Multivariable Calculus 9th Edition by James Stewart solution manual

(c) We shift the graph of  = ln two units upward.
56. (a) The domain of () = ln( − 1) − 1 is   1 and the range is .
(b)  = 0 ⇒ ln( − 1) − 1 = 0 ⇒ ln( − 1) = 1 ⇒
 − 1 =  1 ⇒  =  + 1
(c) We shift the graph of  = ln one unit to the right and one unit downward.
57. (a) ln(4 + 2) = 3 ⇒  ln(4+2) =  3 ⇒ 4 + 2 =  3 ⇒ 4 =  3 − 2 ⇒  =
1
4 (
3
− 2) ≈ 4521
(b)  2−3 = 12 ⇒ ln 2−3 = ln12 ⇒ 2 − 3 = ln12 ⇒ 2 = 3 + ln12 ⇒  =
1
2 (3 + ln12) ≈ 2742
58. (a) log 2 ( 2 −  − 1) = 2 ⇒  2 −  − 1 = 2 2 = 4 ⇒  2 −  − 5 = 0 ⇒
 =
1 ±
 (−1) 2
− 4(1)(−5)
2(1)
=
1 ±
√ 21
2
.
Solutions are  1 =
1 −
√ 21
2
≈ −1791 and  2 =
1 +
√ 21
2
≈ 2791.
(b) 1 +  4+1 = 20 ⇒  4+1 = 19 ⇒ ln 4+1 = ln19 ⇒ 4 + 1 = ln19 ⇒ 4 = −1 + ln19 ⇒
 =
1
4 (−1 + ln19) ≈ 0486
59. (a) ln + ln( − 1) = 0 ⇒ ln[( − 1)] = 0 ⇒  ln[
2 −]
=  0 ⇒  2 −  = 1 ⇒  2 −  − 1 = 0. The
quadratic formula gives  =
1 ±
 (−1) 2
− 4(1)(−1)
2(1)
=
1 ±
√ 5
2
, but we note that ln
1 −
√ 5
2
is undefined because
1 −
√ 5
2
 0. Thus,  =
1 +
√ 5
2
≈ 1618.
(b) 5 1−2 = 9 ⇒ ln5 1−2 = ln9 ⇒ (1 − 2)ln5 = ln9 ⇒ 1 − 2 =
ln9
ln5
⇒  =
1
2
−
ln9
2ln5
≈ −0183
60. (a) ln(ln) = 0 ⇒  ln(ln ) =  0 ⇒ ln = 1 ⇒  =  ≈ 2718
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.
56
¤
CHAPTER 1 FUNCTIONSAND MODELS
(b)
60
1 +  −
= 4 ⇒ 60 = 4(1 +  − ) ⇒ 15 = 1 +  − ⇒ 14 =  − ⇒ ln14 = ln − ⇒
ln14 = − ⇒  = −ln14 ≈ −2639
61. (a) ln  0 ⇒    0 ⇒   1. Since the domain of () = ln is   0, the solution of the original inequality
is 0    1.
(b)    5 ⇒ ln   ln5 ⇒   ln5
62. (a) 1   3−1  2 ⇒ ln1  3 − 1  ln2 ⇒ 0  3 − 1  ln2 ⇒ 1  3  1 + ln2 ⇒
1
3
  
1
3 (1 + ln2)
(b) 1 − 2ln  3 ⇒ −2ln  2 ⇒ ln  −1 ⇒    −1
63. (a) We must have   − 3  0 ⇔    3 ⇔   ln3. Thus, the domain of () = ln(  − 3) is (ln3∞).
(b)  = ln(  − 3) ⇒   =   − 3 ⇒   =   + 3 ⇒  = ln(  + 3), so  −1 () = ln(  + 3).
Now   + 3  0 ⇒    −3, which is true for any real , so the domain of  −1 is .
64. (a) By (9),  ln 300 = 300 and ln( 300 ) = 300.
(b) A calculator gives  ln 300 = 300 and an error message for ln( 300 ) because  300 is larger than most calculators can
evaluate.
65. We see that the graph of  = () =
√  3
+  2 +  + 1 is increasing, so  is 1­1.
Enter  =
  3
+  2 +  + 1 and use your CAS to solve the equation for . You
will likely get two (irrelevant) solutions involving imaginary expressions, as well
as one which can be simplified to
 =  −1 () = −
3
√ 4
6

3
√  − 27 2
+ 20 −
3
√  + 27 2
− 20 +
3
√ 2 
where  = 3
√ 3 √ 27 4
− 40 2 + 16 or, equivalently,
1
6
 23 − 8 − 2 13
2 13
,
where  = 108 2 + 12
√ 48 − 120 2
+ 81 4 − 80.
66. (a) Depending on the software used, solving  =  6 +  4 for  may give six solutions of the form  = ±
√ 3
3
√  − 1, where