young cat2e ssm ch5
DESCRIPTION
Young Solution Manual Chapter 5TRANSCRIPT
241
CHAPTER 5 Section 5.1 Solutions -------------------------------------------------------------------------------- 1. 16 3. 2
1 1255
=
5. ( )23
2238 8 2 4= = = 7. ( ) ( )3 32 2
331
9 9 9 3 27−= = = =
9. 05 1− = − 11. 5.2780 13. 9.7385 15. 7.3891 17. 0.0432 19. 3(3) 3 27f = =
21. ( ) 1116( 1) 16g −
− = = 23. ( ) ( )1 12 21 1
2 16 16 4g −− = = =
25. ( ) 3 19.81ef e = ≈ 27. f ( )1
5 5xy =
Rising a factor of 5 slower than 5x
29. e ( )5xy = −
The reflection of5x over the x-axis
31. b ( )5 1xy −= − +
The reflection of 5 x− over the x-axis and then shifted up 1 unit.
33. y-intercept: 0(0) 6 1f = = , so (0,1). HA: 0y = Domain: ( ),−∞ ∞ Range: ( )0,∞ Other points: (-1,1/6), (1,6)
35. y-intercept: 0(0) 10 1f = = , so (0,1). HA: 0y = Reflect graph of 10xy = over y-axis. Domain: ( ),−∞ ∞ Range: ( )0,∞ Other points: (1, 0.1), (-1,10)
Chapter 5
242
37. y-intercept: 0(0) 1f e= = , so (0,1). HA: 0y = Reflect graph of xy e= over y-axis. Domain: ( ),−∞ ∞ Range: ( )0,∞ Other points: (1,1/e), (-1, e)
39. y-intercept: 0(0) 2 1 0f = − = , so (0,0). HA: 1y = − Shift graph of 2xy = down 1 unit Domain: ( ),−∞ ∞ Range: ( )1,− ∞ Other points: (2,3), (1,1)
41. y-intercept: 0(0) 2 1f e= − = , so (0,1) . HA: 2y = Reflect graph of xe over the x-axis, then shift up 2 units. Domain: ( ),−∞ ∞ Range: ( ), 2−∞ Other points: (1, 2-e), (-1, 2 – 1/e)
43. y-intercept: 0 1(0) 4 4f e e+= − = − , so (0, 4)e − . HA: 4y = − Shift the graph of xe left 1 unit, then down 4 units. Domain: ( ),−∞ ∞ Range:
( )4,− ∞
Other points: (-1,-3), (1, 2 4e − )
Section 5.1
243
45. y-intercept: 0(0) 3 3f e= ⋅ = , so (0,3) . HA: 0y = Expand the graph of xy e= horizon- tally by a factor of 2, then expand vertically by a factor of 3. Domain: ( ),−∞ ∞ Range: ( )0,∞
Other points: (2,3e), (1, 3 e )
47. y-intercept: ( )0 2 212(0) 1 1 2 5f −
= + = + = , so (0,5) . HA: 1y =
Shift the graph of ( )12
xy = right 2 units, then up 1 unit. Domain: ( ),−∞ ∞ Range: ( )1,∞
Other points: (0,5), ( )2,2
49. Use ( )0( ) 2
tdP t P= .
Here, 0 7.1, 88, 48 (2090 2002)P d t= = = − .
Thus, ( )4888(48) 7.1 2 10.4P = ≅ .
So, the expected population in 2090 is approximately 10.4 million.
51. Use ( )0( ) 2tdP t P= .
Here, 0 1500, 5 (doubling time), 30P d t= = = .
Thus, ( )305(30) 1500 2 96,000P = ≅ .
53. Use ( )10 2( )
thA t A=
Here, 0119.77 days, 200 , 30 daysh A mg t= = =
Thus, ( )30
119.7712(30) 200 168A = ≅ .
So, 168 mg remain after 30 days.
55. Use ( )10 2( )
thA t A=
Here, 010 years, 8000, 14 yearsh A t= = =
Thus, ( )14
1012(14) 8000 3031A = ≅ .
So, the value after 14 years is $3031.
Chapter 5
244
57. Use ( )( ) 1 ntrnA t P= + . Here,
3200, 0.025, 4, 3 yearsP r n t= = = = .
Thus, ( )4(3)0.0254(4) 3200 1 3448.42A = + ≅ .
So, the amount in the account after 4 years is $3,448.42.
59. Use ( )( ) 1 ntrnA t P= + . Here,
(18) 32,000, 0.05, 365, 18A r n t= = = = Thus, solving the above formula for P, we
have( )365(18)0.05
365
32,000 13,011.031
P = ≅+
.
So, the initial investment should be $13,011.03.
61. Use ( ) r tA t Pe= . Here,
3200, 0.02, 15P r t= = = . Thus, (0.02)(15)(15) 3200 4319.55A e= ≅ . So, the amount in the account after 15 years is $4319.55.
63. Use ( ) r tA t Pe= . Here, (20) 38,000, 0.05, 20A r t= = = .
Thus, solving the above formula for P, we
have (0.05)(20)
38,000 13,979.42Pe
= ≅ .
So, the initial investment should be $13,979.42.
65. The mistake is that 12 24 4− ≠ . Rather,
12
12
1 1424
− = = . 67. 0.025r = rather than 2.5
69. False. (0, 1)− is the y-intercept. 71. True. ( )1 133 (3 ) xx x− −= =
73.
Note on Graphs: Solid curve is 3xy = and the dashed curve is 3logy x= .
75.
77. y-intercept: 0 1(0)f be a be a− += − = − So, ( )0,be a− .
Horizontal asymptote: For x very large, 1 0xbe− + ≈ , so y a= − is the horizontal asymptote.
Section 5.1
245
79. The domain and range for ( ) xf x b= , where 1b > , are: Domain: ( ),−∞ ∞ Range: [ )1,∞ . Indeed, note that ( )f x is defined piecewise, as follows:
, 0, 0
xx
x
b xb
b x−
⎧ ≥⎪= ⎨<⎪⎩
Recall that the graph of xb− is the reflection of the graph of xb over the y-axis.
81.
Note on Graphs: Solid curve is 1(1 )x
xy = + and dashed curve (the horizontal asymptote) is y e= .
83. The graphs are close on the interval ( )3,3− .
Note on Graphs: Solid curve is xy e= and thin curve is 2 3 4
2 6 241 x x xy x= + + + + .
85. The graphs of f, g, and h are as follows:
The horizontal asymptotes for f, g, and h are 2 4, ,y e y e y e= = = . As x increases,
2 4( ) , ( ) , ( )f x e g x e h x e→ → → .
Chapter 5
246
Section 5.2 Solutions -------------------------------------------------------------------------------- 1. 35 125= 3.
