Given that $9^{2x} = 27^{x^2 - 5}$. Find the possible values of $x$. Given that $9^{2x} = 27^{x^2 - 5}$. Find the possible values of $x$.
I don't know how to approach this question.

As $9=3^2, 9^{2x}=(3^2)^{2x}=3^{4x}$
Similarly, $27^{x^2-5}=3^{3x^2-15}$
As $3^{4x}\ne0,$
$$\implies1=\dfrac{3^{3x^2-15}}{3^{4x}}=3^{3x^2-4x-15}$$
Now like Find all real numbers $x$ for which $\frac{8^x+27^x}{12^x+18^x}=\frac76$,
if $\displaystyle u^m=1,$ 
either $\displaystyle m=0 $
or $\displaystyle u=1$
or $\displaystyle u=-1,m$ is even
But here $u=3\ne\pm1$
