Answer
$x=6$
Work Step by Step
$\log_{9}(x-5)+\log_{9}(x+3)=1$
$\log_{9}[(x-5)(x+3)]=1$
$(x-5)(x+3)=9^{1}$
$x^2+3x-5x-15=9$
$x^2-2x-15-9=0$
$x^{2}-2x-24=0$
$(x-6)(x+4)=0$
$(x-6)=0$ or $(x+4)=0$
$x=6$ or $x=-4$
However, $x=-4$ does not work in the original equation because we can't take the log of a negative number ($\log_{9}(-4-5)$=undefined). So we throw this solution out.