1481 3=
5. 5 1322− = 7. 210 0.01− =
9. 410 10,000= 11. ( ) 3 314 4 64−
= =
13. 1 1ee− = 15. 0 1e =
17. 5xe = 19. zx y= 21. 8log (512) 3= 23. log(0.00001) 5= − 25. 1
225 2log (15) = 27. 25
8125log ( ) 3=
29. 127
13log (3) = − 31. ln 6 x=
33. log y x z= 35. 2log (1) 0=
37. 5log (3125) 5 3125 5xx x= ⇒ = ⇒ =
39. 710log (10 ) 7=
41. ( )14
14log (4096) 4096 4 4096 6x xx x−= ⇒ = ⇒ = ⇒ = −
43. undefined 45. undefined 47. 1.46 49. 5.94 51. undefined 53. 8.11− 55. Must have 5 0x + > , so that the domain is ( )5,− ∞ .
57. Must have 5 2 0x− > , so that the domain is ( )52,−∞ .
59. Must have 7 2 0x− > , so that the domain is ( )72,−∞ .
61. Must have 0x > , so that the domain is ( ) ( ),0 0,−∞ ∪ ∞ .
63. Must have 2 1 0x + > (which always occurs), so that the domain is . 65. b 67. c Reflect the graph of 5logy x= over the x-axis. 69. d Since
5 5log (1 ) 2 log ( ( 1)) 2x x− − = − − − , Reflect the graph of 5logy x= over the y-axis, then shift right 1 unit, and then shift down 5 units.
Section 5.2
247
71. Shift the graph of logy x= right 1 unit. Domain: ( )1,∞ Range: ( ),−∞ ∞
73. Shift the graph of lny x= up 2 units. Domain: ( )0,∞ Range: ( ),−∞ ∞
75. Shift the graph of 3logy x= left 2 units, then down 1 unit. Domain: ( )2,− ∞ Range: ( ),−∞ ∞
77. Reflect the graph of logy x= over the x-axis, then shift up 1 unit. Domain: ( )0,∞ Range: ( ),−∞ ∞
79. Shift the graph of lny x= left 4 units. Domain: ( )4,− ∞ Range: ( ),−∞ ∞
81. Compress the graph of logy x= horizontally by a factor of 2. Domain: ( )0,∞ Range: ( ),−∞ ∞
Chapter 5
248
83. Use 1210log1 10
ID −⎛ ⎞= ⎜ ⎟×⎝ ⎠
.
Here, 6
612
1 1010log 10log(10 )1 10
60log(10) 60
D
dB
−
−
⎛ ⎞×= =⎜ ⎟×⎝ ⎠= =
85. Use 1210log1 10
ID −
⎛ ⎞= ⎜ ⎟×⎝ ⎠.
Here, 0.3
11.712
1
1 1010log 10log(10 )1 10
117 log(10) 117
D
dB
−
−
=
⎛ ⎞×= =⎜ ⎟×⎝ ⎠= =
87. Use ( )4.423 10log EM = .
Here,
( ) ( )
[ ]
17
4.412.62 2
3 310
12.62103
23
1.41 10log log 1.41 10
log(1.41) log (10 )
log(1.41) 12.6 8.5
M ×= = ×
⎡ ⎤= +⎣ ⎦= + ≅
89. Use ( )4.423 10log EM = .
Here,
( ) ( )
[ ]
14
4.49.62 2
3 310
9.62103
23
2 10log log 2 10
log(2) log (10 )
log(2) 9.6 6.6
M ×= = ×
⎡ ⎤= +⎣ ⎦= + ≅
91. Use 10logpH H +⎡ ⎤= − ⎣ ⎦ . Here,
[ ]
410
4
log (5.01 10 )
log(5.01) log(10 )
log(5.01) 4 3.3
pH −
−
= − ×
⎡ ⎤= − +⎣ ⎦=− − ≅
93. Use 10logpH H +⎡ ⎤= − ⎣ ⎦ . Normal Rainwater:
5.610log (10 ) 5.6pH −= − =
Acid rain/tomato juice: 4
10log (10 ) 4pH −= − =
95. Use 10logpH H +⎡ ⎤= − ⎣ ⎦ . Here,
3.610
10
log (10 )( 3.6) log 10 3.6
pH −= −= − − =
97. Use ln
5000.0001216
C
t
⎛ ⎞⎜ ⎟⎝ ⎠= − .
Here, 100 1ln ln500 5 13,236
0.0001216 0.0001216t
⎛ ⎞ ⎛ ⎞⎜ ⎟ ⎜ ⎟⎝ ⎠ ⎝ ⎠= − = − ≅
99. ( )3
3
3
3 1010log 10 log 3 log(10 ) 30 10log(3) 251
dB dB−
−
=−
⎡ ⎤⎛ ⎞× ⎢ ⎥= = + = − + ≅ −⎜ ⎟⎢ ⎥⎝ ⎠ ⎣ ⎦
.
So, the result is approximately a 25 dB loss. 101. 2log 4 x= is equivalent to 2 4x = (not 42x = ). 103. The domain is the set of all real numbers such that 5 0x + > , which is written as ( )5,− ∞ . 105. False. The domain is all positive real numbers. 107. True.
Section 5.2
249
109. Consider ( ) ln( )f x x a b= − − + , where a, b are real numbers. Domain: Must have 0x a− > , so that the domain is ( ),a ∞ . Range: The graph of ( )f x is the graph of ln x shifted a units to the right, then reflected over the x-axis, and then shifted up b units. Through all of this movement, the range of y-values remains the same as that of ln x , namely ( ),−∞ ∞ . x-intercept: Solve ln( ) 0x a b− − + = :
ln( ) 0ln( )
b
b
x a bx a bx a e
x a e
− − + =− =
− =
= +
So, the x-intercept is ( ),0ba e+ .
111.
113. The graphs are symmetric about the line y x= .
Note on Graphs: Solid curve is xy e= and dashed curve is lny x= .
Chapter 5
250
115. The common characteristics are: o x-intercept for both is (1,0). o y-axis is the vertical asymptote for
both. o Range is ( , )−∞ ∞ for both. o Domain is (0, )∞ for both.
Note on Graphs: Solid curve is logy x= and dashed curve is lny x=
117. The graphs of f, g, and h are below. We note that f and g have the same graph with domain ( )0,∞ .
Section 5.3 Solutions -------------------------------------------------------------------------------- 1.
9log 1
9 10
x
x
x
=
==
3. ( )
( )12
12
1 12 2
log
1
x
x
x
=
=
=
5. 8
108
log 10
10 108
x
x
x
=
==
Section 5.3
251
7. 10
3
log 0.001
10 0.001 103
x
x
x
−
=
= == −
9.
32
2
32
log 8
2 8 2x
x
x
=
= ==
11. 8log 58 5=
13. ln( 5) 5xe x+ = +
15. 3
5 53log 2 log 2 35 5 2 8= = = 17.
27 72log 3 log 3 2 1
97 7 3−− −= = =
19. ( ) ( ) ( )
( ) ( )
3 5 3 5log log log
3log 5logb b b
b b
x y x y
x y
= +
= +
21.
( ) ( ) ( )( ) ( )
1 11 13 32 2
1 12 3
log log log
log log
b b b
b b
x y x y
x y
= +
= +
23.
( ) ( )( ) ( )
13 1 1
3 212
1 13 2
log log log
log log
b b b
b b
r r ss
r s
⎛ ⎞= −⎜ ⎟⎜ ⎟
⎝ ⎠= −
25.
