) =
5600
309
(b)
0.5 + 27.5e-°044t
(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 =
300
12,000 •]
(b)
\nB =12,000
10,000
K ' 0.05 + (^2 _ 0.05) e-°-M<12>
5164.
8000
"°= 800°
(c) The population approaches 6000 rabbits.
6000
4000
//"o =50
2000
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,
rate per year
i = 4% per year = -pr- permonth, and k = 12•number
i =
of years, we fill in the table:
we fill in the table:
75. P =
,
, .
,n
Time (years)
Amount
Rate per year
Amount
1
$5203.71
1%
$5256.25
2
$5415.71
2%
$5525.39
3
$5636.36
3%
$5808.08
4
$5865.99
4%
$6104.98
5
$6104.98
5%
$6416.79
6
$6353.71
6%
$6744.25
10,000, r = 0.10, and n = 2. So A(t) = 10,000 (l + ^)2t = 10,000 •1.052t.
(a) A (5) = 10,000 • 1.0510 « 16,288.95, andso thevalue of the investment is $16,288.95.
(b) A (10) = 10,000 • 1.0520 w 26,532.98, and so the value of the investment is $26,532.98.
(c) A (15) = 10,000 • 1.0530 « 43,219.42, and so the 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.
76. P =
A(4) = 4000 •(1.04)16 « 7491.92, and sothe amount due is$7,491.92.
(b) A(6) = 4000 •(1.04)24 « 10,253.22, and so the amount due is$10,253.22.
(c) A (8) = 4000 • (1.04)32 « 14,032.23, and sothe amount due is $14,032.23.
(a)
/
0 09 \rat
/
0 09
=3000 and r= 0.09. Then we have A(t) =3000 f1+
^-^
J ,and so A(5) = 3000 I1+
^^
77. P =
(a)
(b)
(c)
(d)
_
£r—^— per month, and k = 12 • 5 = 60 months,
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.
310
CHAPTER 4 Exponential and Logarithmic Functions
(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)
1+-J
/
7. \;20
,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) is greater than the effective rate in(ii), we can see that the account paying 9^% per year
compounded 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)6v60
«•
100000 = 1.4898P
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.
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 than
method (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 there
are 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
^IHHHj
Logarithmic Functions
V1.
Logarithmic
Exponential
form
form
log88 = l
2.
Logarithmic
Exponential
form
form
81 =8
log, 64 = 3
43 = 64
og8 64 = 2
82=64
log42=±
41'2 = 2
iog84=|
82/3 = 4
43/2 = 8
og8512 = 3
83 = 512
log48=|
log, & = -2
iog41 = -i
log, 3L =-|
o-l _
8 8 = _1
8
"8 64 = ~2
8
1
- 8
o-2 _
1
-64
4~2
= i16
*
4-1/2 - I
1
3. (a) 52 = 25
(b) 5° = 1
4. (a) 10"! =0.1
(b) 83 = 512
5. (a) 81/3 = 2
7. (a) ex = 5
8. (a) e2 = x + 1
(b) e4 = x - 1
9. (a) log5 125 = 3
(b) e5 = y
11. (a)log8l=-l
12. (a) log, 0.125 = -§
(b)log2(|)=-3
(b) log7 343 = 3
15. (a) log33 = 1
(b)log3l=log33° = 0
(c)log332 = 2
17. (a)log636 = log662=2
(b)log981 = log992=2
(c)log7710 = 10
— 2
4-5/2
J.
^
32
6. (a) 34 = 81
(b) 82/3 = 4
(b) 2"3 = I
10. (a) log101000 = 3
(b) log81 9 = I
(b) log10 0.0001 = -4
13. (a) In 2 = a;
14. (a) ln0.5 = .x + l
(b) lni = 0.5x
(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
19. (a)log3(£)=log33-3 = -3
(b) log10 v/10 = log10 101/2 = §
(C)log50.2 = log5(i)=log55-1 = -1
20. (a) log,, 125 = log5 53 = 3
(b)log,97 = log,9491/2 = ±
21. (a) 2log237 = 37
22. (a)e1,,7r = tt
(b) 10log5 =5
(c) 10log87 = 87
(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
(c) log9 s/3 =log9 31/2 =log9 (9I/2) V2 =log9 9
1/4
24. (a) log, s/2 =log, 21/2 =log, (41/2) 1/2"=log, 41'4
_
i
4
(b) log, (1) =log, 2-1 =log, (41'2) ~* =log, 4"1'2
_
_i
1
(C) log, 8=log, 23 =log, (41/2)" =log, 43/2
25. (a) log2 a: = 5 «=> x = 25=32
(b) x = log2 16 = log2 24 = 4
26. (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
28. (a)x = log42 = log441/2 = ±
(b) log,x = 2 <s> x = 42 = 16
29. (a) log10 x = 2 <^ x = 102 = 100
(b) log5 x = 2 «* x = 52 = 25
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 = \
(b)logx8=| «- x3/2 =8 <* x=82/3 =4
<*
x1/2 = 6 & x = 36
(b)logx3 =i «- x1'3 =3 «• x=27
34. (a) log 50 « 1.6990
33. (a) log2« 0.3010
(b) log 35.2 « 1.5465
(b) logy^« 0.1505
(c) log (|) « -0.1761
(c) log (3\/2) « 0.6276
36. (a) In 27 * 3.2958
35. (a) ln5« 1.6094
(b) In 25.3 « 3.2308
(b) In 7.39 « 2.0001
(c) In (1 + V3) « 1.0051
(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 is
y = log2 x.
