# How do I solve #log_2 7 = log_2 (7/16)#?

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I'm not sure if what I'm doing is correct, but I tried expanding it:

#log_2 7 - log_2 7 - log_2 16#

#= log_2 (7/7) - log_2 16#

#=log_2 1 - log_2 16#

#= log_2 (1/16)#

#=log_2 (1/2^4)#

and after that I just got stuck. Can someone help?

I'm not sure if what I'm doing is correct, but I tried expanding it:

and after that I just got stuck. Can someone help?

##### 1 Answer

The equation is the same as 4.

#### Explanation:

I'm guessing that based on your description that it is meant to be

There are two ways of doing this.

The way you did it is almost correct except you forgot to multiply by -1 after expanding

Remember, you are taking the **negative** of **positive** of

You know that

Well,