Multivariable Calculus 9th Edition by James Stewart solution manual


 4
√  2
− 4

= ln 4 − ln( 2 − 4) 12 [Law 2]
= 4ln −
1
2
ln[( + 2)( − 2)] [Law 3]
= 4ln −
1
2 [ln( + 2) + ln( − 2)]
[Law 1]
= 4ln −
1
2
ln( + 2) −
1
2
ln( − 2)
44. (a) ln

3
 − 3
= ln

3
 − 3
 12
=
1
2
ln

3
 − 3

[Law 3]
=
1
2 [ln3 + ln − ln( − 3)]
[Laws 1 and 2]
=
1
2
ln3 +
1
2
ln −
1
2
ln( − 3)
(b) log 2

( 3 + 1)
3
 ( − 3) 2 
= log 2 ( 3 + 1) + log 2
3
 ( − 3) 2
[Law 1]
= log 2 ( 3 + 1) + log 2 ( − 3) 23
= log 2 ( 3 + 1) +
2
3 log 2 ( − 3)
[Law 3]
45. (a) log 10 20 −
1
3
log 10 1000= log 10 20 − log 10 1000 13 = log 10 20 − log 10
3
√ 1000
= log 10 20 − log 10 10 = log 10  20
10

= log 10 2
(b) ln − 2ln + 3ln = ln − ln 2 + ln 3 = ln

 2
+ ln 3 = ln
 3
 2
46. (a) 3ln( − 2) − ln( 2 − 5 + 6) + 2ln( − 3) = ln( − 2) 3 − ln[( − 2)( − 3)] + ln( − 3) 2
= ln
 ( − 2) 3 ( − 3) 2
( − 2)( − 3)

= ln[( − 2) 2 ( − 3)]
(b) log   − log   + log   = log    − log    + log   = log 
   
 

47. (a) log 5 10 =
ln10
ln5
≈ 1430677 (b) log 15 12 =
ln12
ln15
≈ 0917600
c ° 2021 Cengage Learning. All Rights Reserved. May not be scanned, copied, or duplicated, or posted to a publicly accessible website, in whole or in part.
54
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CHAPTER 1 FUNCTIONSAND MODELS
48. (a) log 3 12 =
ln12
ln3
≈ 2261860 (b) log 12 6 =
ln6
ln12
≈ 0721057
49. To graph these functions, we use log 15  =
ln
ln15
and log 50  =
ln
ln50 .
These graphs all approach −∞ as  → 0 + , and they all pass through the
point (10). Also, they are all increasing, and all approach ∞ as  → ∞.
The functions with larger bases increase extremely slowly, and the ones with
smaller bases do so somewhat more quickly. The functions with large bases
approach the ­axis more closely as  → 0 + .
50. We see that the graph of ln is the reflection of the graph of   about the
line  = , and that the graph of log 8  is the reflection of the graph of 8 
about the same line. The graph of 8  increases more quickly than that of   .
Also note that log 8  → ∞ as  → ∞ more slowly than ln.
51. 3 ft = 36 in, so we need  such that log 2  = 36 ⇔  = 2 36 = 68,719,476,736. In miles, this is
68,719,476,736 in ·
1 ft
12 in
·
1 mi
5280 ft
≈ 1,084,5877 mi.
52.
From the graphs, we see that () =  01  () = ln for approximately 0    306, and then ()  () for
306    343 × 10 15 (approximately). At that point, the graph of  finally surpasses the graph of  for good.
53. (a) Shift the graph of  = log 10  five units to the left to
obtain the graph of  = log 10 (+5). Note the vertical
asymptote of  = −5.
 = log 10   = log 10 ( + 5)
(b) Reflect the graph of  = ln about the ­axis to obtain
the graph of  = −ln.
 = ln  = −ln
c ° 2021 Cengage Learning. All Rights Reserved. May not be scanned, copied, or duplicated, or posted to a publicly accessible website, in whole or in part.
SECTION 1.5 INVERSEFUNCTIONSAND LOGARITHMS
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55
54. (a) Reflect the graph of  = ln about the ­axis to obtain
the graph of  = ln(−).
 = ln  = ln(−)
(b) Reflect the portion of the graph of  = ln to the right
of the ­axis about the ­axis. The graph of  = ln||
is that reflection in addition to the original portion.
 = ln  = ln||
55. (a) The domain of () = ln + 2 is   0 and the range is .
(b)  = 0 ⇒ ln + 2 = 0 ⇒ ln = −2 ⇒  =  −2