39. Since the point (3, §) is on the graph, we have 5 = loga 3 & a1/2 =3
«• a = 9. Thus the function is
y = log9 x.
40. Since the point (9,2) isonthe graph, we have 2 = loga 9
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.
«=>
a2 =9
44. IV
•£>
o = 3. Thus the function isy = log3 x.
45. VI
46.1
48. The graph of y = log3 x is obtained from the graph of
y = 3X by reflecting it about the line y = x.
SECTION 4.2 Logarithmic Functions
313
49. / (x) = log2 (x - 4). The graph off isobtained from the 50. / (x) = - log10 x. The graph off isobtained from the
graph of y = log2 x byshifting it to theright 4 units.
graph of y = log10 x by reflecting 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
51. g (x) = log5 (-*). The graph ofg is obtained from the
y
52. g (x) = In (x + 2). The graph ofg is obtained from the
graph of y = log5 x by reflecting it about the y-axis.
graph of y = In x by shifting it to the left 2 units. Domain:
Domain: (-oo,0). Range: (-00,00). Vertical asymptote:
(—2,00). Range: (—00,00). Vertical asymptote: x = —2.
x = 0.
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) - 2
from the graph of y = log3 x by shifting it upward 2 units.
Domain: (0,00). Range: (-00,00). Vertical asymptote:
is obtained from the graph of y = log3 x by shifting it to
theright 1 unit andthen downward 2 units. Domain:
x = 0.
(1,00). Range: (—00,00). Vertical asymptote: x = 1.
314
CHAPTER 4 Exponential and Logarithmic Functions
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 of y = log10 x byreflecting it about the
obtained from thegraph of y = Inx by reflecting it about
x-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 obtained from the
lnx
58. y = In |x|. Note that y =
ifx>0
The
ln(-x) ifx<0
graph of y = In x by reflectingthe part of the graph for
0 < x < 1 about the x-axis. Domain: (0, oo). Range:
graphof y = In |x| is obtainedby reflecting the graph of
[0,oo). Vertical asymptote: x = 0.
y = lnx about the x-axis. Domain: (—oo, 0) U (0, oo).
Range: (—oo, oo). Vertical asymptote: x = 0.
>'4
y*
59. / (x) = log10 (x + 3). We require thatx + 3 > 0
<$•
60. / (x) = logs (8 - 2a0- Tnenwe musthave 8 - 2x > 0
x > -3, so thedomain is (-3, oo).
&
61. g (x) = log3 (x2 - 1). We require that x2 - 1 > 0 <=>
8 > 2x
^
4 > x, and so the domain is (—oo,4).
x2 > 1 ^ x < -1 or x > 1, so the domain is
(-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 from Chapter 1 with the
endpoints 0 and 1, we get the table at right. Thus the domain
is (0,1).
(-00,0)
Interval
Sign of x
Sign of 1 - x
+
Sign of x (1 —x)
63. h (x) = In x + In (2 - x). We require that x>0and2-x>0
domain is (0,2).
«=>
64. h (x) = \/x-2 - log5(10 _ x) • Tnen we mustnave £ - 2 > 0 and 10 - x > 0
2 < x < 10. So the domain is [2,10).
<=>
(l,oo)
+
+
+
+
-
x>0andx<2
(o,D
•«•
-
0 < x < 2, so the
x > 2 and 10 > x
«=>
SECTION 4.2 Logarithmic Functions
65. y = log10 (1 - x2) has domain (-1,1), vertical
315
66. y = In (x2 - x) = In (x (x - 1)) has domain
asymptotesx = -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.
