Question

Solve for X? (Logarithmic Equation)

`log_9(x-5) + log_9(x+3) = 1`

Answer 1

`x=6` and `x = -4`

Explanation:

Recall that `log_{a} AB = log_{a}A + log_{a} B` and `log_{a}a = 1.`
Therefore, `log_{9}(x-5) + log_{9}(x+3) = 1` can be written as `log_{9}{(x-5)(x+3)}=log_{9}9.`
Cancel out `log_{9}` on both sides, we have `(x-5)(x+3)=9`; and solving `(x-5)(x+3)-9=0` for `x` we have `x=6, -4`

Answer 2

`x=6`

Explanation:

Using properties of logarithms we can rewrite the left hand side.

`log(a)+log(b)=log(ab)`

`log_9((x-5)(x+3))=1`

Now rewrite both sides in terms of the base `9`

`9^(log_9((x-5)(x+3)))=9^1`

rewriting the left hand side we have

`(x-5)(x+3)=9`

`x^2-2x+15=9`

`x^2-2x-24=0`

`(x-6)(x+4)=0`

`x-6=0` OR `x+4=0`

`x=6`

If we are restricted to the real numbers, we disregard
`x=-4`