Question

How do you solve `log_4 (x - 7) + log_4 (x + 7) = 2`?

Answer 1

`x=+sqrt65`

Explanation:

`=>log_4((x-7)*(x+7))=2`
`=>x^2-7^2=4^2`
`=>x^2=49+16=65`
`x=+-sqrt65`
negative root not possible as log of negative number is not defined
so `x=sqrt65`

Answer 2

`sqrt 65~~8.062`

Explanation:

`log_b m+log_b n=log_b (mn).`
We can use this here and condense, assuming that x > 7..

`log_4((x-7)(x+7))=2`

Inversely,`(x-7)*(x+7)=4^2`.

`x^2-49=16`

`x=+-sqrt65`

Negative root is inadmissible, as x > 7.