10
-10
67. y = x + In x has domain (0, oo), vertical asymptote
x = 0, and no local maximum or minimum.
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), vertical
x
.horizontal
.
. , asymptote
f/^„ y = 0,
n and, ,„„„,
local „,„„;„,..„,
maximum „,^,nQ7
y « 0..37
asymptote
x = -10,' and local minimum yy « -3.62 at
J v
atx« 2.72.
xw-5.87.
to
•to V,
y
^r
•10
71. The graphofg (x) = y/x growsfaster than the graphof
72. (a)
/ (x) = In x.
(b) From the graph, we see that the solution to the
equation y/x = 1 + In (1 + x) is x «s 13.50.
316
CHAPTER 4 Exponential and Logarithmic Functions
73. (a)
74. (a)
(b) Noticethat / (x) = log (ex) = log c -+- log x, so as c
(b) As c increases, the graphoff (x) = c log (x)
increases, the graph of / (x) = log (ex) is shifted
stretches vertically by a factor of c.
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
(b) y = In (In (lnx))
<&
ev = In (lnx)
•&
«•
&
Inx > 1
ee = lnx
&
&
x > e. Sothedomain off is (e,oo).
ee
= x. Thus the inverse function is
/-1(x) = ce'V
77. (a)/(x) =
2X
1 + 2*" y" = 1+2
y + y2x = 2x
(b)
<&
&
1-x
> 0. Solving this using the methods from
Chapter 1, we start with the endpoints, 0 and 1.
y = 2x-y2x = 2x(l-y)
&
(-oo,0) (0,1) (l,oo)
Interval
2X =
y
<&
x = log2
1-0
r>(,)=log2(^_).
(*)•
Sign of x
Thus
+
Sign of 1 —x
Sign of
1-x
+
+
+
+
Thus the domain off 1(x)is(0,1).
78. Using J=0.7/o we have C=-2500 In (y ) =- 2500 In 0.7 =
891.69 moles/liter.
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. When r = 7% we have i = —— « 9.9 years. And when r = 8% we
0.06
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
and W= 10 we find .he ID to be ^fA{W) = !»«(»• MO/10) _ log20
log 2
5.23
4.32
1.23 times harder.
log 2
log 2
log 2
Sq
sma||er
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 and
5280 feet per mile, there are 12 (5280) = 63,360 inches per mile. So the height ofthe graph is '-g™6 « 264.8, or
about 265 miles.
(b) Since log2 (22'1) = 24, we must be about 224 inchest 265 miles to the right ofthe origin before the height ofthe
graph of y = log2 x reaches 24 inches or 2 feet.
85. log (log 10100) = log 100 = 2
log (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 have
log5 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 integer
where 10""1 < x < 10", then x hasn digits and [logx] = n - 1. Since [logx] = n - 1
•»
n= [logxj + 1, the
numberof digits in x is [logx] + 1.
4.3
Laws of Logarithms
1. 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 = 3
6. log12 9 + log12 16 - log12 (9 •16) = log12 144 = log12 122 = 2
7. 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 =-2
9. 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 =200
13. 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)
17. Iog610 =101og6
16. log., (|) =log5 x- log5 2
18.ln(V5)=ln(zl/2) =±\nz
19. 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 and Logarithmic Functions
2
28. log -^-j=c =log a2 - log (64VE) =2loga- (4log6+±log c)
29. log2 (^=^J=log2x+log2 (x2 +1) -Jlog2 (x2 -1)
30> l0g5 Vfri =2l0g5 (xTl) =*[loS5 (* -1) -log5 (x +1)]
31. \n(xJij =lnx+±ln(^) =lnx+ \(lny - lnz)
3x2
32. ln
(* + l)10
= ln (3x2) - ln(x+ 1)10 = ln3 + 2lnx - 10 ln(x+ 1)
33. log Vx^+F = \ log (x2 + y2)
34. log 3/-_— = logx - log v"l-x = logx - \ log (1 x
35. log
x' + 4
(x2 + 1)(x3 - 7)2
= ±log
x2+4
2*"6 (x2 + 1) (x3 - 7)2
=1[log (x2 +4) - log (x2 +1) (x3 - 7)2]
= 1 [log (x2 + 4) - log (x2 +1) - 2log (x3 - 7)]
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- log2=log (12V7) - log2=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
= logE
"°°
(x-l)(x+l)
= log5(x + l)
X - 1
43. 4logx - i log (x2 + l) + 2log(x- 1) = logx4 - log ^Tl + log(x - l)2
MwTih^'-^M^^-)
44. ln(a + 6) + ln(a-6)-21nc = ln[(a+ 6)(a-6)]-ln(c2) = ln
45.