( ) ( )
( ) ( ) ( )( ) ( ) ( )
log log log
log log log
log log log
b b b
b b b
b b b
x x yzyz
x y z
x y z
⎛ ⎞= −⎜ ⎟
⎝ ⎠= − +⎡ ⎤⎣ ⎦= − −
27.
( ) ( ) ( ) ( ) ( ) ( )122 2 2 1
2log 5 log log 5 log log 5 2log log 5x x x x x x x x+ = + + = + + = + +
29.
( ) ( )( ) ( ) ( )
( )
12
3 23 2 2
2
3 2 2
212
( 2)ln ln ( 2) ln 55
ln ln ( 2) ln 5
3ln( ) 2 ln( 2) ln 5
x x x x xx
x x x
x x x
⎛ ⎞−= − − +⎜ ⎟
+⎝ ⎠
= + − − +
= + − − +
31.
( ) ( )[ ]
2 2
2
2
2 1 ( 1)log log9 ( 3)( 3)
log ( 1) log ( 3)( 3)
2 log( 1) log( 3) log( 3)2 log( 1) log( 3) log( 3)
x x xx x x
x x x
x x xx x x
⎛ ⎞ ⎛ ⎞− + −=⎜ ⎟ ⎜ ⎟− − +⎝ ⎠ ⎝ ⎠
= − − − +
= − − − + +
= − − − − +
Chapter 5
252
33. ( )( )
3 5
3 5
3log 5log log log
log
b b b b
b
x y x y
x y
+ = +
=
35. 5 2
5
2
5log 2log log log
log
b b b b
b
u v u v
uv
− = −
⎛ ⎞= ⎜ ⎟
⎝ ⎠
37. ( )2 21 13 32 21 2
2 3log log log log logb b b b bx y x y x y+ = + =
39.
( )
2 3 2
2 3 2
2
3 2
2 log 3log 2log log log log
log log
log
u v z u v z
u v z
uv z
⎡ ⎤− − = − +⎣ ⎦
= −
⎛ ⎞= ⎜ ⎟
⎝ ⎠
41.
[ ]
2 2 2
2 2
2
2 2
ln( 1) ln( 1) 2ln( 3) ln( 1) ln( 1) ln( 3)ln ( 1)( 1) ln( 3)
1ln( 3)
x x x x x xx x x
xx
+ + − − + = + + − − +
= + − − +
⎛ ⎞−= ⎜ ⎟+⎝ ⎠
43.
( )
1132
1132
12
13
1 12 3ln( 3) ln( 2) ln ln( 3) ln( 2) ln
ln( 3) ln ( 2)
( 3)ln( 2)
x x x x x x
x x x
xx x
⎡ ⎤+ − + − = + − + +⎣ ⎦⎡ ⎤= + − +⎣ ⎦
⎛ ⎞+= ⎜ ⎟⎜ ⎟+⎝ ⎠
45. 5log 7log 7 1.2091log5
= ≅ 47. 12 1
2
log5log 5 2.3219log
= ≅ −
49. 2.7log5.2log 5.2 1.6599log 2.7
= ≅ 51. log10log 10 2.0115logπ π
= ≅
53. 3
log8log 8 3.7856log 3
= ≅
Section 5.3
253
55. Use 1210 log1 10
ID −
⎛ ⎞= ⎜ ⎟×⎝ ⎠
.
In this case, ( )2 2 21 6 1
From music From conversation
(1 10 ) (1 10 ) 1.00001 10W W Wm m mI − − −= × + × = × .
So, 1
12
1.00001 1010log 1101 10
D dB−
−
⎛ ⎞×= ≅⎜ ⎟×⎝ ⎠
that you are exposed to.
57. Use 23 4.4log
10EM ⎛ ⎞= ⎜ ⎟
⎝ ⎠.
Here, the combined energy is 12 8(4.5 10 ) (7.8 10 )× + × joules. The corresponding
magnitude on the Richter scale is 12 8
23 4.4
(4.5 10 ) (7.8 10 )log 5.510
M⎛ ⎞× + ×
= ≅⎜ ⎟⎝ ⎠
.
59. Cannot apply the quotient property directly. Observe that 23log5 log5 3log5 2log5 log5− = − = .
61. Cannot apply the product and quotient properties to logarithms with different bases. So, you cannot reduce the given expression further without using the change of base formula.
63. True. ln 1logln10 ln10
ee = =
65. False. 3ln( ) 3ln( ) 3(ln ln )xy xy x y= = + , which does not equal 3(ln ln )x y+ , in general. 67. Claim: ( )log log logM
b b bN M N= −
Proof: Let log , logb bu M v N= = . Then, ,u vb M b N= = .
Observe that ( ) ( ) ( )log log log log logu
vu vbM
b b b b bN bb u v M N−= = = − = − . ▪
69.
( ) ( )( ) ( ) ( )
6 32 2 6
3 5 3 5 9 15
6 9 15
6 9 15
log log log
log log
log log log
6log 9log 15log
b b b
b b
b b b
b b b
x x xy z y z y z
x y z
x y z
x y z
− − −
−
−
⎛ ⎞ ⎛ ⎞ ⎛ ⎞= =⎜ ⎟ ⎜ ⎟ ⎜ ⎟⎜ ⎟ ⎝ ⎠ ⎝ ⎠⎝ ⎠
= −
⎡ ⎤= − +⎣ ⎦= − +
Chapter 5
254
71. Yes, they are the same graph.
73. No, they are not the same graph.
Note on Graphs: Solid curve is log
log 2xy = and
dashed curve is log log 2y x= − . 75. No, they are not the same graph, even though the property is true.
Note on Graphs: The thin curve plus the dashed curve is 2ln( )y x= (domain all real numbers except 0) and the dashed curve only is 2 lny x= (domain is (0, )∞ ).
77. The graphs of lny x= and loglog
xye
=
do coincide, and are as follows:
Section 5.4 Solutions --------------------------------------------------------------------------------
1.
4
3 813 3
4
x
x
x
=
==
3.
2
1749
7 72
x
x
x
−
=
== −
5. 2
2 4
2
2 16
2 24
2
x
x
xx
=
=
== ±
Section 5.4
255
7. ( )( ) ( ) ( )
1 2723 8
1 3 332 23 2 3
1 34
x
x
xx
+
+ −
=
= =
+ = −= −
9. 2 3 0
32
12 3 0
xe ex
x
+ = =+ =
= −
11. 2 5 3 47 7
2 5 3 4
1
x x
x x
x
− −=− = −
= −
13. 2 12 7
2
2
2 212 7
7 12 0( 4)( 3) 0
3,4
x x
x xx xx x
x
+ =
+ =
− + =− − =
=
15.
( )
2
2
4
2 2 4
2
2
9 3
3 3 3
2 46 0
( 6) 0
0,6
x x x
xx x x
x x xx x
x x
x
−
−
=
= =
= −
− =− =
=
17. 25 1 3
2
2
5 1 35 4 0
( 4)( 1) 0
1,4
x xe ex x
x xx x
x
− +=
− = +
− + =− − =
=
19.
