Question

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

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?

Answer 1

The equation is the same as 4.

Explanation:

I'm guessing that based on your description that it is meant to be `log_2(7)-log_2(7/16)` and not `log_2(7)=log_2(7/16)`

There are two ways of doing this.

The way you did it is almost correct except you forgot to multiply by -1 after expanding `-log_2(7/16)`.

Remember, you are taking the negative of `log_2(7/16)`. The positive of `log_2(7/16)` is `log_2(7)-log_2(16)`, while the negative is `-(log_2(7)-log_2(16))=-log_2(7)+log_2(16)`

`log_2(7)-log_2(7/16)`
`log_2(7)-(log_2(7)-log_2(16))`
`log_2(7)-log_2(7)+log_2(16)`
`log_2(7/7)+log_2(16)`
`log_2(1)+log_2(16)`
`0+log_2(16)`
`=log_2(16)=4`

You know that `log_ab-log_ac=log_a(b/c)`

Well, `log_2(7)-log_2(7/16)`
`=log_2(7/(7/16))`
`=log_2((cancel(7)*16)/cancel(7))`
`=log_2(16)=4`