a2-62
In5 +21nx +31n (x2 +5) =ln (5x2) +ln (x2 +5)3 =ln [5x2 (x2 +5)3]
.2\2
46. 2[log5 x+2log5 y- 3log5 z] =2log5 3L =log5 0^\ =log
5
xV
~6
47. |log(2x+l) + ±[log(x-4)-log(x4-x2-l)] = log V2x~+\ + \ log
x — 4
x4 - x2 - 1
-'-(v^-'J^tt)
LJC
48. logab + clogad-rlogas = loga (bdc) - logasr = loga —
SECTION4.3 Lawsof Logarithms
50. log55 2= r^|
« 0.430677
log5
49. log2 5= J^?
« 2.321928
log 2
51. log316 =
log 16
log 3
53. log7 2.61 =
52. log6 92 = ^?
« 2.523658
log 6
2.523719
log 2.61
log 7
54. log6 532 =
0.493008
55. log4 125 =128125
^ 3482892
log4
57. iog3 x=^
log 532
log 6
3.503061
56. log12
2.5 = t^?
« 0.368743
°"~"
log 12
=_ =_ lnl. The graph of, =^ In* is
shown in the viewing rectangle [—1,4] by [-3,2].
58. Note that log,
319
•"&)••
>
x (by the change of base formula). So
2a = e
the graph of y = logc x is obtainedfrom the graph of y = ln x by
a= 5
—
either shrinking or stretching vertically by a factor of -— depending
0
on whetherln c> 1 or ln c < 1. All of the graphs pass through (1,0)
-2-
/rf/
2
a = 10
4
because logc 1 = 0 for all c.
59. loge =
lne
1
ln 10
ln 10
60. (log25)(log57) =
log 5 log 7 _ log 7
log 2
log 5
log 2
= log2 7
61. -h(.-^m) -,(7Z^=) -.(-^.=±£3) -(5££fJ)
= ln (x + y/x2 - 1)
62. From Example 5(a), P =
Po
(t + lY
Substituting P0 = 80, f. = 24, and c = 0.3 we have P =
80
(24 + 1)0.3
30.5. So the
student 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.
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 species
eightfold.
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 be
their respective magnitudes. Since log is an increasing function, we have log B\ < logB2. Then
logBi<logB2 <*> log Bi - log Bo < log B2 - 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
320
CHAPTER 4 Exponential and Logarithmic Functions
66. (a) False; log(x/y) = logx - log y ^ (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; the equation is an identity: log 2~ = z log2.
(e) False; logP + logQ = log(PQ) ^ (log P) (logQ).
(f) False; loga - logb = log(a/6) ^ (loga) / (log/;).
(g) False; x log2 7 = log2 T # (log2 7)J:.
(h) True; theequation is an 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. The error is on 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/ shrunk
horizontally by a factor of|, whereas the graph of4/ (x) is the same as the graph of/ (x) stretched vertically by a factor
of 4.
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 the
sameas 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
\ isthe same asa vertical shift upward by ln2.
