1. Logarithms and Equation Solving
Chapter 2 Exponents and Logarithms
2.7A
Logarithms
and
Equation Solving
MATHPOWERTM 12, WESTERN EDITION
2.7A.1

2. Solving Log Equations 1. log272 = log2x + log212 2. 2x = 8
log272 - log212 = log2x
x = 8
log 2x = log 8
xlog2 = log 8
x = 3
x = 6 3.
x = 3.79
xlog7 = 2log40
2.7A.2

3. 4. log7(2x + 2) - log7(x - 1) = log7(x + 1)
Solving Log Equations
log7(2x + 2) - log7(x - 1) = log7(x + 1)
2x + 2 = (x- 1)(x + 1)
2x + 2 = x2 - 1
0 = x2 - 2x - 3
0 = (x - 3)(x + 1)
x - 3 = 0 or x + 1 = 0
x = x = -1
Therefore, x = 3.
Check:
log7(2x + 2) - log7(x - 1) = log7(x + 1)
log7(2x + 2) - log7(x - 1) = log7(x + 1)
log7(2(3) + 2) - log7(3 - 1) = log7(3 + 1)
log7(2(-1) + 2) - log7(-1 - 1) = log7(-1 + 1)
log70 - log7(-2) = log7(0)
Negative logarithms and
logs of 0 are undefined.
log74 = log74
2.7A.3

4. Solving Log Equations 5. log7(x + 1) + log7(x - 5) = 1
log7[(x + 1)(x - 5)] = log77
(x + 1)(x - 5) = 7
x2 - 4x - 5 = 7
x2 - 4x - 12 = 0
(x - 6)(x + 2) = 0
x - 6 = 0 or x + 2 = 0
x = x = -2
6. log4(4x) - log4(x - 2) = 3
64(x - 2) = 4x
64x = 4x
60x = 128
x = 2.13
x = -2 is extraneous.
Therefore, x = 6.
2.7A.4

5. Solve: log5(x - 6) = 1 - log5(x - 2)
Solving Log Equations
Express 12 as a power of 2:
Solve: log5(x - 6) = 1 - log5(x - 2)
log5(x - 6) + log5(x - 2) = 1
log5(x - 6)(x - 2) = 1
log5(x - 6)(x - 2) = log551
(x - 6)(x - 2) = 5
x2 - 8x + 12 = 5
x2 - 8x + 7 = 0
(x - 7)(x - 1) = 0
x = 7 or x = 1
xlog2 = log12
x = 3.58
23.58 = 12
Since x > 6, the value of x = 1
is extraneous. Therefore, the
solution is x = 7.
2.7A.5

6. Solving Log Equations 7. 3x = 2x + 1 log(3x) = log(2x + 1)
x log 3 = x log log 2
x log 3 - x log 2 = log 2
x(log 3 - log 2) = log 2
x - 1 = 32x - 1
(3x - 1) log 2 = (2x - 1) log 3
3x log 2 -1 log2 = 2x log3 - log3
3xlog2 - 2xlog3 = log2 - log3
x(3log2 - 2log3) = log2 - log3
x = 3.44
x = 1.71
2.7A.6

7. Applications of Logarithms
1. Carbon 14 has a half-life of 5760 years. Find the age of a
specimen with 24% C-14 relative to living matter.
t =
Therefore, the specimen is
years old.
2.7A.7

8. Applications of Logarithms
Find the time period required for $7000 invested at
10%/a compounded semi-annually to grow to $
A(t) = P(1 + i) 2n
= 7000(1.05)2n
log10 - log7 = 2nlog1.05
7.31 = 2n
3.66 = n
It would take 3.66 years for the investment to grow to $
2.7A.8

9. Applications of Logarithms
3. The value of an investment is given by f(x) = (1.052)x,
where x is the number of 6-month periods. Find the
number of complete periods until the investment is
worth at least $600.
f(x) = (1.052)x
600 = (1.052)x
= (1.052)x
log = x log 1.052
x = 18.28
Therefore, after 19 periods
the investment would be worth
at least $600.
2.7A.9

10. 36 = t Applications of Logarithms
4. Cell population doubles every 3 h. How long would
it take 4 cells to reach a count of ?
It would take 36 h to
reach cells.
36 = t
2.7A.10

11. Applications of Logarithms
5. For every metre below the water surface, light
intensity is reduced by 5%. At what depth is light
intensity 40% of that at the surface?
Id = Io( )d
40 = 100(0.95)d
0.4 = 0.95d
log 0.4 = dlog0.95
d = 17.86
Therefore, at a depth of m
the light intensity would be 40%.
2.7A.11

12. More Applications - Comparing Intensities of Sound
For any intensity, I, the decibel level, dB, is defined as follows:
where Io is the intensity
of a barely audible sound
6. The sound at a rock concert is 106 dB. During
the break, the sound is 76 dB. How many times
as loud is it when the band is playing?
Louder
Softer
Comparison
Thus, it would be
1000 times as loud.
I = Io
I = Io
2.7A.12

13. More Applications - The Richter Scale
I = Io(10)m
where m is the measure on the scale
7. Compare the intensities of the Japan earthquake of 1933,
which measured 8.9 on the Richter Scale, to the earthquake
of Turkey in 1966, which measured 6.9 on the scale.
Therefore, the earthquake in Japan is
100 times as intense as the one in Turkey.
2.7A.13

14. More Applications - The Richter Scale
8. The magnitude of earthquakes is given by
where I is the quake intensity and Io is the reference
intensity.
How many times as intense is a quake of 8.1
compared to a quake with a magnitude of 6.4?
Comparison
Therefore, a quake of 8.1 is 50.1 times as great.
2.7A.14

15. More Applications - The Richter Scale
9. Earthquake intensity is given by I = Io (10)m, where
m is the magnitude and Io is the relative intensity.
A quake of magnitude 7.9 is 120 times as intense as
a tremor. What is the magnitude of the tremor?
Iq = Io (10)7.9
It = Io (10)m
log 120 = (7.9 - m) log 10
log 120 = (7.9 - m)
The magnitude of
the tremor is 5.8.
m = log 120
m = 5.8
2.7A.15

16. Assignment Suggested Questions: Pages 113 and 114 A 1-13 odd,
21-25, 40 a-e
B , 40 f-h,
42-44
Page 99 46
2.7A.16

17. Logarithmic Identities
Chapter 2 Exponents and Logarithms
2.7B
Logarithmic
Identities
MATHPOWERTM 12, WESTERN EDITION
2.7B.17

18. Some equations are solved for all values of the variable for
Proving Identities
Some equations are solved for all values of the variable for
Which both sides of the equation are defined. These equations
are called identities.
= logax -1
= -1logax
-logax
= -logax
L.S. = R.S.
Therefore, the identity
is proved.
L.S. = R.S.
2.7B.18

19. Proving Identities Prove: = Conclusion: This suggests that L.S. = R.S.
2.7B.19

20. Proving Identities
Verify:
L.S. = R.S.
2.7B.20

21. Assignment
Suggested Questions:
Page 106
44-46
Page 113
33-38
2.7B.21