### Document

```Section 2.5
141
Logarithmic Functions
50. (a) y = mx + b
2.5
b = 0.275
Use the point (21,291) to find m.
291 = m(21) + 0.275
290.725 = 21m
Logarithmic Functions
1. 53 = 125
Since ay = x means y = loga x, the equation in
logarithmic form is
log5 125 = 3.
290.725
m
21
« 13.844
y = 13.844a; + 0.275
2. 72 = 49
Since ay = x means y = logn .t, the equation in
logarithmic form is
y = ax2 + 6
log7 49 = 2.
b = 0.275
Use the point (21,291) to find a.
291 = a(21)2 + 0.275
3. 34 = 81
The equation in logarithmic form is
290.725 = 441a
log3 81 = 4.
290.725
441
« 0.6592
y = 0.6592.T2 + 0.275
4. 27 = 128
Since aV = x means y = logrt x, the equation in
logarithmic form is
y = abx
log2 128 = 7.
a = 0.275
Use the point (21,291) to find 6.
291 = 0.275621
5 3-2-5.
3 -g
The equation in logarithmic form is
*• = 0.275
291
6 =
log3 i =-2.
21/29T
.275
V0.275
» 1.3932
y = 0.275(1.3932)*
-2
Since ay = x means y = log„ x, the equation in
logarithmic form is
(b)
300
.log4/5 Jq
25 = -2.
9
(>•= 13.844* +0.275]
7. log2 32 = 5
Since y = logft re means ay = rr, the equation in
exponential form is
25 = 32.
y - 0.275(1.3932)* is the best fit.
(c) y = 0.275(1.3932)25 « 1095.6 million
8. log3 81 = 4
Since y = logn x means ay = re, the equation in
exponential form is
(d) The regression equation isy = 0.247(1.3942)x.
This is close to the function found in part (b).
34 = 81.
Chapter 2
142
9. In -
NONLINEAR FUNCTIONS
18. log3 — = re
= -1
81
e
The equation in exponential form is
1
3* =
81
e-=I.
e
3T = 3-4
re = -4
10. log2 | =-3
19.1og2 tf{=x
The equation in exponential form is
2X =
>-3
-
-4
11.
1/3
2X =
log 100,000 = 5
log10 100,000 = 5
(A)
2T = 2"2/3
105= 100,000
2
When no base is written, logj0 is understood.
12.
log 0.001 = -3
logl0 0.001 = -3
a:="3
20. iog8 yl=x
1/4
10"3 = 0.001
When no base is written, log10 is understood.
13. Let log8 64 = a:.
(23)x = 2"1/4
3re= - 4
Then, 8X = 64
8X = 82
1_
x
re = 2.
Thus, log8 64 = 2.
14. Let log9 81 = re.
=
— •
12
21. In e = re
Recall that In means log(,.
Then, 9* = 81
9r = 92
ex = e
x = l
a; = 2.
Thus, log9 81 = 2.
22. In e3 = re
Recall that In means logc .
15. log4 64 = x
4* = 64
ex = e3
4X = 43
re = 3
a: = 3
16. log3 27 = x
23. In e5/3 = x
ex _ e5/3
3* =27
5
3*=33
T=3
a: = 3
24. In 1 = x
17' l°g2 16 = X
ex =
1
ex = eo
1
1 "16
2T = 2"4
x =
—4
x = 0
25. The logarithm to the base 3 of 4 is written log3 4.
The subscript denotes the base.
Section 2.5
143
Logarithmic Functions
27. log5 (3fc) = log5 3 + logg *
36. log6 (962) = log6 9 + log6 62
= log6 32 + log,, b2
28. log9 (4m) = log9 4 + log9 m
= 2 log,, 3 + 2 log6 b
= 2c+2(1)
29. log3 ^3p
= 2c + 2
= log3 3p - log3 5/e
= (log3 3 + log3 p) - (log3 5 + log3 k)
= 1 + log3 p - log3 5 - log3 fc
In 30
37. log5 30 =
In 5
3.4012
15p
30. log7 -^
1.6094
2.113
= log7 15p - log7 ly
= (log7 15+ log7 p) - (log7 7 4- log7 y)
= log7 15 4- log7 p - log7 7 - log7 2/
= log7 15 + log7 p - 1 - log7 y
In 210
38. logi2 210 =
In 12
2.152
31. In
3\/5
^6
= ln3\/5- In ^6
In 0.95
39. log!.2 0.95 =
In 1.2
= ln3-51/2-ln61/3
-0.281
= ln3 + ln51/2-ln61/3
= ln3 + iln5-iln6
32. In
In 0.12
40. log28 0.12 =
In 2.8
9^5
-2.059
= ln9^5-ln^3
= ln9-51/3-ln31/4
= ln9 + ln51/3-ln31/4
41.