4.4
1. 10J' = 25
2. lO"* = 4
Exponential and Logarithmic Equations
&
&
loglOJ" = log25 & x log10 = log25
logl0-a! = log4 <^ -x = log4 <=>
^
x « 1.398
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 585Q
log 2
6. 32-r"1=5 ^ logS2'-1 =log5 <=> (2x-l)log3 =log5 ^ 2x - 1= ^log 3 «• 2x = 1+ ^log 3
x=i(l + 52lf J^i.2325
2 V~ ' log3,
7. 3c1 = 10
«•
ex = f
«•
8. 2e12* = 17 ^
e12a! = ¥
9. <' •'•'
I
2 o
!./•
10. 4(l + 105x) = 9 <=>
x = ln(f) « 1.2040
<=> 12x = ln(f)
In 2 •=•
4a:
1 + 105x = f
<* x= £ [hi(f)] »0.1783
I I-In 2 <>
«• 105x = f
.*•
-' "4 — =0.0767
<*>
5x = log (|)
<*
X=\ [log5-log4] « 0.0194
11. 4+35x =8 <* 35r =4 «• log35x = log4 <£> 5xlog3 = log4 *> 5x = }^
log 3
x=J°ll«
0.2524
5 log 3
<3>
SECTION 4.4 Exponential and Logarithmic Equations
321
12. 23x =34 <£> log23x =log34 «• 3xlog2 =log 34 <=> x=j^—^ ~16958
13. 804x =5 «*• log8°-4x =log5 & 0.4xlog8 =log5 <S> 0.4x =|^| e> x=Q°^ g»1.9349
log 0.1
14. 3I/14=0.1 <S> log3x/14 =log0.1 ^ (^)log3 =log0.1 <S> x= 14log
3
-29.3426
15. 5"x/100 =2 «- Iog5-x/100 =log2 ^> --^-log5
=log2 <*> z=-10|01(?2
^-43.0677
100
log 5
16. e3_5x = 16 <=> 3- 5x =ln 16 <=>
-5x =ln 16 - 3 <S> x=-^ (ln 16 - 3) « 0.0455
o
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; =-^«-3.1144
log 4
19. 5X = 4X+1
<=>
log5x =log4x+1
<£>
xlog5 = (x + l)log4 = xlog4 + log4
«=>
xlog5-xlog4 = log4
* x(log5-log4) =log4 ^ X=_M1_« 6.2126
20. 101_X=6X
<S>
loglO1-1 = log6x
<=>
l-x = x(log6)
«*•
l=x(log6) + x
^
l=x(log6 + l)
<* x=,log 6*+1 »0.5624
21. 23x+1 = 3X"2
&
log23x+1 = log3x_2
log2 = xlog3-2log3
<^
<*>
(3x + l)log2 = (x-2) log3
3xlog2-xlog3 = -log2-21og3
<=>
&
3xlog2 +
x(31og2 - log3) = - (log2+ 21og3)
log2 + 2log3
31og2-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
1 + e-1
<=>
50 = 4 + 4e-x
&
46 = 4e_x
&
11.5 = e_x
<S>
4 = e_x
<&
lnll.5 = -x
&
x = -lnll.5 « -2.4423
24. ——— =2
&
10 = 2 + 2e_x
«*-
8 = 2e_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„
14.0065
2 log 1.04
26. (1.00625)12' = 2
<S>
log 1.0062512' = log2
«.
12r.log 1.00625 = log2
<S>
t = 12 log1^
« 9.2708
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 the
only solutions are x = ±1.
28. x210x - xl0x = 2(10x)
«•
x210x - xl0x - 2(10x) =0
&
10x (x2 - x - 2) = 0 =» 10x = O(never) or
x2 - x - 2 = 0. If x2 - x - 2 = 0, then (x - 2) (x + 1) = 0 =• x = 2, -1. Sothe onlysolutions are x = 2, -1.
29. 4x3e_3x - 3x4e"3x =0
3x = 4
<f>
«*•
x3e_3x (4 - 3x) = 0 => x = 0 or e_3x = 0 (never) or 4 - 3x = 0. If4 - 3x = 0, then
x = 4. So the solutions are x = 0 and x = 4.
322
CHAPTER 4 Exponential and Logarithmic Functions
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
31. e2x - 3ex + 2 = 0 <*
(ex - 1)(ex - 2) = 0 => ex - 1 = 0 ore1 - 2 = 0. Ifex - 1 = 0, then ex = 1 &
x = ln 1 = 0. If ex - 2 = 0, then ex = 2
32. e21 - ex - 6 = 0 <*
&
.
2
<*
x = ln 2 « 0.6931. So the solutions are x = 0 andx « 0.6931.
(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, since
e2x > 0 for all x. But we can solve e2x = 3 <=> 2x = ln3 <=>• x = ±ln3 « 0.5493. So the only solution is
x 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
«»
36. ln (2 + x) = 1
37. logx = -2
x = e10w 22026
<*
<*
2 + x = e1
<s>
x = e - 2 « 0.7183
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 «31.6667
40. log3 (2 - x) = 3
2 - x = 33 = 27
&
-x = 25
2 = ln(3-x)
<s>
e2 = 3-x
&
41. 2-ln(3-x)=0
«»
42. log2 (x2-x-2)=2
<*
*>
x = -25
<=>
x = 3 - e2 * -4.3891
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) <=>
(x —4) (x - 2) = 0
«=•
log(x2) = log(6x - 8) «•
x2=6x-8
<=>
x2-6x + 8 = 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 is
undefined. Thus the only solution is x = 5.
