How find this value of $x$ such $\log_{\frac{1}{12}}{(x^2+2x-3)}>x^2+2x-16$ if $$\log_{\frac{1}{12}}{(x^2+2x-3)}>x^2+2x-16$$
Find the  value of  $x$  
My idea: since $$x^2+2x-3>0\Longrightarrow x>1 ,or, x<-3$$
$$x^2+2x-3<\left(\dfrac{1}{12}\right)^{x^2+2x-16}$$
let
$$(x+1)^2=y\ge 4$$
so
$$\left(\dfrac{1}{12}\right)^{y-17}>y-4$$
Then I can't.Thank you

Let, $x^2+2x-16=y$. So, $x^2+2x-3=y+13$.  
Now, we have, $y+13<\bigg(\dfrac1{12}\bigg)^{y} \implies 12^y(y+13)<1$  
From this we can conclude, when $y<-1$ or $x^2+2x-15<0$ or $x\in[-5,3]$ there exists solution.
