5.1 SOLUTIONS 285 CHAPTER FIVE
[Pages:79]CHAPTER FIVE
5.1 SOLUTIONS 285
Solutions for Section 5.1
Skill Refresher
S1. Since 1,000,000 = 106, we have = 6.
S2. Since 0.01 = 10-2, we have = -2.
S3. Since 3 =
3 12 = 32, we have
= 3.
2
S4. Since 100 = 1, we have = 0. Similarly, it follows for any constant , if
= 1, then = 0.
S5. Since can never equal zero, the equation = 0 has no solution. It similarly follows for any constant , can never
equal zero.
S6. Since 1 = -5, we have 3 = -5. Solving 3 = -5, we have = -5 .
5
3
S7. Since 9 =
9
2 , we have
9
2=
7. Solving the equation 9
= 7, we have
= 14 .
2
9
S8. We have 10- = 10-1
=1 10
. For any constant
> 0,
> 0 for all
. Since
1 10
>
0, it follows 10-
=
1 10
=
-1000 has no solution.
S9. Since 4 0.1 = 0.114 =
10-1 14 = 10-14, we have 2 = - 1 . Solving for , we therefore have 4
= -1. 8
S10.
3 = 3 5 3 = 53 3 = 53
= 59.
Exercises
1. The statement is equivalent to 19 = 101.279.
2. The statement is equivalent to 4 = 100.602.
3. The statement is equivalent to 26 = 3.258.
4. The statement is equivalent to 0.646 = -0.437.
5. The statement is equivalent to = 10 .
6. The statement is equivalent to = .
7. The statement is equivalent to 8 = log 100,000,000.
8. The statement is equivalent to -4 = ln(0.0183).
9. The statement is equivalent to = log .
10. The statement is equivalent to = ln .
11. We are looking for a power of 10, and log 1000 is asking for the power of 10 which gives 1000. Since 103 = 1000, we know that log 1000 = 3.
12.
Using the rules of logarithms, we have log 1000 = log 100012 =
1 log 1000 =
2
1 2
3
=
1.5.
13. We are looking for a power of 10, and log 1 is asking for the power of 10 which gives 1. Since 1 = 100, log 1 = 0.
286 Chapter Five /SOLUTIONS
14.
We are looking for a power of 10, and log 0.1 is asking for the power of 10 which gives 0.1. Since 0.1 =
1 10
= 10-1, we
know that log 0.1 = log 10-1 = -1.
15. We can use the identity log 10 = . So log 100 = 0. We can check this by observing that 100 = 1, and we know that log 1 = 0.
16. Using the identity log 10 =
, we
have
log 10
=
log 1012
=
1.
2
17. Using the identity log 10 = , we have log 105 = 5.
18. Using the identity log 10 = , we have log 102 = 2.
19. Using the identity 10log = , we have 10log 100 = 100.
20. Using the identity 10log = , we have 10log 1 = 1.
21. Using the identity 10log = , we have 10log 0.01 = 0.01.
22. Since 1 = 0, ln 1 = 0.
23. Using the identity ln = , we get ln 0 = 0. Or we could notice that 0 = 1, so ln 0 = ln 1 = 0.
24. Using the identity ln = , we get ln 5 = 5.
25. Recall that = 12. Using the identity ln = 26. Using the identity ln = , we get ln 2 = 2.
, we get ln = ln 12 = 1 . 2
27. Using the identity log 10 = , we have log 1 = log 10-12 = - 1 .
10
2
28. Since 1 = -12, ln 1 = ln -12 = - 1 .
2
29. We are solving for an exponent, so we use logarithms. We can use either the common logarithm or the natural logarithm. Since 23 = 8 and 24 = 16, we know that must be between 3 and 4. Using the log rules, we have
2 = 11
log(2 ) = log(11)
log(2) = log(11)
=
log(11) log(2)
=
3.459.
If we had used the natural logarithm, we would have
=
ln(11) ln(2)
=
3.459.
30. We are solving for an exponent, so we use logarithms. We can use either the common logarithm or the natural logarithm. Using the log rules, we have
1.45 = 25 log(1.45 ) = log(25) log(1.45) = log(25)
log(25) = log(1.45) = 8.663.