46. logs x + loS5 (ar + 1) = log5 20
(x + 5) (x - 4) = 0
«=>
&
log5 (x2 + x) = log5 20
&
x2 + x = 20
&
x2 + x - 20 = 0 «•
x = -5 or x = 4. Since log5 (-5) is undefined, theonly solution isx = 4.
47. log5(x+l)-log5(x-l) =2 «. log5 f^l
=2 ^ ^x —1 =52 & x+1=25x - 25 ^
\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 only
solution.
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
SECTION 4.4 Exponential and Logarithmic Equations
323
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>/3
sox = 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
<=>
T _ c-i/2 - _i_
log222/'og5x = log2(^)
log5x
= -4
<=>
log5x = -1
<*
: 0.4472
54. log2 (logs x) = 4 «55. lnx = 3 - x
<*
&•
log3 x = 24 = 16
<s>
x = 316 = 43,046,721
lnx + x-3 = 0. Let
/ (x) = ln x + x —3. Weneed to solvethe equation
/ (x) = 0. Fromthe graph off, 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 «
57. x3 - x = log10 (x + 1)
<=>
1.47.
58. x = ln (4 - x2)
«*•
x - ln(4 - x2) = 0. Let
x3 - x - log10 (x + 1) = 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
0
-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 equation
we get x ss -0.57.
/ (x) = 0. From the graph of /, we get x «s 1.38.
o
324
CHAPTER 4 Exponential and Logarithmic Functions
61. 4_x = ^x
<=•
4-x - y/x = 0. Let
62. ex -2 = x3-x
/ (x) = 4_x - y/x. We need to solvetheequation
«»
ex -2-x3 + x = 0. Let
/ (x) = ex —2 - x3 + x. We need to solve the equation
/ (x) = 0. From thegraph off, we get x « 0.36.
/ (x) = 0. From the graph off, 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 ofalogarithm
is positive we must have 0 < -x2 + llx - 18
<s>
0 < (x- 2) (9 - x). Using the methods from Chapter 1with the
endpoints 2,4, 7, 9 forthe intervals, we make the following table:
(~oo,2)
Interval
(2,4)
(4,7)
(7,9)
Sign of x —7
+
Sign of x —4
Sign of x - 2
+
+
+
+
+
Sign of 9 - x
+
+
+
+
Signof(x-7)(x-4)
+
+
Signof (x-2) (9 - x)
+
-
+
-
+
+
Thus the solution is (2,4) U (7,9).
64. 3 < log2 x < 4
65. 2 < 10x < 5
<&
23<x<24
&
(9,oo)
<&
log 2 < x < log 5
8 < x < 16.
<s>
+
+
1
-
0.3010 < x < 0.6990. Hence the solution to the inequality is
-
approximately 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 1with
the endpoints - \/2 and y/2, noting that ex > 0 for all x. We make a table:
Interval
Sign of ex
Sign of (x- V2)
Sign of (x + y/2)
Sign ofex (x - y/2) (x+ y/2)
(-oo,-v^)
(-V2,y/2)
(V5,oo)
+
+
+
+
+
+
-
+
+
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.
( 0 085 \ 4t
1+ ^—J = 5000 (1.021254*) & 2= 1.021254t «*•
(b)
t =
log 2
71—,, --,-_ « 8.24 years. Thus the investment will double in about 8.24 years.
4 log 1.02125
J
log 2= 4t log 1.02125 «*>
SECTION 4.4 Exponential and Logarithmic Equations
325
68. (a) A{2) = 6500e006(2) « $7328.73
(b) 8000 =6500eoo6t & g =e006t «. ln (If) =0.06t <* t= -L ln (if) « 3.46. So the
investment doubles in about 3| years.
69.8000 =5000^1 +^^") =5000(1.018754t) «- 1.6 =1.018754* <s> log 1.6 =4tlog 1.01875 ^
t=
^—'•
4 log 1.01875
« 6.33 years. The investment will increase to $8000 in approximately 6years and 4 months.
70. 5000 =4000 (l +2^21^
_ —logl .25
& 1.25 = (1.04875)2' & log 1.25 =2tlog 1.04875 «•
s .^
b
2
2log 1.04875
71. 2 = e0085t
<&
save $5000
3'
ln 2 = 0.085*
4*
In 9
t = -—— « 8.15 years. Thus the investment will 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 only
5 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) =
» '
y '
x + 4e—
ttzt*; = 7.337, so there are approximately 7337 fish after 3 years.
' '
(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 9 months.
—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 intensity
drops to 25 lumens at 86.6 ft.