( )2 3
2 3
3 log(81)2
10 81
log 10 log(81)
2 3 log(81)2 3 log(81)
2.454
x
x
xx
x
−
−
+
=
=
− == +
= ≅
21. 1
3
3
3 5log (5) 1
log (5) 1 0.465
x
x
x
+ == +
= − ≈
23.
[ ]
3 1
2
123
27 2log (27) 3 1
log (27) 1 1.918
x
x
x
−== −
= + ≈
25. 3 8 7
3 155
ln 5 1.609
x
x
x
eee
x
− =
=
=
= ≈
27. 0.1
0.1
0.1
9 2 12 8
40.1 ln 4
13.863
x
x
x
eee
x
x
− =
=
==
≈
Chapter 5
256
29. ( )
( )
3
2 3 11 9
2 3 20
3 10
log (10) 2.096
x
x
x
x
− =
=
=
= ≈
31.
( )3 4
3 4
4 ln(22)3
22
ln ln(22)
3 4 ln(22)3 4 ln(22)
0.303
x
x
e
e
xx
x
+
+
− +
=
=
+ == − +
= ≅ −
33.
( )
2
2
2
ln 62
3 186
ln ln(6)
2 ln(6)0.896
x
x
x
ee
e
xx
=
=
=
== ≅
35. Note that 2 7 3 0x xe e+ − = is equivalent to ( ) ( )27 3 0x xe e+ − = . Let xy e= and
solve 2 7 3 0y y+ − = using the quadratic formula: 27 7 4(1)( 3) 7 612 2
y− ± − − − ±
= =
So, substituting back in for y, the following two equations must be solved for x: 7 61
2xe − += and 7 61
2xe − −=
Since 7 61 02
− −< , the second equation has no real solution. Solving the first one
yields 7 61ln 0.9042
x⎛ ⎞− +
= ≅ −⎜ ⎟⎜ ⎟⎝ ⎠
.
37.
( )23 3 0
3 3 03 3
0
x x
x x
x x
x xx
−
−
−
− =
− =
== −=
39. 2 1
52 5
7ln(7) 1.946
x
x
x
ee
ex
=−
= −
== ≈
41.
2
2
2
2
20 46
20 24 44 4
12 0
x
x
x
x
ee
ee
x
=−
= −
=
==
Section 5.4
257
43.
( )2
2
2
10
4 210 7
4 2 10 14
10 92 log (9)
0.477
x
x
x
xx
=−
= −
==≈
45.
2
log(2 ) 210 2100 250
xxx
x
=
===
47. 3
4
log (2 1) 4
2 1 32 80
40
x
xxx
+ =
+ ===
49. 2
3
189
32
log (4 1) 3
2 4 14 1
x
xxx
−
− = −
= −= +
=
51. 2
2
2
ln ln 9 0ln ln 9
93
xxxx
− =
=
== ±
53.
( )5 5
5
2
log ( 4) log 1log ( 4) 1
( 4) 54 5 0
( 5)( 1) 0
5, 1
x xx xx x
x xx x
x
− + =
− =
− =
− − =− + =
= −
55.
2
2
log( 3) log( 2) log(4 )log( 3)( 2) log(4 )
( 3)( 2) 46 4
5 6 0( 6)( 1) 0
6, 1
x x xx x xx x x
x x xx x
x x
x
− + + =− + =− + =
− − =
− − =− + =
= −
57. ( )4 4 4
44
4
log (4 ) log 3
4log 3
log (16) 3
x
x
x
x
− =
⎛ ⎞=⎜ ⎟
⎝ ⎠=
Since the last line is a false statement, this equation has no solution.
61. 2
2 5
5
ln 5
12.182
xx e
x e
=
=
= ± ≈ ±
59.
258
log(2 5) log( 3) 12 5log 1
32 5 10
32 5 10 30
8 25
x xx
xx
xx x
xx
− − − =
−⎛ ⎞ =⎜ ⎟−⎝ ⎠−
=−− = −
==
63.
2
log(2 5) 22 5 10
2 9547.5
xx
xx
+ =
+ ===
Chapter 5
258
65. ( )2
2 4
2 4
4
ln 1 4
11
1 7.321
x
x ex e
x e
+ =
+ =
= −
= ± − ≈ ±
67.
2
212
ln(2 3) 22 3
3 1.432
xx e
x e
−
−
+ = −
+ =
⎡ ⎤= − + ≈ −⎣ ⎦
69. ( )( )( )( ) 1.5
2
2
log(2 3 ) log 3 2 1.5
log (2 3 ) 3 2 1.5
(2 3 ) 3 2 10 31.622
6 13 6 31.6226 13 25.622 0
x x
x x
x x
x xx x
− + − =
− − =
− − = ≅
− + ≅
− − ≅
Now, use the quadratic formula: 13 783.93 , so that 3.42
12x x±= ≅ , 1.25−
71.
4
2 4
4
ln ln( 2) 4ln ( 2) 4
( 2)2 0
2 4 4(1)( )2
6.456
x xx xx x e
x x e
ex
+ − =− =
− =
− − =
± − −=
≈ − , 8.456
73. 7 7 7
7 7
2
log (1 ) log ( 2) log ( )1log log ( )
21
21 ( 2)
3 1 0
x x xx x
xx x
xx x x
x x
− − + =
−⎛ ⎞ =⎜ ⎟+⎝ ⎠−
=+− = +
+ − =
There are no rational solutions since neither 1 nor 1− work. So, graph to find the real roots:
So, the solution is approximately 0.3028.
Section 5.4
259
75.
1 1 12 2 2
2
2
ln 4 ln 2 ln 1ln( 4) ln( 2) ln( 1)
4ln ln( 1)24 124 2
2 6 0
2 4 4(1)( 6)1 7
21.646
x x xx x x
x xxx xxx x x
x x
x
+ − − = ++ − − = +
+⎛ ⎞ = +⎜ ⎟−⎝ ⎠+
= +−+ = − −
− − =
± − −= = ±
≈ − , 3.646
77. Use ( )( ) 1 ntrnA t P= + .
Here, 0.035, 1r n= = .
In order to triple, if P is the initial investment, then we seek the time t such that ( ) 3A t P= . So, we solve the following equation:
( )1( )0.0351
1.035
3 1
3 (1.035)log 3
t
t
P P
t
= +
==
So, it takes approximately 31.9 years to triple.
79. Use ( )( ) 1 ntrnA t P= + .
Here, 20,000, 0.05, 4, 7500A r n P= = = = .
So, we solve the following equation: ( )4( )0.05
4
4
1.0125
11.01254
20,000 7500 1
2.667 (1.0125)log 2.667 4log 2.667
t
t
tt
= +
==
=
So, it takes approximately 19.74 years.
81. Use 23 4.4log
10EM ⎛ ⎞= ⎜ ⎟
⎝ ⎠.