If we had used the natural logarithm, we would have
=
ln(25) ln(1.45)
=
8.663.
5.1 SOLUTIONS 287
31. We are solving for an exponent, so we use logarithms. Since the base is the number , it makes the most sense to use the natural logarithm. Using the log rules, we have
0.12 = 100
ln( 0.12 ) = ln(100)
0.12 = ln(100)
=
ln(100) 0.12
=
38.376.
32. We begin by dividing both sides by 22 to isolate the exponent:
10 22
=
(0.87)
.
We then take the log of both sides and use the rules of logs to solve for :
log
10 22
=
log(0.87)
log
10 22
=
log(0.87)
log 10
=
22
log(0.87)
=
5.662.
33. We begin by dividing both sides by 17 to isolate the exponent:
48 17
= (2.3)
.
We then take the log of both sides and use the rules of logs to solve for :
log
48 17
=
log(2.3)
log
48 17
=
log(2.3)
log 48
=
17
log(2.3)
=
1.246.
34. We take the log of both sides and use the rules of logs to solve for :
log
2 7
=
log(0.6)2
log
2 7
=
2
log(0.6)
log 2 7 =2
log(0.6)
log
2 7
=
log(0.6)
2
= 1.226.
Problems
35. (a)
Since log 10 = , then
log 100 = log(102) = log 102 .
log 102 = 2 .
288 Chapter Five /SOLUTIONS
(b)
Since 10log = we know that (c)
1000log = (103)log = (10log )3
(10log )3 = ( )3 = 3.
log 0.001
= log
1 1000
= log(10-3)
= log 10-3
= -3 .
36. (a) Using the identity ln = , we get ln 2 = 2 .
(b) Using the identity ln = , we get ln(3 +2) = 3 + 2.
(c) Since 1 = -5 , we get ln 1 = ln -5 = -5 .
5
5
(d) Since
=(
)12 =
1 2
, we have ln
= ln
1 2
=1
.
2
37. (a) log(10 100) = log 1000 = 3
log 10 + log 100 = 1 + 2 = 3
(b) log(100 1000) = log 100,000 = 5
log 100 + log 1000 = 2 + 3 = 5
(c)
log 10 = log 1 = log 10-1 = -1
100
10
log 10 - log 100 = 1 - 2 = -1
(d)
log 100 = log 1 = log 10-1 = -1
1000
10
log 100 - log 1000 = 2 - 3 = -1
(e) log 102 = 2
2 log 10 = 2(1) = 2
(f) log 103 = 3
3 log 10 = 3(1) = 3 In each case, both answers are equal. This reflects the properties of logarithms.
38. (a) Patterns:
log( ) = log + log log = log - log
log = log (b) Using these formulas, we rewrite the expression as follows:
log
= log
= (log(
39. True. 40. False. log cannot be rewritten.
log
41. False. log log = log log
, not log
42. True. 43. True. = 12 and log 12 = 1 log .
2
44. False. log = (log )12.
+ log .
) - log ) = (log + log - log ).
45. Using properties of logs, we have
log(3 2 ) = 8 log 3 + log 2 = 8
log 2 = 8 - log 3 8 - log 3
= log 2 = 24.990.
5.1 SOLUTIONS 289
46. Using properties of logs, we have
ln(25(1.05) ) = 6
ln(25) + ln(1.05) = 6
ln(1.05) = 6 - ln(25)
=
6 - ln(25) ln(1.05)
=
57.002.
47. Using properties of logs, we have
ln( ) = ln + ln =
ln = =
- ln
- ln ln
.
48. Using properties of logs, we have
log( ) =
log + log =
log = - log
- log
= log
.
49. Using properties of logs, we have
ln(3 2) = 8 ln 3 + 2 ln = 8
2 ln = 8 - ln 3 ln = 8 - ln 3 2 = (8-ln 3)2 = 31.522.
Notice that to solve for , we had to convert from an equation involving logs to an equation involving exponents in the last step.