P
79. (a) ln (—\
= --k «» ^=e_,,/fc «*• P = Poe_,,/fc. Substituting k=7and P0 = 100 we get
\Po J
"
Po
P = 100e_,,/7.
(b) When /i = 4 we have P = 100e_4/7 » 56.47 kPa.
80- (3) lD (w) =-O.iu
«.
T~20 =e-011t
200
<* T - 20 = 200e-011t
(b) When t = 20we have T = 20+ 200e-°n(20) = 20+ 200e-22 « 42.2° F.
^
T = 20 + 200e"0 nt-
326
CHAPTER 4 Exponential and Logarithmic Functions
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
&
•--V
Ce~kl =M-P
&
(c)
* -"-m —-4
M-P
in
(b) P(t) = 20 - 14e-°02". Substituting M = 20,C = 14,fc = 0.024,
and /' = 12 into /. = -- ln ( ——— ), we have
t =
1
, /20-12
— ———- ln
0.024
\
14
~ 23.32. So it takes about 23 months.
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 to 9* = 100must be between 2 and 3 (because 100 is between 81 and 729).
84. Notice that log (x1/lo*x)
= r—
logx = 1, so 1/loBX = 101 for all x >0. So
v
/
log X
x1/ logr = 5 has no solution, and x1/ logx = khas asolution only when k = 10.
This isverified by the graph of/ (x) = xx/logx.
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=
11
x
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
.5
In 2
J
Modeling with Exponential and
Logarithmic Functions
1. (a) n (0) =500.
(b) The relative growth rate is 0.45 = 45%.
(c) n (3) = 500e°',5(3) « 1929.
(d) 10,000 = 500e°-lr"
40 minutes.
«•
20 = e045'
&
0.45* = In20
<=>•
i = ^In 20 « 6.66 hours, or 6 hours
0.45
<=>
SECTION 4.5 Modeling with Exponential and Logarithmic Functions
2. (a) The relative growth rate is 0.012 = 1.2%.
327
(d)
(b) n(5) = 12e0012(5) = 12e006 « 12.74 million fish.
(C) 30= 12e0012t <=> 2.5 = e0012t <=> 0.012* = ln2.5
ln2.5
* =
0.012
76.36. Thus the fish population reaches 30 million after
about 76 years
3. (a) r = 0.08 and n (0) = 18000. Thusthe population is given by the
(c)
formulan(*) = 18,000e008t.
(b) t = 2008 - 2000 = 8. Then we have
n(8) = 18000e008(8) = 18000e064 « 34,137. Thus there should
be 34,137 foxes in the region by the year 2008.
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, the
projected population will beapproximately 233 million people by theyear 2020.
(b) r = 0.02;n(25) = 110,000,000e002(25) = 110,000,000e0-50 « 181,359,340. Thus ata 2% growth rate, the
projected population will be approximately 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 the
population will reach 600 frogs in about 11 years.
7. (a) The deer population in 1996 was 20,000.
(b) Using the model n (*) = 20,000ert and the point (4,31000), we have 31,000 = 20,000e4r
4r = ln 1.55 ^ r = \ ln 1.55 « 0.1096. Thus n(i) = 20,000eolO96t
«•
1.55 = e4r
(c) n (8) = 20,000eolo96(8) w 48,218, so the projected deer population in 2004 is about 48,000.
(d) 100,000 = 20,000eulUi
„
*•
„5 =_ e^
^
„
<^
„.
0.1096* = ln5
..
<&
.t
=
In5
Q1Qg6
14.63. Since
1996 + 14.63 = 2010.63, the deer populationwill reach 100,000 during the year 2010.
8. (a) Since the population grows exponentially, the population is represented by n (*) = noer\ with no = 1500
and 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
<*
328
CHAPTER 4 Exponential and Logarithmic Functions
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 number
0.1508
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 =
noe2r
'
400
=64
<*
e4r = 64
<*•
4r = ln64
<«•
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 the
1.04
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 have
t = 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. Then
ra(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. So
0.0268
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 be
approximately 40,429,000.
13. n(t) = n0e2t. When n0 = 1, the critical level is ra(24) = e2(24) = e48.We solve the equation
e48 = n0e2t, where ra0 = 10. This givese48 = 10e2t
* = 1 (48- ln 10) « 22.85 hours.
«»
48 = ln 10 + 2*
«•
2* = 48 - ln10 <=>
In O
14. From theformula forradioactive decay, wehave m (*) = moe~rt, where r = —r~.
h
In 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 4 mg.
