Intermediate Algebra
Name _________________________
Problem Set 8.5
Solutions to Every
Odd-Numbered Problem
Date _________________________
8.5 Common Logarithms and Natural Logarithms
1.
3.
5.
7.
9.
11.
13.
15.
17.
19.
21.
Evaluating the logarithm:
Evaluating the logarithm:
Evaluating the logarithm:
Evaluating the logarithm:
Evaluating the logarithm:
Evaluating the logarithm:
Evaluating the logarithm:
Evaluating the logarithm:
Evaluating the logarithm:
Evaluating the logarithm:
Solving for x:
log x = 2.8802
log 378 ! 2.5775
log 37.8 ! 1.5775
log 3, 780 ! 3.5775
log 0.0378 ! –1.4225
log 37, 800 ! 4.5775
log 600 ! 2.7782
log 2, 010 ! 3.3032
log 0.00971 ! –2.0128
log 0.0314 ! –1.5031
log 0.399 ! –0.3990
23.
25.
x = 10 2.8802 ! 759
Solving for x:
log x = 3.1553
29.
x = 10 3.1553 ! 1,430
Solving for x:
log x = !7.0372
33.
x = 10 !7.0372 " 0.0000000918
Solving for x:
log x = !10
37.
x = 10 !10
Solving for x:
log x = !2
x = 10 !2 =
41.
x = e!1 =
27.
x = 10 !2.1198 " 0.00759
Solving for x:
log x = !5.3497
31.
x = 10 !5.3497 " 0.00000447
Solving for x:
log x = 10
35.
x = 1010
Solving for x:
log x = 20
39.
1
100
Solving for x:
ln x = 1
43.
1
e
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Solving for x:
log x = !2.1198
x = 10 20
Solving for x:
log x = log 2 8
log x = 3
x = 10 3 = 1,000
Solving for x:
log x = 2 log 5
log x = log 5 2
x = 25
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45.
47.
49.
51.
53.
55.
57.
59.
61.
63.
Intermediate Algebra
Name _________________________
Problem Set 8.5
Solutions to Every
Odd-Numbered Problem
Date _________________________
Solving for x:
ln x = !3ln 2
ln x = ln 2 !3
1
x=
8
Simplifying the logarithm:
Simplifying the logarithm:
Simplifying the logarithm:
Simplifying the logarithm:
ln e = ln e1 = 1
ln e5 = 5
ln e x = x
log10, 000 = log10 4 = 4
1
Simplifying the logarithm: ln 3 = log e!3 = !3
e
3
2
Using properties of logarithms: ln10e3t = ln10 + ln e3t = ln10 + 3t
Using properties of logarithms: ln Ae!2t = ln A + ln e!2t = ln A ! 2t
3t
Using properties of logarithms: log !"100 (1.01) #$ = log10 2 + log1.013t = 2 + 3t log1.01
Simplifying the logarithm: log 1000 = log10 3/2 =
(
)
65.
Using properties of logarithms: ln Pert = ln P + ln ert = ln P + rt
67.
Using properties of logarithms: ! log 4.2 " 10 !3 = ! log 4.2 ! log10 !3 = 3 ! log 4.2
69.
Evaluating the logarithm: ln15 = ln ( 3 • 5 ) = ln 3 + ln 5 = 1.0986 + 1.6094 = 2.7080
1
Evaluating the logarithm: ln = ln 3!1 = ! ln 3 = –1.0986
3
Evaluating the logarithm: ln 9 = ln 32 = 2 ln 3 = 2 (1.0986 ) = 2.1972
71.
73.
75.
77.
(
)
Evaluating the logarithm: ln16 = ln 2 4 = 4 ln 2 = 4 ( 0.6931) = 2.7724
For the 1906 earthquake:
logT = 8.3
T = 10 8.3 = 1.995 ! 10 8
For the atomic bomb test:
logT = 5.0
T = 10 5.0 = 1 ! 10 5
1.995 ! 10 8
" 2000 . San Francisco earthquake was approximately
Dividing the two values:
1 ! 10 5
2,000 times greater.
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This worksheet may be printed and used for educational purposes
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Intermediate Algebra
Name _________________________
Problem Set 8.5
Solutions to Every
Odd-Numbered Problem
Date _________________________
(1 + x)1/ x
2
x
1
79.
81.
0.5
0.1
It appears to approach e. Completing the table:
0.01
0.001
0.0001
0.00001
Substituting s = 15:
5 ln x = 15
2.25
2.5937
2.7048
2.7169
2.7181
2.7183
ln x = 3
83.
85.
x = e3 ! 20
Approximately 15% of students enrolled are in the age range in the year 2009.
Computing the pH: pH = ! log 6.50 " 10 !4 # 3.19
(
)
Finding the concentration:
4.75 = ! log "# H + $%
!4.75 = log "# H + $%
87.
"# H + $% = 10 !4.75 & 1.78 ' 10 !5
Finding the magnitude:
5.5 = logT
89.
Finding the magnitude:
8.3 = logT
T = 10 5.5 ! 3.16 " 10 5
91.
T = 10 8.3 ! 2.00 " 10 8
Location
Date
Magnitude (M ) Shockwave (T )
Moresby Island
January 23
4.0
1.00 ! 10 4
Vancouver Island
April 30
5.3
1.99 ! 10 5
Completing the table:
Quebec City
June 29
3.2
1.58 ! 10 3
Mould Bay
November 13
5.2
1.58 ! 10 5
St. Lawrence
December 14
3.7
5.01 ! 10 3
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This worksheet may be printed and used for educational purposes
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93.
Intermediate Algebra
Name _________________________
Problem Set 8.5
Solutions to Every
Odd-Numbered Problem
Date _________________________
Finding the rate of depreciation:
1
4500
log (1 ! r ) = log
5
9000
log (1 ! r ) " !0.0602
95.
1 ! r " 10 !0.0602
Finding the rate of depreciation:
1
5750
log (1 ! r ) = log
5
7550
log (1 ! r ) " !0.0237
1 ! r " 10 !0.0237
r = 1 ! 10 !0.0602
r = 1 ! 10 !0.0237
r " 0.129 = 12.9%
Solving the equation:
5 ( 2 x + 1) = 12
10 x + 5 = 12
10 x = 7
7
x=
= 0.7
10
100, 000
= 3.125
99. Using a calculator:
32, 000
1 " !0.6931 %
+ 3' ( 1.2575
101. Using a calculator: $
&
2 # 1.4289
97.
r " 0.053 = 5.3%
103. Rewriting the logarithm: log1.05 t = t log1.05
105. Simplifying: ln e0.05 t = 0.05t
Copyright © 2010 MathTV.com, Inc.
All rights reserved.
Videos at http://www.mathtv.com
This worksheet may be printed and used for educational purposes
only. It cannot be used for commercial purposes without the written
consent of MathTV.com, Inc. Created by Ross Rueger.