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!
 A: $\frac{4^{2x-2}}{2^{x-2}} = \frac{(2^2)^{2x-2}}{2^{x-2}} = 2^{(4x-4)-(x-2)} = 2^{3x-2}$
A: You are doing the right thing. Just be careful at the end. Remember that: $$(a^b)^c = a^{bc}$$
So $$4^{2x - 2} = (4^2)^{x-1} = (2^4)^{x - 1} = 2^{4x - 4}$$This happens because $2^4 = 4^2 = 16$. In general $a^b \neq b^a$. And you can write $\frac{1}{2^{x - 2}} = 2^{2 - x}$. This gives: $$2^{4x - 4} \cdot 2^{2 -x} = 2^{3x - 2}$$
Seeing your attempt, I'm sure you can go on your own now. But if you need a little more help, say.
A: $\frac {12^{x-2}.4^{x}} {6^{x-2}} = 2^{3x-2}$
$\frac {4^{x-2}.3^{x-2}.4^{x}}{3^{x-2}.2^{x-2}} = 2^{3x-2}$
$\frac {4^{2x-2}}{2^{x-2}} = 2^{3x-2}$
$\frac{2^{2x-2}}{2^{x-2}} 2^{2x-2} = 2^{3x-2}$
Then use the fact that : $\frac{a^{b}}{a^{c}} = a^{b-c}$ with $a\ne{0}$
So : $2^{2x-2-(x-2)} 2^{2x-2} = 2^{3x-2}$
$\Rightarrow$ $2^{x}2^{2x-2} = 2^{3x-2}$
A: $$\large \frac {12^{x-2}.4^{x}} {6^{x-2}}$$
$$\begin{align}
& \implies \large \frac {(3\times4)^{x-2}.2^{2x}} {(3\times2)^{x-2}} \\
& \implies \large \frac {(3)^{x-2}\times(2)^{2(x-2)}\times2^{2x}} {(3\times2)^{x-2}} \\
& \implies \large \frac {(3)^{x-2}\times(2)^{2(x-2)}2^{2x}} {(3)^{x-2}\times(2)^{x-2}} \\
& \implies \large (2)^{2(x-2)+2x-(x-2)} \\
& \implies \large (2)^{2x-4+2x-x+2}\\
& \implies \large (2)^{3x-2} \\
& \end{align}$$