logx 36 = -2
x~2 = 36
(x-2)"l/2 = 36-l/2
1
= ln9 + iln5-i
ln3
3
4
X= 6
33. log6 32 = log,, 25
= 5 log,, 2
42. logg 27 = m
9m = 27
= 5a
(32)™ = 33
34. log,, 18
32m = 33
= log,, (2-9)
2?n = 3
= log6(2-32)
= log,, 2 + log,, 32
3
m
=
-
= logb 2 + 21og6 3
= o + 2c
43.
35. log,, 726 = log6 72 + log,, b
= log,, 72 + 1
= log623-33 + l
= log6 23+logb 32 + l
= 3 log,, 2 + 2 log6 3 + 1
= 3a + 2c + 1
log8 16 = z
82 = 16
(23)2 = 24
232 = 24
3z = 4
4
144
Chapter 2
44.
50.
log,, 8 =
4
y3'4 = 8
(y3/4)4/3 = 84/3
y = (81/3)4
NONLINEAR FUNCTIONS
log (re + 5) + log (re + 2) = 1
log[(.r + 5)(rc + 2)] = l
(re + 5)(re + 2) = 101
a;2 + 7x + 10 = 10
x2 + 7x = 0
= 24 = 16
.r(a; + 7) = 0
.-e = 0
45. log,. 5 = -
or
x = —7
x = —7 is not a solution of the original equa
tion because if x = -7, x + 5 and .-e + 2 would be
r1/2 = 5
(ri/2)2 = 52
negative, and the domain of y = log a: is (0, oo).
Therefore, x = 0.
r = 25
51. log3 (re - 2) + log3 (:c + 6) = 2
log3[(.r-2)(.T + 6)] = 2
(rr-2)(re + 6)=32
46. log4 (5a: + 1) = 2
42 = 5rc + 1
16 = 5rc + 1
re2 + 4a: - 12 = 9
a:2 + 4.t - 21 = 0
5rc = 15
re = 3
(x + 7){x - 3) = 0
a: = —7
47. log5 (9re - 4) = 1
51 = 9re - 4
or
x = 3
x = —7 does not check in the original equation.
The only solution is 3.
9 = 9re
l=re
52. log3(.e2 + 17) - log3(.T + 5) = 1
48. log4 rc-log4 (re + 3) = -l
log4
log3
-1
—
a:2 + 17
=
x + 5
rc + 3
4-1
a:2 + 17
31
X
re+ 5
=
re + 3
1
1
3.r + 15 = x2 + 17
x
Ax = .t + 3
0
= a;2 - 3.T + 2
0
= (a:-l)(a:- 2)
re= 1 or x = 2
3x = 3
x= 1
53.
log2(n:2 - 1) - log2(re+l) =
.
49. logg m - logg (m - 4) = -2
in — 4
re2 - 1
l0g2.x-+l = 2
m
!ogs
2
= -2
22 =
x2 - 1
m —4
1
77?.
81
m - 4
4 =
-4 = 80m
54.
2X = 6
In T
This value is not possible since log9 (-0.05) does
not exist.
Thus, there is no solution to the original equation.
(x -l)(re + 1)
.r + 1
m - 4 = 81m
-0.05 = 771
+ 1
X
m
9"2 =
= In 6
re In 2 = In 6
_ In 6
X~ln~2
2.585
4 =
x
re =
5
-
-1
Section 2.5
55.
145
Logarithmic Functions
5(0.10)* = 4(0.12)*
60.
5* = 12
ln^O.lO)*] = 11114(0.12)*]
re log 5 = log 12
In 5 + reIn 0.10 = In 4 + reIn 0.12
log 12
rc(ln0.12 - In 0.10) = In 5 - In 4
~ log 5
In5-ln4
« 1.544
x
=
In0.12-ln0.10
1.224
efe-1 = 6
In ek~l = In 6
56.
61.