Here, 7.4M = . So, substituting in, we can solve for E:
15.5
23 4.4
4.4
11.14.4
11.1 4.4 15
10
7.4 log10
11.1 log10
1010
10 10 3.16 10
E
E
E
E=
⎛ ⎞= ⎜ ⎟⎝ ⎠
⎛ ⎞= ⎜ ⎟⎝ ⎠
=
× ≈ × =
So, it would generate 153.16 10× joules of energy.
83. Use 1210log10
ID −⎛ ⎞= ⎜ ⎟⎝ ⎠
. Here, 120D = . So, substituting in, we can solve for I:
12
12
1212
12 12
120 10log10
12 log10
1010
1 10 10
I
I
I
I
−
−
−
−
⎛ ⎞= ⎜ ⎟⎝ ⎠
⎛ ⎞= ⎜ ⎟⎝ ⎠
=
= × =
So, the intensity is 21Wm .
Chapter 5
260
85. Use 0.50
tA A e−= . Here, 00.10A A= , where 0A is the initial amount of anesthesia. So, substituting into the above equation, we can solve for t:
0.50 0
0.5
10.5
0.10
0.10ln(0.10) 0.5ln(0.10)
t
t
A A e
et
t
−
−
=
== −
− =
So, it takes about 4.61 hours until 10% of the anesthesia remains in the bloodstream.
87. Use 0.2
2001 24 tN
e−=+
. Here, 100N = . So, substituting into the above equation, we
can solve for t:
( )
( )( )
0.2
0.2
0.2
0.2
0.2 124
124
1 10.2 24
2001001 24
100 1 24 200
1 24 224 1
0.2 ln
ln 15.89
t
t
t
t
t
ee
eee
t
t
−
−
−
−
−
=+
+ =
+ =
=
=
− =
= − ≈
So, it takes about 15.9 years.
89. Use 23 4.4log
10EM ⎛ ⎞= ⎜ ⎟
⎝ ⎠.
For P waves: Here, 6.2M = . So, substituting in, we can solve for E:
13.7
23 4.4
4.4
9.34.4
9.3 4.4
10
6.2 log10
9.3 log10
1010
10 10
E
E
E
E=
⎛ ⎞= ⎜ ⎟⎝ ⎠
⎛ ⎞= ⎜ ⎟⎝ ⎠
=
× =
For S waves: Here, 3.3M = . So, substituting in, we can solve for E:
9.35
23 4.4
4.4
4.954.4
4.95 4.4
10
3.3 log10
4.95 log10
1010
10 10
E
E
E
E=
⎛ ⎞= ⎜ ⎟⎝ ⎠
⎛ ⎞= ⎜ ⎟⎝ ⎠
=
× =
So, the combined energy is ( )13.7 9.3510 10+ joules. Hence, the reading on the Richter
scale is: 9.35 13.7
23 4.4
10 10log 6.210
M⎛ ⎞+
= ≈⎜ ⎟⎝ ⎠
Section 5.4
261
93. 5x = − is not a solution since log( 5)− is not defined.
91. ln(4 ) 4xe x≠ . Should first divide both sides by 4, then take the natural log:
94
9494
4 9
ln( ) ln( )ln( )
x
x
x
ee
ex
=
=
=
=
95. True.
97. False. Since lnln10log xx = , ( )
1ln 1ln10ln10 ln10log lnxx xe e e x x= = = ≠ .
99.
2
1 13 2
3 21 13 2
( 1)
23
2
2 2
log ( ) log ( 2 1) 2
log ( ) log ( 1) 2log log ( 1) 2
log ( ( 1)) 2
( 1)0
b bx
b b
b b
b
x x x
x xx x
x x
b x xx x b
= −
⋅
+ − + =
+ − =+ − =
− =
= −
− − =
Now, use the quadratic formula to find the solutions:
2
2 2
1 1 4( )2
1 1 4 1 1 4,2 2
bx
b b
± − −=
+ + − +=
This is negative, for any valueof . So, it cannot be a solution.b
101.
( )
( )( )
0.2
0.2
0.2
0.2
30000.22
30002
30002
30001 2
1 2 3000
2 30002 3000
0.2 ln
5ln
t
t
t
t
yty
yy
yy
ye
y e
y yeye y
e
t
t
−
−
−
−
−−
−
−
=+
+ =
+ =
= −
=
− =
= −
Chapter 5
262
103. Consider the function 2
x xe ey−+
= , for 0, 1x y≥ ≥ . (Need this restriction in order
for the function to be one-to-one, and hence have an inverse.) Solve 2
x xe ey−+
= for x.
( )
2
2
( ) 11
( ) 1
2
2
2
2 22
2 ( ) 1
( ) 2 1 0
x
x x
x
x
x x
exe e
ee
x x
x x
e ey
ey
y
ye e
e y e
−
+
+
+=
+= =
=
= +
− + =
Now, solve using the quadratic formula:
2 2
2
2 ( 2 ) 4 2 4 42 2
2 2 1 22 1
y y y yx
y y
e
y y
± − − ± −
± −
= =
= = ± −
Since 1( ) (0, ) ( )dom f rng f −= ∞ = , we
use only 2 1y y+ − here. Now to solve for x, take natural log of both sides:
( )2
2
1
ln 1
xe y y
x y y
= + −
= + −
Thus, inverse function is given by
( )1 2( ) ln 1f x x x− = + − .
105. Observe that 2
2
2
3 9 4 3 52 2
ln(3 ) ln( 1)3 1
3 1 0
x xx x
x x
x ± − ±
= +
= +
− + =
= =
These solutions agree with the graphical solution seen to the right. Note on Graphs: Solid curve is ln(3 )y x= and the thin curve is 2ln( 1)y x= + .
Section 5.5
263
107.
Note on Graphs: Solid curve is 3xy = and the thin curve is 5 2y x= + .
109. The graph of ( )2
x xe ef x−+
= is given
below:
The domain is ( ),−∞ ∞ , and the graph is symmetric about the y-axis. There are no asymptotes.
