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08-035_08_AFSM_C08_001-026.qxd
7/23/08
11.
10:46 AM
Page 6
y
8
6
4
2
x
–4 –3 –2 –1 0
–2
–4
–6
–8
1
2 3 4
8.3 Evaluating Logarithms, pp. 466–468
1
5 22
36
1
53
e) log 13
27
1
f) log8 2 5
3
1 23
d) a b 5 216
6
1. a) log4 16 5 2
d) log6
b) log3 81 5 4
c) log8 1 5 0
2. a) 23 5 8
e) 6 2 5 !6
1
25
c) 34 5 81
3. a) 5x 5 5; x 5 1
b) 7x 5 1; x 5 0
1
c) 2x 5 ; x 5 22
4
1
b) 522 5
f) 100 5 1
d) 7x 5 !7; 7x 5 7 2; x 5
1
1
2
2 x
8
e) a b 5 ; x 5 3
3
27
3
2
f) 2x 5 "
1
5 23
1
x5
3
1
; 10x 5 1021; x 5 21
10
b) 10x 5 1; 10x 5 100; x 5 0
c) 10x 5 1 000 000; 10x 5 106; x 5 6
d) 10x 5 25; x 5 1.40 (typing log (25) into the
calculator); 101.40 8 25
e) 100.25 5 1.78; x 5 1.78
f) 1022 5 0.01; x 5 0.01
5. a) 6x 5 !6
1
6x 5 6 2
1
x5
2
4. a) 10x 5
8-6
b) log5 125; 5x 5 125; x 5 3
log5 25; 5x 5 25; x 5 2
log5 125 2 log5 25 5 3 2 2 5 1
c) log3 81; 3x 5 81; x 5 4
log4 64; 4x 5 64; x 5 3
log3 81 1 log4 64 5 4 1 3 5 7
1
1
d) log2 ; 2x 5 ; x 5 22
4
4
log3 1; 3x 5 1; x 5 0
1
log2 2 log3 1; 22 2 0 5 22
4
1
3
x
e) 5 5 "
5; x 5
3
f) 3x 5 !27;
3x 5 "33
3
3x 5 32
3
x5
2
6. a) 53 5 x; x 5 125
b) x 3 5 27; x 5 3
1
1
c) 4x 5 ; 4x 5 3 ; 4x 5 423; x 5 23
64
4
22
1
d) a b 5 x; 42 5 x; x 5 16
4
e) 52 5 x; x 5 !5
f) 41.5 5 8
7. f(x) 5 3x can be written as log3 f(x).
a) f(x) 5 17. The point (2.58, 17) is on the graph.
log3 17 8 2.58.
b) f(x) 5 36. The point (3.26, 36) is on the graph.
log3 36 8 3.26.
c) f(x) 5 112. The point (4.29, 112) is on the
graph. log3 112 8 4.29.
d) f(x) 5 143. The point (4.52, 143) is on the
graph. log3 143 8 4.52.
8. a) 4x 5 32; by guess and check x 8 2.50
b) 6x 5 115; by guess and check x 8 2.65
c) 3x 5 212; by guess and check x 8 4.88
d) 11x 5 896; by guess and check x 8 2.83
9. a) log3 35; 3x 5 35; x 5 5
b) Given that log5 25 5 2, 5log5 (25) 5 52 5 25
1
c) Given that log4 161 5 22, 4log4 16 5 422 5 161
d) The base m log of a value is the exponent you
need to raise m to in order to get that value. You
need to raise m to the n power to get mn. Therefore,
logm m n 5 n.
e) The expression loga b means what power do you
need to raise a to in order to get b. If you substitute
that answer into the expression a loga b the result is b.
1
Chapter 8: Exponential and Logarithmic Functions
7/19/08
8:41 PM
Page 7
f) logn 1 5 0 for all non zero values for n.
1
1
10. log2 163 5 log2 (2(4) )(3 )
4
4
log2 23 5
3
11. 40(10x ) 5 2000
10x 5 50
log10 50 5 x
x 8 1.7 weeks
n
12. To determine amount of decay, use ( 12) 1620 where
n is the number of years the radium has been
decaying.
150
1 1620
a) 5a b 5 4.68 g
2
n
1 1620
b) 5a b 5 4 g
2
n
1 1620
4
a b 5
2
5
4
n
log0.5 5
5
1620
4
1620 log0.5 5 n
5
n 5 522 years
13. A: s(0.0625) 5 0.159 1 0.118 log (0.0625)
Slope 5 0.017
B: s(1) 5 0.159 1 0.118 log (1)
Slope 5 0.159
B has a steeper slope
14. a) s(50) 5 93 log (50) 1 65
s(50) 8 233 mph
b) 250 5 93 log d 1 65
185 5 93 log d
1.99 5 log d
101.99 5 d
d 5 98 miles
15. If log 365 5 32 log 150 2 0.7, then Kepler’s
equation gives a good approximation of the time
it takes for Earth to revolve around the sun.
log 365 5 2.562
3
log 150 2 0.7 5 2.564
2
3
16. a) log 2854 2 0.7 8 83 years
2
3
b) log 4473 2 0.7 8 164 years
2
Advanced Functions Solutions Manual
x
17. a) y 5 100(2)0.32
b)
y
60 000
50 000
40 000
30 000
20 000
10 000
–1 0
x
1 2 3 4
y
c)
4
3
2
1
0
x
10 000
20 000
30 000
40 000
50 000
60 000
08-035_08_AFSM_C08_001-026.qxd
d) y 5 0.32 log2 Q 100 R ; this equation tells how
many hours, y, it will take for the number of
bacteria to reach x.
e) Evaluate the inverse function for x 5 450.
