$(x^2-9)^{(3x+5)}=(x-3)^{(x-1)}(x+3)^{(x-1)}$ What is $x$? $(x^2-9)^{(3x+5)}=(x-3)^{(x-1)}(x+3)^{(x-1)}$
$(x^2-9)^{(3x+5)}=(x^2-9)^{(x-1)}$
$3x+5=x-1$
$x=-3$
But when I try to use WolfarmAlpha the integer solution is $3$ instead of $-3$. The numerical solution is $x\approx 3.1622776600...$
So I tried to do other things, such as making sure the base is larger than $0$.
$x^2-9>0$
$x<-3$ or $x>3$
But I still didn't obtain the $x$.
So what is $x$?

The domain is $x\ge3$ due to the power functions involved, and it may easier to examine the equation with $x=t+3$, which leads to
$$(t^2+6t)^{t+2}[( t^2+6t)^{2(t+6)} -1]=0$$
The first factor produces the solution $t=0$, or $x=3$, and the second factor reduces to $t^2+6t=1$,
which yields $t=\sqrt{10}-3$, or $x=\sqrt{10}$.
