Answer
$x=3$
Work Step by Step
$\log_{3}(x-2) + \log_{3}(x+6)=2$
$\log_{3}[(x-2)(x+6)]=2$
$x^{2} + 4x -12=3^{2}$
$x^{2}+4x -15=0$
$x=3$ and $x=-7$
Since $x=-7$ is not a valid value for x in the original equation, the only solution is $x=3$.