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
|
loga(MN)=logaM+logaN
|
|
loga(M/N)=logaM-logaN
|
|
loga(Mp)=plogaM |
|
Exponential Properties
|
axay=ax+y |
|
ax/ay=ax-y |
|
(ax)p=apx |
|
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: