Prove that $ \frac {12^{x-2}.4^{x}} {6^{x-2}} = 2^{3x-2} $ Can someone please help me with this question?
$ \large \frac {12^{x-2}.4^{x}} {6^{x-2}} = 2^{3x-2} $
My steps so far:
$ \large \frac {4^{x-2}.3^{x-2}.4^{x}}{3^{x-2}.2^{x-2}} = 2^{3x-2} $
$ \large \frac {4^{x-2}.4^{x}}{2^{x-2}} = 2^{3x-2} $
$ \large \frac {4^{2x-2}}{2^{x-2}} = 2^{3x-2}$
I get stuck here but I am assuming I need to get that 4 to a 2 somehow so I can combine them... so:
$ \large \frac {(2^2)^{2x-2}}{2^{x-2}} = 2^{3x-2}$
I feel like I am not on the right track here as I have no idea where to go now. Could anyone help please? Thank you!

$\frac {12^{x-2}.4^{x}} {6^{x-2}} = 2^{3x-2}$
$\frac {4^{x-2}.3^{x-2}.4^{x}}{3^{x-2}.2^{x-2}} = 2^{3x-2}$
$\frac {4^{2x-2}}{2^{x-2}} = 2^{3x-2}$
$\frac{2^{2x-2}}{2^{x-2}} 2^{2x-2} = 2^{3x-2}$
Then use the fact that : $\frac{a^{b}}{a^{c}} = a^{b-c}$ with $a\ne{0}$
So : $2^{2x-2-(x-2)} 2^{2x-2} = 2^{3x-2}$
$\Rightarrow$ $2^{x}2^{2x-2} = 2^{3x-2}$