An alternate way to solve the original equation is to begin by converting from an equation involving logs to an equation involving exponents:
ln(3 2) = 8 3 2= 8
8
2= 3
= 83 = 31.522.
Of course, we get the same answer with both methods.
290 Chapter Five /SOLUTIONS
50. Using properties of logs, we have
log(5 3) = 2
log 5 + 3 log = 2
3 log log
= 2 - log 5
=
2
- log 5 3
= 10(2-log 5)3 = 2.714.
Notice that to solve for , we had to convert from an equation involving logs to an equation involving exponents in the last step.
An alternate way to solve the original equation is to begin by converting from an equation involving logs to an equation involving exponents:
log(5 3) = 2 5 3 = 102 = 100 3 = 100 = 20 5 = (20)13 = 2.714.
Of course, we get the same answer with both methods. 51. (a)
log 6 = log(2 3) = log 2 + log 3 = +.
(b)
log
0.08
=
log
8 100
= log 8 - log 100
= log 23 - 2
= 3 log 2 - 2
= 3 - 2.
(c)
log 3 = log
3
12
2
2
=
1 2
log
3 2
= 1 (log 3 - log 2) 2
= 1( 2
-
).
(d)
log
5
=
log
10 2
= log 10 - log 2
=1- .
52. (a) log 3 = log 15 = log 15 - log 5 5 (b) log 25 = log 52 = 2 log 5 (c) log 75 = log(15 5) = log 15 + log 5
5.1 SOLUTIONS 291
53. (a) The initial value of is 25. The growth rate is 7.5% per time unit. (b) We see on the graph that = 100 at approximately = 19. We can use graphing technology to estimate as accurately as we like. (c) We substitute = 100 and use logs to solve for :
= 25(1.075)
100 = 25(1.075)
4 = (1.075)
log(4) = log(1.075 )
log(1.075) = log(4)
=
log(4) log(1.075)
=
19.169.
54. (a) The initial value of is 10. The quantity is decaying at a continuous rate of 15% per time unit. (b) We see on the graph that = 2 at approximately = 10.5. We can use graphing technology to estimate as accurately as we like. (c) We substitute = 2 and use the natural logarithm to solve for :
= 10 -0.15
2 = 10 -0.15
0.2 = -0.15
ln(0.2) = -0.15
=
ln(0.2) -0.15
=
10.730.
55. To find a formula for , we find the points labeled ( 0, 1) and ( 1, 0) in Figure 5.1. We see that 0 = 4 and that 1 = 27. From the graph of , we see that
0 = (4) = 5.1403(1.1169)4 = 8.
To find 1 we use the fact that ( 1) = 27:
5.1403(1.1169) 1 = 27
1.1169
1
=
27 5.1403
1
=
log(275.1403) log 1.1169
= 15.
We have (4) = 27 and (15) = 8. Using the ratio method, we have
Now we can solve for :
15
=
(15)
4
(4)
11 = 8 27
=
8 27
111
0.8953.
(0.8953)4 = 27 = 27 (0.8953)4 42.0207.
so ( ) = 42.0207(0.8953) .
292 Chapter Five /SOLUTIONS
27
( 0, 1)
() ( 1, 1)
( 0, 0)
( 1, 0) ()
4
Figure 5.1
56. To find a formula for , we find the points labeled ( 0, 0) and ( 1, 1) in Figure 5.4. We see that 1 = 8 and that 1 = 50. From the graph of , we see that
0 = (8) = 117.7181(0.7517)8 = 12.
To find 0 we use the fact that ( 0) = 50:
117.7181(0.7517) 0 = 50
(0.7517)
0 0
= =
50 1lo1g7(.570181117.7181)
log 0.7517
= 3.
We have (3) = 12 and (8) = 50. Using the ratio method, we have
Now we can solve for :
8
=
(8)
3
(3)
5 = 50 12
=
50 12
15
1.3303.
(1.3303)3 = 12 = 12 (1.3303)3 5.0969.
so ( ) = 5.0969(1.3303) .
50
( 0, 1)
() ( 1, 1)
( 0, 0)
( 1, 0) ()
8
Figure 5.2
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