Homework Assignment 7 (3.2,3.3)
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Section 3.2 (1,4,6,10,11,14,15,18,21,22,27,30,38,39,42,55)
= log2 218 = log2 2−8 = −8
1
6. log8 2 = log8 8 3 = 13
1
10. log 1000 = log 10−3 = −3
1 6.3
1
= log8 8 3 ×6.3 = log8 82.1 = 2.1
14. log8 26.3 = log8 8 3
18. Find t such that log2 t = 8.
Solution: By denition, log2 t = 8 means 28 = t. So t = 28 = 256.
22. logb 64 = 2
Solution: By denition, b2 = 64 and b > 0. So b = 8.
30. Find x such that log4 (3x + 1) = −2.
5
Solution: By denition, 4−2 = 3x + 1 and 3x + 1 > 0. So x = (4−2 − 1)/3 = − 16
.
x
38. Find the inverse function of f (x) = 4.7 .
4. log2
1
256
Solution:
y
=
4.7x
log4.7 y
=
x
x =
f
−1
(y)
=
log4.7 y
log4.7 y
42. Find the inverse function of f (x) = 5x − 3.
Solution:
2
y
=
5x − 3
y+3
=
5x
log5 (y + 3) =
x =
x
log5 (y + 3)
f −1 (y)
log5 (y + 3)
=
Section 3.3 (2,5,9,12,16,19,22,24,30,33,36,38)
2. x = 0.4 and y = 3.5. Evaluate
(a) log (x + y) = 1.3610
(b) log (x) + log (y) = 0.3365
1
12. Suppose m and n are positive integers. log m ≈ 41.3 and log n ≈ 12.8. How many
digits does mn have?
Solutoin: log mn = log m + log n ≈ 41.3 + 12.8 ≈ 54.1. So 55 digits.
16. Suppose log a = 203.4 and log b = 205.4. Evaluate
Solution: log ab = log b − log a = 2. So
b
a
= 100.
b
a.
22. Given log4 u = 3.2 and log4 v = 1.3.
log4
u
8v
=
log4 u − log4 8 − log4 v
=
3.2 − log4 4 2 − 1.3
=
3.2 − 1.5 − 1.3
=
0.4
3
24. Given log4 u = 3.2 and log4 v = 1.3.
log4
√
1
u =
log4 u 2
1
=
log4 u
2
1
=
× 3.2
2
= 1.6
30. Given log4 u = 3.2 and log4 v = 1.3.
log4
u2
v3
=
2 log4 u − 3 log4 v
=
2 × 3.2 − 3 × 1.3
=
2.5
36. log5 (x + 4) + log2 (x + 2) = 2.
Solution: First of all, x + 4 > 0 and x + 2 > 0. So x > −2.
log5 (x + 4) + log2 (x + 2)
=
2
log5 (x + 4) (x + 2)
=
2
(x + 4) (x + 2)
=
25
x + 6x + 8
=
25
x2 + 6x − 17
=
2
x
x
So
x
0
√
−6 ± 36 + 68
=
√2
−6 ± 104
=
2
√
−6 + 104
=
2√
= −3 + 26
2
38.
log9 (13x)
log9 (4x)
=2
Solution: First of all, x > 0.
log9 (13x)
= 2
log9 (4x)
log9 (13x) = 2 log9 (4x)
log9 (13x) = log9 16x2
So x = 0 or x =
16
13 .
13x
=
16x2
x (16x − 13)
=
0
But x > 0. So x =
16
13 .
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