# How do you evaluate #log_sqrt7 49#?

##### 1 Answer

Apr 4, 2016

#### Answer:

#### Explanation:

We should attempt to write

First, simply rewrite

#log_sqrt7 49=log_sqrt7(7^2)#

This can be rewritten using the rule:

#log_a(b^c)=c*log_ab#

Thus, we have

#log_sqrt7(7^2)=2log_sqrt7 7#

Now, we can write that

#2log_sqrt7(sqrt7)^2#

Rewrite using the rule we found earlier.

#2*2log_sqrt7sqrt7#

Note that

#log_aa=1#

So we are just left with

#2*2log_sqrt7sqrt7=4#

We can test this by using our rules of logarithms. We said that:

#log_sqrt7 49=4#

This means that

#sqrt7 xx sqrt7xxsqrt7xxsqrt7=7xx7=49#