There are three big important properties of logarithms that each come from a
property of exponents. Here are the properties with their corresponding exponent
properties.
Logarithm Properties
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loga(MN)=logaM+logaN
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loga(M/N)=logaM-logaN
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loga(Mp)=plogaM
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Exponential Properties
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axay=ax+y |
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ax/ay=ax-y |
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(ax)p=apx
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We can get each of the logarithm properties from the corresponing
exponential property by applying the exponential property to x=logaM
and y=logaN, taking logs on both sides and using the The Inverse
Properties of Logarithms.
Property 1
alogaMalogaN=alogaM+logaN
By one of the Inverse
Properties of Logarithms the left side is MN. (You can also simply see this
by saying to yourself, for example in the first factor, that you are
raising a to the power that it needs to be raised to in order to get M, so what
you get must be M. See my article The Inverse
Properties of Logarithms for further explanation.) So this gives us
MN=alogaM+logaN.
Then taking logs on both sides it becomes
loga(MN)=loga(alogaM+logaN),
and the right side of this is logaM+logaN by the other
Inverse
Properties of Logarithms (You can also see this by noticing that the
right side is asking what power can you raise a to in order to get a to the
logaM+logaN-th power. See my article The Inverse
Properties of Logarithms for further explanation.) So
loga(MN)=logaM+logaN.
Property 2
alogaM/alogaN=alogaM-logaN
By one of the Inverse
Properties of Logarithms the left side is M/N, so this gives us
M/N=alogaM-logaN.
Then taking logs on both sides it becomes
loga(M/N)=loga(alogaM-logaN),
and the right side of this is logaM-logaN by the other
Inverse
Properties of Logarithms, so
loga(M/N)=logaM-logaN.
Property 3
(alogaM)p=aplogaM
By one of the Inverse
Properties of Logarithms the left side is Mp, so this gives us
Mp=aplogaM.
Then taking logs on both sides it becomes
loga(Mp)=loga(aplogaM),
and the right side of this is plogaM by the other Inverse
Properties of Logarithms, so
loga(Mp)=plogaM.
A good way to get familiar with how to use these properties is to do two
kinds of exercises, one where you need to go from the log of some big
complicated expression to an expression involving individual logs, and the other
where you do the reverse. Here are some worked out examples of both kinds.
Type 1
For these kinds of problems the instruction is to write the
expression in terms of individual logarithms and simplify.
Example 1:
loga(x3y5)
Solution:
loga(x3y5)=loga(x3)+loga(y5)=3logax+5logay
Example 2:
Solution:
Example 3:
Solution:
Type 2
For these the instruction is to write the expressionas a single
logarithm and then simplify.
Example 1:
Solution:
Example 2:
Solution: