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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!

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$\frac{4^{2x-2}}{2^{x-2}} = \frac{(2^2)^{2x-2}}{2^{x-2}} = 2^{(4x-4)-(x-2)} = 2^{3x-2}$

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    $\begingroup$ Thank you very much. I was obviously doing that last step incorrectly and it was throwing me off. Thanks for your time to help me! $\endgroup$
    – Dani
    Aug 5 '14 at 0:20
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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.

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  • $\begingroup$ Understand completely now, thank you so much for your help! I was just going wrong with that bit at the end as you said, and I can see the obvious mistake I've made. Again, thank you! $\endgroup$
    – Dani
    Aug 5 '14 at 0:22
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$\frac {12^{x-2}.4^{x}} {6^{x-2}} = 2^{3x-2}$

$\frac {4^{x-2}.3^{x-2}.4^{x}}{3^{x-2}.2^{x-2}} = 2^{3x-2}$

$\frac {4^{2x-2}}{2^{x-2}} = 2^{3x-2}$

$\frac{2^{2x-2}}{2^{x-2}} 2^{2x-2} = 2^{3x-2}$

Then use the fact that : $\frac{a^{b}}{a^{c}} = a^{b-c}$ with $a\ne{0}$

So : $2^{2x-2-(x-2)} 2^{2x-2} = 2^{3x-2}$

$\Rightarrow$ $2^{x}2^{2x-2} = 2^{3x-2}$

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$$\large \frac {12^{x-2}.4^{x}} {6^{x-2}}$$

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

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