Answer
The solution set is $\{6\}$.
Work Step by Step
The given equation is
$\log _3(x-5) +\log _3(x+3)=2$
Use product rule.
$\log _3 [(x-5)(x+3)]=2$
Rewrite in exponential form.
$(x-5)(x+3)= 3^2$
Use the distributive property on the left.
$x^2+3x-5x-15=9$
Subtract $9$ from both sides.
$x^2+3x-5x-15-9=9-9$
Simplify.
$x^2-2x-24=0$
Factor.
$x^2-6x+4x-24=0$
$x(x-6)+4(x-6)=0$
$(x-6)(x+4)=0$
Set both factors equal to $0$.
$x-6=0$ or $x+4=0$
Isolate $x$.
$x=6$ or $x=-4$
$-4$ is not in the domain of a logarithmic function.
The solution is $x=6$.