In 1.5 + reIn 1.05 = In 2 + x In 1.01
(k - 1) In e = In 6
rc(ln1.05 - In 1.01) = In 2 - In 1.5
In 6
k-l =
1.5(1.05)T = 2(1.01)x
Infl^l-OS)*] = ln[2(1.01)x]
In 2-In 1.5
In e
re =
In 1.05-In 1.01
In 6
fc-l =
1
« 7.407
k = 1 + In 6
62. /(re) = log (5 - re)
» 2.792
5 - re > 0
-x > -5
x < 5
57.
e2* = 15
The domain of / is re < 5.
In e2*/ = In 15
63. /(re) = In (re2 - 9)
Since the domain of f(x) = In x is (0,oo), the
domain of f(x) = In(a:2 - 9) is the set of all real
2y In e = In 15
2y{\) = In 15
In 15
y =
numbers re for which
2
re2 - 9 > 0.
1.354
To solve this quadratic inequality, first solve the
58.
2e5«+i2 = 10
e5n+12 = 5
re2 - 9 = 0
(re + 3)(re-3) = 0
In e5a+12 = In 5
£ + 3=0
(5a + 12) In e = In 5
x — —3
5a + 12 = In 5
or
re-3=0
or
x = 3
These two solutions determine three intervals on
In 5
a
12
the number line: (-co,-3), (-3,3), and (3,oo).
=
5
If re =-4, (-4 + 2)(-4-2)>0.
Ifre = 0, (0 + 2)(0-2)>0.
If a: = 4, (4 + 2)(4-2)>0.
-2.078
The domain is re < -3 or re > 3, which is written
59.
lOe32"7 = 100
In 10e32-7 = In 100
in interval notation as (-co, -3) U (3,co).
64. logA-logS = 0
In 10 + In e32"7 = In 100
In 10 + (3* - 7) In e = In 100
32 - 7 = In 100 - In 10
32 = In 100 - In 10 + 7
In 100 - In 10 + 7
Z=
« 3.101
3
log^=0
| =10O =l
A = B
A-B = 0
146
Chapter 2
Thus, solving log A - log B = 0 is equivalent to
NONLINEAR FUNCTIONS
For 7- = 0.08, we use the rule of 72.
solving A - B = 0.
72
x
65. Let 77i = logrt -, n = logfl re, and p = log„ y.
Then am = -, an = re, and a? = «.
fiS
Substituting gives
x
9 years
100(0.08)
(a)
In 2
/, =
In (1 + r)
an
In 2
a™ = - = — = a""P.
y
qp
/. =
So m = n —p.
Therefore,
In (1 + 0.07)
i ss 10.24
It will take 11 years for the compound amount to
log0 '- = loga x - loga y.
be at least double.
(h\ I=
ln 3
1 } ' ln (1 + 0.07)
66. Let m = log,, re'' and n = logax.
Then, am = xr and an = x.
Substituting gives
I »
16.24
It will take 17 years for the compound amount to
am = xr = (an)r = anr.
at least triple.
Therefore, rn = nr, or
(c) The rule of 72 gives
lognrer = rlog„re.
72
100(0.07)
67. From Example 7, the doubling time t in years
when m = 1 is given by
= 10.29
years as the doubling time.
In 2
/. =
ln(l+r)
69. A = Pert
1200 = 500er14
(a) Let r = 0.03.
2.4 = e14r
In 2
ln(2.4) = lne14r
t =
In 1.03
ln(2.4) = 14r
= 23.4 years
ln(2.4)
(b) Let r = 0.06.
-14- = r
In 2
0.0625 w r
t =
In 1.06
The interest rate should be 6.25%.
= 11.9 years
(c) Let r = 0.08.
70.
r (sec)
In 2
t =
In 1.08
(d) Since 0.001 < 0.03 < 0.05, for r = 0.03, we
ln(l+r)
use the rule of 70.
70
100(0.03)
0.05
0.08
0.12
693.5
35
14.2
9.01
6.12
700
35
14
8.75
5.83
720
36
14.4
9
6
70
lOOr
70
0.02
ln 2
= 9.0 years
lOOr
0.001
= 23.3 years
72
lOOr
Since 0.05 < 0.06 < 0.12, for r = 0.06, we use the
For 0.001 < r < 0.05, the Rule of 70 is more
rule of 72.
accurate. For 0.05 < r < 0.12, the Rule of 72 is
72
72
lOOr
100(0.06)
= 12 years
more accurate. At r = 0.05, the two are equally
accurate.
Section 2.5
Logarithmic Functions
71. After x years at Humongous Enterprises, your salary
would be 45,000 (1 +0.04)* or 45,000 (1.04)*.