Section 5.5 Solutions--------------------------------------------------------------------------------- 1. c (iv) 3. a (iii) 5. f (i) 7. Use 0
r tN N e= . Here, 0 80, 0.0236.N r= = Determine N when 7t = (determined by 2010 – 2003): 0.0236(7)80 94N e= ≈ million. 9. Use 0
r tN N e= . Here, 0 103,800, 0.12.N r= = Determine t such that 200,000N = :
( )0.12 0.12200,000 200,0001103,800 0.12 103,800200,000 103,800 ln 5.5t te e t= ⇒ = ⇒ = ≈
So, the population would hit 200,000 sometime in the year 2008. 11. Use 0
r tN N e= . Here, 0 487.4, 0.165.N r= = Determine N when 3t = (corresponds to the number of cell phone subscribers in 2010): 0.165(3)487.4 799.6N e= ≈ There are approximately 799.6 subscribers in 2010. 13. Use 0
r tN N e= . Here, 0 185,000, 0.30.N r= = Determine N when 3t = : 0.30(3)185,000 455,027N e= ≈
The amount is approximately $455,000. 15. Use 0
r tN N e= (t measured in months). Here, 0 100 (million), 0.20.N r= =
Determine N when 6t = : ( )0.20 6100 332 millionN e= ≈ . 17. Use 0
r tN N e= (t measured in months). Here, 0 1 (million), 0.025.N r= =
Determine N when 7t = : ( )0.025 71 1.19 millionN e= ≈ . 19. Use 0.5100 tA e−= . Observe that 0.5 (4)(4) 100 13.53 mlA e−= ≈
Chapter 5
264
21. Use 0r tN N e= (1). We know that
102(5,730)N N= (2). If 0 5N = (grams),
find t such that ( ) 2N t = . To do so, we must first find r. To this end, substitute (2) into (1) to obtain:
( )(5,730)1 1 12 5,730 2lnre r= ⇒ =
Now, solve for t: ( )
( ) ( )( )( )
1 15,730 2ln
2 1 15 5,730 2
25
1 15,730 2
2 5ln ln
ln7575 years
ln
tet
t
=
=
= ≈
23. Use 0r tN N e= (1). We know that
9 102(4.5 10 )N N× = (2). Find t such that
0( ) 0.98N t N= . To do so, we must first find r. To this end, substitute (2) into (1) to obtain:
( )9
9(4.5 10 )1 1 1
2 24.5 10lnre r×
×= ⇒ =
Now, solve for t:
00.98 N 0N=( )
( )( )
1 19 24.5 10
9
ln
1 124.5 10
ln 0.98131,158,556 years old
ln
te
t
×
×
= ≈
25. Use ( )0k t
s sT T T T e−= + − . Here, 0 325, 72, and (10) 200sT T T= = = . Find (30)T . To do so, we must first find k. Observe
( )
( )
(10)
10
128110 253
200 72 325 72
128 253ln
k
k
e
ek
−
−
= + −
=
= −
Now, ( )( )128110 253ln 30(30) 72 253 105 FT e− − ⋅
= + ≈ . 27. Use ( )0
k ts sT T T T e−= + − (1).
Assume 0t = corresponds to 7am. We know that (0) 85, (1.5) 82, 74sT T T= = = . (2)
Find t such that ( ) 98.6T t = . We first use (2) in (1) to find k and 0T . ( )( )
( )
00 0 0
1.5 1.5
811.5 11
85 74 74 85
82 74 85 74 8 11
ln
k k
T e T T
e e
k
− −
= + − = ⇒ =
= + − ⇒ =
⇒ = −
Now, solve for t: ( )( )
( )
( )( )
811.5 11
811.5 11
ln
ln
24.611
811.5 11
98.6 74 (85 74)
24.6 11ln
3.8ln
t
t
e
e
t
− −= + −
=
− ≈ =
So, the victim died approximately 3.8 hours before 7am. So, by 8:30am, the victim has been dead for about 5.29 hours.
Section 5.5
265
29. Use 0r tN N e−= (1).
We have 0(0) 38,000
(1) 32,000N NN
= ==
(2)
In order to determine (4)N , we need to first determine r. To this end, substitute (2) into (1) to obtain:
( )32,00038,00032,000 38,000 lnre r−= ⇒ = −
Thus, ( )( )32,00038,000ln 4
(4) 38,000 19,100N e− − ⋅
= ≈ . The book value after 4 years is approximately $19,100.
31. Use 2
100,0001 10 tN
e−=+
.
a. 2(2)
100,000(2) 84,5201 10
Ne−= ≈
+
b. 2(30)
100,000(30) 100,0001 10
Ne−= ≈
+
c. The highest number of new convertibles that will be sold is 100,000 since the smallest that 21 10 te−+ can be is 1.
33. Use 0r tN N e−= (1).
Assuming that 0t = corresponds to 1997, we have
0(0) 2,422(6) 7,684
N NN
= ==
(2)
In order to determine (13)N , we need to first determine r. To this end, substitute (2) into (1) to obtain:
( )(6) 7.68416 2,4227,684 2,422 lnre r−= ⇒ = −
Thus, ( )( )7.6841
6 2,422ln 13(13) 2,422 29,551N e
− − ⋅= ≈ cases.
35. Find t such that 1.56
10,000 50001 19 te− =+
.
( )
( )
1.56
1.56
1.56
1.56 119
1 11.56 19
10,000 5000 1 19
2 1 191 19
ln 1.89 years
t
t
t
t
e
ee
e
t
−
−
−
−
= +
= +
=
=
= − ≈
37. Consider 2
( ) rI r e−= , whose graph is below. Note that the beam is brightest when 0r = .
Chapter 5
266
39. Consider 2
2( 75)25( ) 10
x
N x e−−
= . a. The graph of N is as follows:
b. Average grade is 75.
c. 2
2(50 75)25 1(50) 10 10 4N e e−
−−= = ≈
d. 2
2(100 75)25 1(100) 10 10 4N e e−
−−= = ≈
41. Use ( )( )
ln 1ln 1
PrnR
rn
tn
−= −
+.
a. Here, 80,000, 0.09, 12, 750P r n R= = = = . So, ( )
( )
80,000(0.09)12(750)
0.0912
ln 118 years
12ln 1t
−= − ≈
+.
b. Here, 80,000, 0.09, 12, 1000P r n R= = = = . So, ( )
( )
80,000(0.09)12(1000)
0.0912
ln 110 years
12ln 1t
−= − ≈
+.
43. r = 0.07, not 7 45. True 47. False (Since there is a finite number of students at the school to which the lice can spread.) 49. Take a look at a couple of graphs for increasing values of c. For definiteness, let a = k = 1, and take c = 1 and c = 5, respectively. The graphs are:
As c increases, the model reaches the carrying capacity in less time.
Chapter 5 Review
267
51. a. The graphs are below: For the same periodic payment, it will take Wing Shan fewer years to pay off the loan if she can afford to pay biweekly.
b. 11.58 years c. 10.33 years d. 8.54 years, 7.69 years, respectively. Chapter 5 Review Solutions ----------------------------------------------------------------------- 1. 17,559.94 3. 5.52 5. 24.53 7. 5.89
9. 4 ( 2.2) 6.22 2 73.52− − = ≅ 11. ( ) ( )121 6( ) 22 2
5 5 6.25− −= =
13. b y-intercept 14(0, ) 15. c y-intercept (0,11)
17. y-intercept: (0, 1)− HA: 0y =
Reflect the graph of ( )16
x over the x- axis.
19. y-intercept: (0, 2) . HA: 1y =
Shift the graph of ( )1100
x up 1 unit.
Chapter 5
268
21. y-intercept: (0,1) . HA: 0y =
23. y-intercept: (0,3.2) . HA: 0y =
25. Use ( )( ) 1 ntrnA t P= +
Here, 4500, 0.045, 2, 7P r n t= = = = .
Thus, ( )2(7)0.0452(7) 4500 1 6144.68A = + ≅ .
So, the amount in the account after 7 years is $6144.68.
27. Use ( ) r tA t Pe= . Here, 13,450, 0.036, 15P r t= = = . Thus,
(0.036)(15)(15) 13,450 23,080.29A e= ≅ . So, the amount in the account after 15 years is $23,080.29.
29. 34 64= 31. 2 110010− =
33. 6log 216 3= 35. ( )213
4169log 2=
41. 1.51 37.
