Saying that
*log _{a}M =x *means exactly the same thing as saying

In other words:

Let's use this property to understand logarithms of products, quotients, and powers.

For example:
What is* log _{5}(5*125)*?

Notice that *5=5 ^{1 }*and 125

But this means that

*log _{5}(5*125)=log_{5}(5^{4}) = 4 =
1+3=*

*log _{5}(5^{1})+log_{5}(5^{3})=log_{5}(5)+log_{5}(125)*

*In other words, the l og of the product 5*125
equals the sum of the logs of 5 and 125.*

Apply this idea to the following. Of course, there is nothing special about the base 5.

So what is the general rule?

We can understand the logarithm of a quotient the same
way. For example: How is *log _{2}(128/16)* related to

Well, 12*8=2 ^{7 }*and 16

But this means that

*log _{2}(128/16)=log_{2}(2^{3})
= 3 =
7-4=*

*log _{2}(2^{7})-log_{2}(2^{4})=log_{2}(128)-log_{2}(16)*

*In other words, the l og of the quotient 128/16
equals the difference of the logs of 128 and
16. *

Apply this idea to the following.

Think of the general rule and check by clicking the button.

Finally, let's look at the logarithm of a power using

Here's one way to start understanding this. If
*p * is a positive whole number, then

*M ^{p} = M M M M M .... M *(the
number of

*log _{a}(M ^{p}) =
log_{a}(M)+log_{a}(M)+log_{a}(M)+.....+log_{a}(M)=
p log_{a}(M)*

since we are adding *log _{a}(M) *a
total of

*log _{3}(9)= 2 *so

Now, it turns out that the same rule works
even if *p* is not a whole positive number --- in fact,
it works for any value of *p *. So, once again, guess the
general rule for logs of powers and click below to check.

*Precalculus Tutorials, B. Kaskosz and L. Pakula, 2002.
*

* *