After re years at Crabapple Inc., your salary would
be 30,000 (1 + 0.06)* or 30,000 (1.06)*.
First we find when the salaries would be equal.
45,000(1.04)* = 30,000(1.06)*
(1.04)*
(1.06)*
30,000
45,000
147
74. // = -[Pi ln Pj + P2 In P2
+ P3 ln P3 + P4 ln P4]
H = -[0.521 ln 0.521+0.324 ln 0.324
+ 0.081 ln 0.081+ 0.074 ln 0.074]
H = 1.101
75. (a) 3 species, \ each:
A=F2=P3 = 5
2
U06 )
3
H = -(P1lnP1+P2lnP2+P3lnP3)
l0g O =bg (I)
xlog (ifi) =log (0
-(S-S)
--'"5
log
re =
« 1.099
(i)
log {m)
(b) 4 species, \ each:
Pi = P2 = Ps = Pa = t
x « 21.29
2009 + 21.29 = 2030.29
Therefore, on July 1, 2031, the job at Crabapple,
H
Inc., will pay more.
-(5-5)
72. If the number N is proportional to m-0,6, where m
is the mass, then AT = km~06, for some constant
of proportionality k.
Taking the common log of both sides, we have
logAT = log(A:m-0-6)
= logk + logm-0-6
= (Pi 1n Pi + P21n P2 + P3 In P3 + P4 ln P4
» 1.386
(c)
Notice that
- ln - = h^-1)-1 = ln3 « 1.099
= log k —0.6 log m.
«3
This is a linear equation in logm. Its graph is a
straight line with slope —0.6 and vertical intercept
log A:.
and
-In- =ln(4"1)-1 =ln4« 1.386
73. (a) The total number of individuals in the com
munity is 50 + 50, or 100.
50
Let Px = —
by Property c of logarithms, so the populations
are at a maximum index of diversity.
= 0.5, P2 = 0.5.
H = -l[P1 In Px+P2 In P2)
= -l[0.5 ln 0.5 + 0.5 ln 0.5]
« 0.693
(b) For 2 species, the maximum diversity is ln 2.
(c) Yes, ln2«0.693.
76. mX + N =
=
=
=
777 log,, re + log,, n
log,, xm + log,, n
log,, nxm
log6 y
= Y
Thus, Y = 777X + N.
Chapter 2
148
NONLINEAR FUNCTIONS
Substituting this value into 4.3 = a(0.3)6,
77. C(t) = C0e~kt
When t = 0, C(t) = 2, and when t = 3, C(t) = 1.
4.3 = a(0.3)°-7028
2 = C0e-fc(°)
4.3
a
C0 =2
1 = 2e"3fc
1
=
10.02.
(0.3)0.7028
Therefore, y = 10.02re07028.
-3fc
2=C
(d) If x = 0.5,
-3fc = ln - = ln 2"1 = - ln 2
y = 10.02(0.5)07028
ln2
fc =
T =
r
=
«6.16.
1 .
c2
:fclnc:
1 1 !5Ci
: In 2 ln "
3
3 In 5
T =
In 2
We predict that the oxygen consumption for a
guinea pig weighing 0.5 kg will be about 6.16 ml/min.
79. (a) h{t) = 37.79(1.021)'
Double the 2005 population is 2(42.69) = 85.38
million
T& 7.0
85.38 = 37.79(1.021)'
The drug should be given about every 7 hours.
85.38
78. (a) From the given graph, when re = 0.3 kg
y « 4.3 ml/ min, and when x = 0.7 kg
y w 7.8 ml/ min.
37.79
= (1.021)'
/ 85.38 \
!°Si 021
1,37.79/ =
(b) If y = axb, then
t =
ln y = ln (aa:fc)
t
111 V37.79/
In 1.021
39.22
= ln a + 6 ln a:.
Thus, there is a linear relationship between In y
and ln x.
(c)
4.3 = a(0.3)6
7.8 = a(0.7)6
4.3 _ a(0.3)b
7.8 ~ a(0.7)b
The Hispanic population is estimated to double
their 2005 population in 2039.
(b) h(t) = 11.14(1.023)'
Double the 2005 population is 2(12.69) = 25.38
million
25.38 = 11.14(1.023)'
7.8 " \0.7J
25.38
11.14
log 1.023
= (1.023)'
/ 25.38 \
/, =
6 m(«)
ln 1.023
36.21
i»(lt»)
6 « 0.7028
The Asian population is estimated to double their
2005 population in 2036.
Section 2.5
149
Logarithmic Functions
(f) /0 = 0.0002 microbars
80. N(r) = -5000 ln r
l,200,000,000/o = 1,200,000,000(0.0002)
(a) 7V(0.9) = -5000 ln (0.9) « 530
(b) AT(0.5) = -5000 ln (0.5) « 3500
= 240,000 microbars
895,000,000,000/0 = 895,000,000,000(0.0002)
(c) N(0.3) = -5000 ln (0.3) « 6000
(d) N(0.7) = -5000 ln (0.7) « 1800
(e) -5000 ln r = 1000
= 179,000,000 microbars
83. Let I\ be the intensity of the sound whose decibel
rating is 85.
1000
ln r =
(a)
-5000
10 log ^-=85
Jo
log £-=8.5
ln r = —5
Jo
r = e-l/5
log I\ - log /o = 8.5
log/i = 8.5 +log/o
r « 0.8
81.
C=£log2(- +l)
Let /2 be the intensity of the sound whose decimal
rating is 75.
10 log ^=75
§-*(= +0
Jo
2c/b = £ + i
logT=7b
n
Jo
£ = 2C'D - 1
log /2 - log /0 = 7.5
log/0 =log/2 -7.5
n
82. Decibel rating: 10 log —
Substitute for /0 in the equation for log /j.
•'o
(a) Intensity, / = 1157o
log h = 8.5 + log /o
= 8.5 + log I2- 7.5
= l + log/2
= 10 • log 115
log h - log h = 1
«21
1logJ»
(b) / = 9,500,000/q
10 log
(9-5X/o106/°) =10 log 9.5 x10°
«70
(c) I = l,200,000,000/o
10 log
= 1
l
Then j1 = 10, so /2 = ^/j. This means theinten
sity of the sound that had a rating of 75 decibels
is jq as intense as the sound that had a rating of
85 decibels.
1.2 x 109/0
(12x/o10"f'
) =10 log 1.2 x10»
91
84. R(I) = log Jo
(a) P(1,000,000/o)
,
(d) / = 895,000,000,000/0
= log
10 1og(8-95XJo10ll/o)=10 1og8.95xlon
1,000,000 /0
-
Jo
= log 1,000,000
= 6
120
(e) / = 109,000,000,000,000/o
10 log (109XJo10H/°) =10 log 1.09x10"
140
(b) R{100,000,000 /o)
,
= log
100,000,000 /o
-
Jo
= log 100,000,000
= 8
150
Chapter 2
NONLINEAR FUNCTIONS
(g) Find the energy of a magnitude 6.7 earth
(c) P(/) = logJo
quake. Using the formula from part f,
6.7 = log —
9
Jo
F*
6J=3l0gTo
106-7 =^
E
log — = 10.05
/o
Eq
/ « 5,000,000/0
E
_ JQ10.05
Eo
(d) /?(/) = log f
E = /?01010-05
Jo
8.1 =log-f
For an earthquake that releases 15 times this much
Jo
energy, E = £0(15)101005.
108-1 =j/o
/ «126,000,000/0
, x 1985 quake
(e)
K' 1999 quake
=
=\ log(15 •101005)
126,000,000/0
-
«
5,000,000/0
o
o^
«7.5
The 1985 earthquake had an amplitude more than
25 times that of the 1999 earthquake.
(f) P(£) =|log|-
So, it's true that a magnitude 7.5 earthquake re
leases 15 times more energy than one of magnitude
6.7.
85. pH = - log [H+]
For the 1999 earthquake
C7
21
(a) For pure water:
7=-log[H+]
-7 = log [11+]
E
6J=3l0g^
10.05 = log
E_
10"7 = [11+]
Eq
For acid rain:
_ = 101005
Eo
4 =-log [11+]
-4 = log [H+]
E = 101005£b
10-4 = [H+]
10 -4
For the 1985 earthquake,
10-
The acid rain has a hydrogen ion concentration
81 =f"*i
1000 times greater than pure water.
12.15 =log -IEo
E
Eo
= 103 = 1000
(b) For laundry solution:
ll = -log[H+]
10"n =[H+]
= 101215
E = 101215£0
For black coffee:
5 = -log[H+]
[11+]
The ratio of their energies is
10 - 5
10,215£0
101005£0
= 10
2.1
126