Thomas Calculus 13th [Solutions]

40 Chapter 1 Functions
56. (a)
(b)
(c)
0.01t
0.01t
e 1000 ln e ln1000 ( 0.01t)ln e = ln 1000 0.01t = ln 1000 t = 100 ln 1000
kt 1 kt 1
e ln e ln10 kt ln e ln10 kt = ln 10 t ln10
10
k
(ln 2) t 1 ln 2 t 1 t 1
e ( e ) 2 2 2 t = 1
2
57.
t 2 t 2 2
e x ln e ln x t 2ln x t 4(ln x)
58.
2 2 2
x 2x 1 t x 2x 1 t x 2x 1 t 2
e e e e e ln e ln e t x 2x
1
log
59. (a)
5 7
log
5 7 (b)
8 2
log
8 2 (c)
3 75
1.3 75
2
(d) log4 16 log4 4 2log4
4 2 1 2
1/2
(e) log 1 1 1
3 3 log3 3 log
2 3 3 1 0.5
2 2
1
(f) log 1
4 log
4 4 4 1log4
4 1 1 1
log
60. (a)
2 3
log
2 3 (b)
10 (1/2)
10 1
(c)
2
2
(d) log11121 log1111 2log11
11 2 1 2
1/2
(e) log 1 1 1
12111 log121121 log
2 121 121 1
2 2
(f)
1
2
log3 log
9 3 3 2log3
3 2 1 2
log 7
7
z 2z z 2 z
61. (a) Let z log4
x 4 x 2 x (2 ) x 2 x
z z 2 2 2z 2 z 2
(b) Let z log3
x 3 x (3 ) x 3 x 9 x
(ln 2)sin x sin x
(c) log 2( e ) log2
2 sin x
2 z 2 z 4
62. (a) Let z log 5(3 x ) 5 3x 25 9x
x
(b) log e( e ) x
(c)
x x x
e sin x ( e sin x)/2 log sin
4(2 ) log4 4
e x
2
log2
x
63. (a) ln x ÷ ln x ln x ln 3 ln 3
log x ln 2 ln 3 ln 2 ln x ln 2
(c)
3
ln x ln ln ln ln
2
2ln
ln
x
2 ln 2
ln x ln ln ln
a a ÷ a a x x 2
a x x a x
log2
x
(b) ln x ÷ ln x ln x ln 8 3ln 2 3
log x ln 2 ln8 ln 2 ln x ln 2
8
log9
x
64. (a) ln x ÷ ln x ln x ln 3 1
log x ln 9 ln 3 2ln 3 ln x 2
(b)
(c)
3
log x
ln 2
ln x ÷ ln x ln x
ln 2
log ln 10 ln 2 ln10 ln x ln10
10
x
1
2 2
loga
b ln b ÷ ln a ln b ln b ln b
log a ln a ln b ln a ln a ln a
b
1
2
2
65. (a)
6
(b)
4
(c)
3
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