Answer
$x=5$
Work Step by Step
We have to find x:
$$\log_{3}(x-4)=2-\log_{3}(x+4)$$
$$\log_{3}(x-4)+\log_{3}(x+4)=2$$
Use the Product Property of logarithm
$$\log_{3}(x-4)(x+4)=2$$
$$(x-4)(x+4)=3^2=9$$
$$x^2-16=9$$
$$x^2=9+16$$
$$x^2=25$$
$$x_{1}=5 \text{ and } x_{2}=-5$$
But $x_{2}=-5 $ is not suitable for us because the expression $\log_{a}x$ is defined only for $x>0$ and when we replace $x=-5$ in the given equation we get $\log_3 (-5-4)=\log_3(-9)$ which is not possible. For $x=5$ the logarithms are defined.
So the only solution of the equation is:
$$x=5$$
