<s>[INST] $(x^2-9)^{(3x+5)}=(x-3)^{(x-1)}(x+3)^{(x-1)}$ 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 $x$?
 [/INST] 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.
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