(c) We have to solve for*in theequation 18 = 22e-°-°oo433t j^ gives 18 _ 22e-0000433t
i
«• -0.000433* = ln (^)
& t=
/
9 \
Jl"'* « 463.4, so it takes about 463 years.
-0.000433
ii
SECTION4.5 Modeling with Exponential and LogarithmicFunctions
In 9
In 0
/I
oU
329
15. (a) Usingm(*) = m0e-rt with m0 = 10and h = 30, we have r = — = —- « 0.0231. Thusm(t) = lOe-00231'.
(b) m(80) = i0e-00231(8°) « 1.6grams.
<=.
(11 =-0.0231*
= _n.n9.3H
/. =
(c) 2=10e-00231t <^ 1=e-°-023U' &
lnIn (1)
<*_> t=
ln|L « 70years.
-0.0231
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 takes
about 50 days.
(c) We need to solve for *in the equation 20 = 40e-°0277t. We have 20 = 40e_0-0277t
-0.0277* = ln 1
2
<=•
*=
«•
e"0277' = \
<»
lnl
2„„ « 25.02. Thus the half-life of thorium-234 is about 25 days.
-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 (§§) <*
28
t = -— •ln (§§) « 18.03, so ittakes about 18 years.
In2
18. 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~00231t
oU
<*
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
ln (0.8) =
h
-48
<=>>
h=
In 0.8
<s>
0.8 = e-^n2)/h],4S
<*
• 48 ss 149.1 hours. So the half-life is approximately 149 hours.
In 9
20. 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 hin 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
_^t =ln0<2
t=
=e-K1"2)/3-82)* &
=ln0.2 ^
<* *=_3-8^"0-2
~8.87. So it takes roughly 9days for a
o.oZ
3.82
In Z
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 the formula forradioactive decay, we have 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 &
*>
-0.000121* = ln 0.59
**•
*=
InO 59
'
~~ U • VJUU X £» _
» 4360.6. So it will take about 4360 years.
c-o.ooomt = Q59
330
CHAPTER 4 Exponential and Logarithmic Functions
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 of Cooling: T (*) = Ts + D0e~kt with k = 0.1947, T8 = 60,and D0 = 98.6- 60 = 38.6
SoT (*) = 60+ 38.6e-°1947t.
(b) Solve T(t) = 72. So 72 = 60 + 38.6e-°1947t
-0.1947t =ln(iy „ t._aiM7
1
38.6e-°1947t = 12
In
(—)
\38.6J
<^
e"0194'
12
38.6
**
6.00, and the time of death was about 6 hours ago.
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>
* = 30
Mil
ln(l§)
116.1. So the temperature will be 100° F after 116 minutes.
26. We useNewton's Law of Cooling: T (*) = Ts + D0e~kt, with Ts = 20 and
Do = 100 - 20 = 80. So T (*) = 20+ 80e_fct. Since T (15) = 75, we have
20 + 80e-15fc = 75
-15fc = ln(li)
<*
«.
80e"15fc = 55
&
e~15k = 11
«*•
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 cheese
is 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=log20 +log J0 - log5. Then
Mi - 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. So
t
-|p|8.3
~T = 1fl4 9 = in3 4* 2511.9, and so the San Francisco earthquake was 2500 times more intense than the Japan
earthquake.
35. Let the subscript A represent the Alaska earthquake and S represent the San Francisco earthquake. Then
MA =log (^\ =8.6 <* Ia =S- 1086 ;also, Ms =log f^-\ =8.3
t
Q • in®-®
Is
o • 108--5
<=>
Is = S • 1086.So
T" = r, ^^o o = 100'3 « 1.995, and hence the Alaskan earthquake was roughly twice as intense as the San Francisco
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
J
107'2
In
10do
&
Ik = S- 10'"*. So
I- = ma » = lft° 4 ^ 2.51, and so the Kobe, Japan earthquake was 2.5 times more intense than theNorthridge,
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^o - log%= MT - MM. Therefore
1m
1m
Im/o
b
Mt = Mm + log 1.26 « 8.1 + 0.1 = 8.2. Thus the magnitude of the Tangshan earthquake was roughly 8.2.
38. 0=101og(£) =Mkgf^QxiO-") =101°g(2x 10?) =10 (log2+log107) =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
in12
Jc •1012 = lO120. So y= —^
= 101,4 « 25.12, and so the ratio ofintensity is roughly 25.
41. (a) fa =10log (J±\ and ^=A^ =i0lOg f^-) =10 log (j-\ -21ogd1 =10log (|-) _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, and
so the intensity level at 10 m is approximately 106 dB.