7log 1
7 10
x
x
x
=
==
39.
( )16
416
log 1296
1296 64
x
x
x
=
= =
= −
43. 2.08−
45. Must have 2 0x + > , so that the domain is ( 2, )− ∞ . 47. Since 2 3 0x + > , for all values of x, the domain is ( , )−∞ ∞ .
49. b 51. d Shift the graph of 7log x left 1 unit, then down 3 units. Also, VA is
1x = − .
Chapter 5 Review
269
53. Shift the graph of 4log x right 4 units, then up 2 units.
55. Reflect the graph of 4log x over the x-axis, then shift down 6 units.
57. Use 10logpH H +⎡ ⎤= − ⎣ ⎦ . Here,
[ ]
710
7
log (3.16 10 )
log(3.16) log(10 )
log(3.16) 76.5
pH −
−
= − ×
⎡ ⎤= − +⎣ ⎦=− −
≅
59. Use 1210log1 10
ID −
⎛ ⎞= ⎜ ⎟×⎝ ⎠.
Here, 7
512
1 1010log 10log(10 )1 10
50log(10) 50
D
dB
−
−
⎛ ⎞×= =⎜ ⎟×⎝ ⎠= =
61. 1 63. 6
65. ( ) ( ) ( )
( ) ( )log log log
log log
a b a bc c c
c c
x y x y
a x b y
= +
= +
67.
( ) ( )( ) ( ) ( )
33log log log
log log 3log
j j j
j j j
rs rs tt
r s t
⎛ ⎞ = −⎜ ⎟⎝ ⎠
= + −
69.
( ) ( )( ) ( ) ( )( ) ( ) ( )
12 3 21
52 23 2
52
3 2152 2
31 22 2 5
log log log
log log log
log log log
a a b cb c
a b c
a b c
⎛ ⎞= −⎜ ⎟⎜ ⎟
⎝ ⎠⎡ ⎤= − +⎣ ⎦
= − −
71.
8log 3log 3log80.5283
=
≅
73. log1.4log 1.4log
0.2939
π π=
≅
75.
4
14256
4 44
x
x
x
−
=
== −
77. 3 4 0
43
13 4 0
xe ex
x
− = =− =
=
Chapter 5
270
79. ( )
( )
213
21 4
2 4
81
3 3
3 32 4
6
x
x
x
xx
+
+−
− −
=
=
=− − =
= −
81. 2 3
2 3
3 ln132
3 1013
2 3 ln130.218
x
x
eex
x
+
+
− +
− =
=+ =
= ≈ −
83. Note that 2 6 5 0x xe e+ + = is equivalent to ( ) ( ) ( )( )26 5 5 1 0x x x xe e e e+ + = + + = .
Neither No solution No solution
5 0 or 1 0x xe e+ = + = has a real solution. So, the original equation has no
solution. 85.
( )( )( ) ( )2 2
2 2
2 2
2 2 2 2 0
2 2 0
2 2 02 22 2
0
x x x x
x x
x x
x x
x xx
− −
−
−
−
− + =
− =
− =
== −=
87.
2
1003
log(3 ) 210 3100 3
xxx
x
=
===
89.
( )
( )( )
4 4
24
2 8
2 812
412
log ( ) log 2 8
log 2 8
2 4
4
4
128 2 , 128 2
x x
x
x
x
x
x
+ =
=
=
=
= ±
= −
91. 2
2 2.2
2.2
ln 2.2
3.004
xx e
x e
=
=
= ± ≈ ±
Chapter 5 Review
271
93.
( )3 3 3
23 33
23
2
4 16 82
log (2 ) log ( 3) log ( )log log ( )
2 ( 3)4 2 0
2 6, 2 6
xx
xx
x x xx
xx x x
x x
x
x
−+
−+
− ± +
− − + =
=
=
− = +
+ − =
=
= − + − −0.449≈
95. Use ( ) r tA t Pe= . Here,
30,000, 0.05, 1A r t= = = Substituting into the above equation, we can solve for P:
0.05
0.05(1)
30,000
30,00028,536.88
e
PeP
=
≅ =
So, the initial investment in 1 year CD should be approximately $28,536.88.
97. Use ( )( ) 1 ntrnA t P= +
Here, we know that 0.042, 4, 2r n A P= = = . We can substitute these in to find t.
( )( )
4( )0.0424
4
1.0105
11.01054
2 1
2 1.0105log 2 4
16.6 log 2
t
t
P P
tt
= +
=
=
≅ =
So, it takes approximately 16.6 years until the initial investment doubles. 99. Use 0
r tN N e= . Here, 0 2.62, 0.035.N r= = Determine N when 6t = : 0.035(6)2.62 3.23 millionN e= ≈
The population in 2010 is about 3.23 million. 101. Use 0
r tN N e= (1). We have
0(0) 1000(3) 2500
N NN
= ==
(2)
In order to determine (6)N , we need to first determine r. To this end, substitute (2) into (1) to obtain ( )(3) 25001
3 10002500 1000 lnre r= ⇒ = .
Thus, ( )250013 1000ln 6(6) 1000 6250 bacteriaN e ⋅= ≈ .
Chapter 5
272
103. Use 0r tN N e−= .
We know that 102(28)N N= . Assuming
that 0 20N = , determine t such that 5N = . To do so, we first find r:
( )(28)1 1 10 02 28 2lnrN N e r−= ⇒ = −
Now, solve: ( )( )
( )
( )( )
1 128 2
1 128 2
14
1 128 2
ln
ln14
ln
ln
5 20
56 years
t
t
e
e
t
− −=
=
= ≈
105. Use 0r tN N e−= .
Assuming that 2003 occurs at 0t = , we know that
0 5600, (1) 2420N N= = . Find (7)N . To do so, we first find r:
( )(1) 242056002420 5600 lnre r−= ⇒ = −
Now, solve: ( )( )2420
5600ln (7)(7) 5600 16 fishN e− −= ≈
107. Use ( )0.0351000 1 tM e−= − . Since 0t = corresponds to 1998, we have
( )0.035 (12)(12) 1000 1 343 miceM e−= − ≈ .
109. The graph is as follows:
Using the calculator to compute the functional values for large values of x suggests that the HA is about 4.11y = . (Note: The exact equation of the HA is
2y e= .)
111. Let 2.4 0.81 log (3 1), 2 log ( 1) 3.5y x y x= − = − + .
The graphs are as follows:
The coordinates of the point of intersection are about (2.376, 2.071).
113. The graphs agree on ( )0,∞ , as seen below.
Chapter 5 Practice Test
273
115. The graph is below. Domain: ( , )−∞ ∞ . Symmetric about the origin. Horizontal asymptotes:
1 (as )y x= − → −∞ , 1 (as )y x= →∞
117. a. Using 0r tN N e= with (0,4) and
(18,2), we need to find r: ( )18 1 1
18 22 4 ln 0.35508re r= ⇒ = ≈ − So, the equation of dosage is given by
0.0385084 4(0.9622)t tN e−= ≈ . b. 4(0.9622)tN = c. Yes, they are the same.
Chapter 5 Practice Test----------------------------------------------------------------------------- 1.
