$(x^2-9)^{(3x+5)}=(x-3)^{(x-1)}(x+3)^{(x-1)}$ 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 $x$?
 A: The equation:
$$(x^2-9)^{(3x+5)}=(x^2-9)^{(x-1)}
$$
obviously holds if $x^2-9=0$ provided that the exponents are non-negative. This is the case for $x=3$.
If $x^2-9\ne0$ we can divide both sides by $(x^2-9)^{(x-1)}$ obtaining:
$$(x^2-9)^{(2x+6)}=1.$$
This equation holds if $2x+6=0$ or $x^2-9=1$. The roots of the equations are $x=-3$ and $x=\pm\sqrt{10}$, respectively.
The substitution in the original equation shows that
$x=3,\sqrt{10}$ satisfy the equation whereas the roots $x=-3,-\sqrt{10}$ do not suit it.
A: Hint
I assume you are looking for real solution.
First, check the cases:
a) $(x^2-9)\in \{-1,0,1\}$. Test each case in the inital equation;
b) When $(x^2-9)>0$ you do $3x+5=x-1$, and take the intersection.
An observation, if you try $x=-3$ in the first equation, the term $(x-3)^{x-1}$ is equal to $(-6)^{-4}$. Is it a solution?
EDIT:
Plotting both functions (RHS and LHS) it seems we have two intersection. One at $x=3$ an other close to it. @user showed how to find the second solution.

A: 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 $x=\sqrt{10}$.