450
y 5 0.32 log2 a
b
100
y 8 0.69 h
log 5
18. a) log5 5 5
log 5
5 1.0000
log 10
b) log2 10 5
log 2
5 3.3219
log 45
c) log5 45 5
log 5
5 2.3652
log 92
d) log8 92 5
log 8
5 2.1745
log 0.5
e) log4 0.5 5
log 4
5 20.5000
log 325
f) log7 325 5
log 7
5 2.9723
x
8-7
08-035_08_AFSM_C08_001-026.qxd
7/19/08
8:41 PM
Page 8
19. a) positive for all values x . 1
b) negative for all values 0 , x , 1
c) undefined for all values x # 0
20. a) 33 1 103 5 27 1 1000 5 1027
b) 51.292 2 33.24 5 8 2 35.14 5 227.14
21. a) y 5 x 3
"2
b)
3
x22
c)
"0.5
x22
3
d) 2 1 3
22. a)
y
8
6
4
y = 3log (x + 6)
2
function: y 5 22 log5 3x
D 5 5xPR0 x . 06
R 5 5yPR6
asymptote: x 5 0
inverse: y 5
D 5 5xPR6
R 5 5yPR0 y . 06
asymptote: y 5 0
10
8
6
(x22)
——— 4
y 5 10 3
2
x
2 4 6 8
x
y = 10 3 – 6
function: y 5 2 1 3 log x
D 5 5xPR0 x . 06
R 5 5yPR6
asymptote: x 5 0
x
inverse: y 5 10 3 26
D 5 5xPR6
R 5 5yPR0 y . 266
asymptote: y 5 26
8-8
(x 2 2)
inverse: y 5 10 3
D 5 5xPR6
R 5 5yPR0 y . 06
asymptote: y 5 0
y
–4 –2 0
–2
–4
–6
–8
x
–8 –6 –4 –2 0 2 4 6 8
–2
–4
–6 y 5 2 1 3 logx
–8
function: y 5 3 log (x 1 6)
D 5 5xPR0 x . 266
R 5 5yPR6
asymptote: x 5 26
8
6
4
2
y
c)
–8 –6 –4 –2 0
–2
–4
–6
–8
b)
1 2 x2
(5 )
3
x
x
2 4 6 8 10 12
y 522 log53x
y
d)
2—
y 5 31 (5 2 )
6
y 520(8)x 4
2
x
–6 –4 –2 0 2 4 6
–2
x
—
(
–4 y5log8 (–20
–6
Chapter 8: Exponential and Logarithmic Functions
08-035_08_AFSM_C08_001-026.qxd
7/19/08
8:41 PM
function: y 5 20(8)x
function: y 5 25x 2 3
D 5 5xPR6
R 5 5yPR0 y , 236
asymptote: y 5 23
inverse: y 5 log 8 Q R
20
inverse:
y 5 log5 (2x 2 3)
D 5 5xPR0 x , 236
R 5 5yPR6
asymptote: x 5 23
D 5 5xPR6
R 5 5yPR0 y . 06
asymptote: y 5 0
x
D 5 5xPR0 x . 06
R 5 5yPR6
asymptote: x 5 0
e)
Page 9
23. Given the constraints, two integer values are
possible for y, either 1 or 2. If y 5 3, then x must
be 1000, which is not permitted.
y
16
12
8
y 52(3)x+2
4
x
–8 –4 0 4 8 12 16
–4
x
–8 y 5log3 (—
2 ( –2
function: y 5 2(3)x12
D 5 5xPR6
R 5 5yPR0 y . 06
asymptote: y 5 0
inverse: y 5 log 3 Q R 22
2
x
D 5 5xPR0 x . 06
R 5 5yPR6
asymptote: x 5 0
y
f)
2
–10 –8 –6 –4 –2 0
–2
y 525x23 –4
–6
–8
–10
y5log5(2x23) –12
–14
x
2
Advanced Functions Solutions Manual
8.4 Laws of Logarithms, pp. 475–476
1. a) log 45 1 log 68
b) logm p 1 logm q
c) log 123 2 log 31
d) logm p 2 logm q
e) log2 14 1 log2 9
f) log4 81 2 log4 30
2. a) log (7 3 5) 5 log 35
4
b) log3 5 log3 2
2
c) logm ab
x
d) log
y
e) log6 (7 3 8 3 9) 5 log6 504
(10 3 12)
f) log4 a
b 5 log4 6
20
3. a) 2 log 5
b) 21 log 7
c) q logm p
1
d) log 45
3
1
e) log7 36
2
1
f) log5 125
5
135
4. a) log3 135 2 log3 5 5 log3
5
5 log3 27
53
b) log5 10 1 log5 2.5 5 log5 (10 3 2.5)
5 log5 25
52
8-9