Examples
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Solution:
We have
ln(3x^{3})^{1/2} = 1/2
ln(3x^{3})
(Property 1)
= 1/2ln3 + 1/2lnx^{3}
(Property 2)
= 1/2ln3 + 3/2lnx.
(Property 1)
Exercises: Expand the following:

log[(x^{2}(x  4)^{5})/100]

log_{3}(sqrt(x^{5}/9))
Example:
Write the following with only one logarithm:
3log_{4}x  5log_{4}(x^{2} + 1) +
2log_{4}x^{2}
Solution:
We use the properties:
log_{4}x^{3} 
log_{4}(x^{2} + 1)^{5} +
log_{4}(x^{2})^{2 }
(Property 1)
= log_{4}[x^{3}/(x^{2} +
1)^{5}] + log_{4}(x^{4})
(Property 3)
=
log_{4}[x^{3}x^{4}/(x^{2}
+ 1)^{5}]
(Property 2)
= log_{4}[x^{7}/(x^{2} +
1)^{5}]
(A Property of Exponents)
Exercises:
Write the following with only one logarithm:

2log_{3}x  2log_{3}sqrt(x) + 5log_{3}1/x

logx  2log(x  1) + log(x + 1)
Application
The Rictor scale for earthquakes is as follows: if I is the intensity
of an earthquake and I_{0} is the intensity of the shaking without
an earthquake, then the magnitude R of an earthquake is defined by
R = log[I/I_{0}]
The Loma Prieta quake measured 7.1 on the Rictor scale and the Hokkaido quake
measured 8.2. How many times more intense was the Hokkaido quake?
Solution
Let
I_{L} =
The intensity of the Loma Prieta quake
and
I_{H} =
The intensity of the Hokkaido quake
We write
log(I_{H}/I_{L})
= log(I_{H}/I_{0 }/ I_{L}/I_{0})
= log(I_{H}/I_{0})
 log(I_{L}/I_{0})
= 8.2  7.1 = 1.1
By exponentiating both
sides with base ten, we get
I_{H}/I_{L}
= 10^{1.1} = 12.6
We can conclude that
the Hokkaido quake was more than 12 times more intense than the Loma Prieta
quake.