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!

$$\large \frac {12^{x-2}.4^{x}} {6^{x-2}}$$
$$\begin{align}
& \implies \large \frac {(3\times4)^{x-2}.2^{2x}} {(3\times2)^{x-2}} \\
& \implies \large \frac {(3)^{x-2}\times(2)^{2(x-2)}\times2^{2x}} {(3\times2)^{x-2}} \\
& \implies \large \frac {(3)^{x-2}\times(2)^{2(x-2)}2^{2x}} {(3)^{x-2}\times(2)^{x-2}} \\
& \implies \large (2)^{2(x-2)+2x-(x-2)} \\
& \implies \large (2)^{2x-4+2x-x+2}\\
& \implies \large (2)^{3x-2} \\
& \end{align}$$
