Solving Log Equations
1.
log272 = log2x + log212
log272 - log212 = log2x
72 

log 2
 log 2 x
12 
2.
72   x
12 
2x = 8
log 2x = log 8
xlog2 = log 8
log8
x
log2
x=3
x=6
3.
7
1
x
2
 40
1
x log7  log 40
2
xlog7 = 2log40
2log 40
x
log 7
x = 3.79
Solving Log Equations
4. log7(2x + 2) - log7(x - 1) = log7(x + 1)
2x  2

log 7
 log 7 (x  1)
 x  1 
2x  2  (x  1)
 x  1 
Check:
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=3
x = -1
log7(2x + 2) - log7(x - 1) = log7(x + 1)
log7(2(3) + 2) - log7(3 - 1) = log7(3 + 1)
8
log 7   log 7 4
2 
log74 = log74
Therefore, x = 3.
log7(2x + 2) - log7(x - 1) = log7(x + 1)
log7(2(-1) + 2) - log7(-1 - 1) = log7(-1 + 1)
log70 - log7(-2) = log7(0)
Negative logarithms and
logs of 0 are undefined.
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=6
x = -2
x = -2 is extraneous.
Therefore, x = 6.
6. log4(4x) - log4(x - 2) = 3
4x 

log 4
 log 4 4 3
x  2 
 4x   43
x  2 
64(x - 2) = 4x
64x - 128 = 4x
60x = 128
x = 2.13
Solving Log Equations
Express 12 as a power of 2:
2 x  12
xlog2 = log12
log12
x
log 2
x = 3.58
23.58 = 12
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
Since x > 6, the value of x = 1
is extraneous. Therefore, the
solution is x = 7.
Solving Log Equations
3x = 2x + 1
8.
23x - 1 = 32x - 1
log(3x) = log(2x + 1)
(3x - 1) log 2 = (2x - 1) log 3
x log 3 = (x + 1)log 2
3x log 2 -1 log2 = 2x log3 - log3
x log 3 = x log 2 + 1 log 2 3xlog2 - 2xlog3 = log2 - log3
x log 3 - x log 2 = log 2
x(3log2 - 2log3) = log2 - log3
log 2  log 3
x(log 3 - log 2) = log 2
x
log 2
3log 2  2 log 3
x
log3  log 2
x = 3.44
x = 1.71
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
h
1 

A(t)  Ao  
2
t
1 5760

24  100 
2
t
5760
1 
0.24  
2 
t
1 

log 0.24 
log
5760 2 
1
5760 log0.24  t log 
2 
5760 log0.24
t
1
log
2
t = 11 859.23
Therefore, the specimen is
11 859.23 years old.
Applications of Logarithms
2. Find the time period required for $7000 invested at
10%/a compounded semi-annually to grow to $10 000.
A(t) = P(1 + i) 2n
10 000 = 7000(1.05)2n
10
 1.052 n
7
10
log    2n log1.05
 7 
log10 - log7 = 2nlog1.05
log 10  log 7
 2n
log1.05
7.31 = 2n
3.66 = n
It would take 3.66 years for the investment to grow to $10 000.
Applications of Logarithms
3. The value of an investment is given by f(x) = 237.50(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) = 237.50(1.052)x
600 = 237.50(1.052)x
2.5263 = (1.052)x
log 2.5263 = x log 1.052
log2.5263
x
log1.052
x = 18.28
Therefore, after 19 periods
the investment would be worth
at least $600.
Applications of Logarithms
4. Cell population doubles every 3 h. How long would
it take 4 cells to reach a count of 16 384?
A(t)  Ao (2)
t
d
t
3
16 384  4(2)
t
3
4096  (2)
t
log 4096  log(2)
3
3 log4096  t log(2)
3 log4096
t
log 2
36 = t
It would take 36 h to
reach 16 384 cells.
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(1 - 0.05)d
40 = 100(0.95)d
0.4 = 0.95d
log 0.4 = dlog0.95
log 0.4
d
log 0.95
d = 17.86
Therefore, at a depth of 17.86 m
the light intensity would be 40%.
More Applications - Comparing Intensities of Sound
For any intensity, I, the decibel level, dB, is defined as follows:
 I  where I is the intensity
o
dB  10 log 
Io  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?
Comparison
Louder
Softer
 I 
dB  10 log  
Io 
 I 
106  10log  
Io 
 I 
10.6  log  
 Io 
I
 1010.6
Io
I=
1010.6 I
o
 I 
dB  10 log  
Io 
 I 
76  10log  
Io 
 I 
7.6  log  
Io 
I
 10 7.6
Io
I = 107.6 Io
Ilouder 1010.6 Io
 7.6
Isofter 10 Io
Ilouder
 103
Isofter
Thus, it would be
1000 times as loud.
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.
IJapan 10 8.9 Io
 6.9
ITurkey 10 Io
IJapan
ITurkey
 10
2
Therefore, the earthquake in Japan is
100 times as intense as the one in Turkey.
More Applications - The Richter Scale
 I 
8. The magnitude of earthquakes is given by m  log  
Io 
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?
 I 
m  log  
Io 
I1 
8.1  log 
Io 
8.1
10
I1 
  
Io 
I2 
6.4  log 
Io 
6.4
10
Comparison
I1 
8.1
10   
Io 
6.4
10
I2 
  
Io 
I2 
  
Io 
I1 
10   
I2 
Therefore, a quake of 8.1 is 50.1 times as great.
1.7
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
Iq
 107.9 m
It
120  10 7.9 m
log 120 = (7.9 - m) log 10
log 120 = (7.9 - m)
m = 7.9 - log 120
m = 5.8
The magnitude of
the tremor is 5.8.