3 3 3log10 log10x x x= = 3.
( )1
3
4 13
log 81
3 34
x x
x
x
−
=
= =
− =
5. 2 1
2
2
421 ln 42
1 ln 42
1 ln 42 2.177
xex
x
x
− =
− =
= +
= ± + ≈ ±
7.
30027
0.2 1
0.2 1 30027
30027
30027
1 ln( )0.2
27 300
0.2 1 ln( )0.2 1 ln( )
7.04
x
x
ee
xx
x
+
+
− +
=
=
+ =
= − +
= ≅
9.
2
2
3ln( 4) 6ln( 4) 2
44 11.389
xxx e
x e
− =− =
− =
= + ≈
11. ln(ln ) 1
ln15.154e
xx ex e
==
= ≈
Chapter 5
274
13.
( )6 6
6
2
2
log log ( 5) 2log ( 5) 2
5 365 36 0
( 9)( 4) 0
4
x xx x
x xx xx x
x
+ − =
− =
− =
− − =− + =
= − ,9
15.
( )2
2
ln ln( 3) 1ln ( 3) 1
33 0
3 9 4( )2
0.729, 3.729
x xx x
x x ex x e
ex
+ + =
+ =
+ =
+ − =
− ± − −=
≈ −
17.
( )12
12 61 2
12 6 126 12
ln 0.693
x
x
x
ee
ex
=+
= +
=
= ≈ −
19. Must have 2 10x
x −> and 2 1 0x − ≠
CPs 1,0,1−
| | |1 0 1
− + − +
−
So, the domain is ( ) ( )1,0 1,− ∪ ∞ .
21. y-intercept: 0(0) 3 1 2f −= + = . So, (0,2). x-intercept: None Domain: ( , )−∞ ∞ Range: (1, )∞ Horizontal Asymptote: y = 1 The graph is as follows:
23. y-intercept: None x-intercept: Must solve ln(2 3) 1 0.x − + =
11 3ln(2 3) 1 2 3
2e
ex x x +− = − ⇒ − = ⇒ =
So, 13 ,02
e+⎛ ⎞⎜ ⎟⎝ ⎠
.
Domain: 32( , )∞ Range: ( , )−∞ ∞
Vertical Asymptote: 32x =
The graph is as follows:
25. Use ( )( ) 1 ntr
nA t P= + . Here, 5000, 0.06, 4, 8P r n t= = = = . Thus, (8)A =
( )4(8)0.0645000 1 8051.62+ ≅ . So, the amount in the account after 8 years is $8051.62.
27. Use 1210log1 10
ID −
⎛ ⎞= ⎜ ⎟×⎝ ⎠. Here, 3
912
1 1010log 10log(10 ) 90log(10) 901 10
D dB−
−
⎛ ⎞×= = = =⎜ ⎟×⎝ ⎠
.
Chapter 5 Practice Test
275
29. Use ( )4.423 10log EM = . Here,
For 5M = :
( )( )
11.9
4.4
4.4
4.4
23 10
10
7.510
7.5 4.4
10
5 log
7.5 log
10
10 10
E
E
E
E=
=
=
=
× =
So, the energy here is 11.9 1110 7.9 10≈ × joules.
For 6M = :
( )( )
13.4
4.4
4.4
4.4
23 10
10
910
9 4.4
10
6 log
9 log
10
10 10
E
E
E
E=
=
=
=
× =
So, the energy here is 13.4 1310 2.5 10≈ × joules.
Thus, the range of energy is 11 137.9 10 2.5 10E× < < × joules. 31. Use 0
r tN N e= (1). We have
0(0) 200(2) 500
N NN
= ==
(2)
In order to determine (8)N , we need to first determine r. To this end, substitute (2) into (1) to obtain:
( )(2) 50012 200500 200 lnre r= ⇒ =
Thus, ( )50012 200ln 8(8) 200 7800 bacteriaN e ⋅= ≈ .
33. Solve 0.4
200010001 3 te−=+
.
( )( )
0.4
0.4
0.4
0.4 13
13
1 10.4 3
1000(1 3 ) 20001 3 2
3 1
0.4 ln
ln 3 days
t
t
t
t
eeee
t
t
−
−
−
−
+ =
+ =
=
=
− =
= − ≈
35. The graph is below. Domain: ( , )−∞ ∞ . Symmetric about the origin. No asymptotes
Chapter 5
276
Chapter 5 Cumulative Review --------------------------------------------------------------------
1. 1 2
3 58 51 21
3 5 5 6281
52
2x y x x y y x yx y−− = =
3. 25 4 3 0
4 16 4(5)( 3)2(5)
2 195
x x
x
− − =
± − −=
±=
5.
( ]
4 35
4 1519
, 19
x
xx
+≤ −
+ ≤ −≤ −
−∞ −
7. The line 4 3 6x y+ = is equivalent to 43 2y x= − + and so, has slope 4
3− . A line perpendicular to it must have slope 3
4 . Since the desired line must pass through (7,6), we have
3 3 34 4 46 ( 7)y x y x− = − ⇒ = +
9. a. 1 b. 5 c. 1 d. undefined e. Domain: ( 2, )− ∞ Range: (0, )∞ f. Increasing on (4, )∞ Decreasing on (0,4) Constant on (-2,0) 11. Yes, f is one-to-one since
( ) ( ) 4 44 4
f x f y x yx yx y
= ⇒ − = −
⇒ − = −⇒ =
13.
( )( )
2
2
2
2
( ) 4 8 5
4 2 5
4 2 1 5 4
4( 1) 1
f x x x
x x
x x
x
= − + −
= − − −
= − − + − +
= − − −
So, the vertex is (1,-1). 15. Rewrite as: ( )4 3 24 3 7 20 ( ( 4))x x x x x− − + + − ÷ − − Synthetic division then gives
4 1 4 3 7 204 0 12 20
1 0 3 5 0
− − − −−
− −
So, 3( ) 3 5, ( ) 0Q x x x r x= − + − = . 17. Vertical asymptote: x = 3 Horizontal asymptote: None Slant asymptote: 3y x= + since
( )
2
2
33 0 7
0 9
16
xx x x
x x
+− + +
− + −
Chapter 5 Cumulative Review
277
19. ( )3 32 2 31
25 25 5 125−= = = 21. ( )5
3 3log 243 log 3 5= =
23.
( )
22log(4 9) log(4 9) 2
14
10 10 (4 9) 1214 9 11
4 9 11
9 11 5
x x xx
x
x
+ += = + =+ = ±
= − ±
= − ± = − , 0.5
25. Use r tA Pe= . We know that 8500P = and 0.04r = . Determine t such that 12,000A = .
( )0.04
12,00010.04 8,500
12,000 8500
ln 8.62 years
te
t
=
= ≈
27. a. Using 0r tN N e= with (0,6) and (28,3), we need to find r:
( )28 1 128 23 6 ln 0.247553re r= ⇒ = ≈ −
So, the equation of dosage is given by 0.2475536 6(0.9755486421)t tN e−= ≈ .
b. (32) 2.72 gramsN ≈
