# $2^{10 - x} \cdot 2^{10 - x} = 4^{10-x}$

$$2^{10 - x} \cdot 2^{10 - x} = 4^{10-x}$$ Is that correct?

I would've done

$$2^{10 - x} \cdot 2^{10 - x}\;\; = \;\; (2)^{10 - x + 10 - x} \; = \; (2)^{2 \cdot (10 - x)} \;=\; 4^{10 - x}\tag{1}$$

Is that allowed?

If so, can I say that

$$\frac{4^x}{2^y} = 2^{x - y} \tag{2}$$

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$$\text{Yes:}\;\;\;\large 2^{10 - x} \cdot 2^{10 - x} =\; 4^{10-x}\tag{1.a}$$

$$\underbrace{\large 2^{10 - x} \cdot 2^{10 - x} = (2)^{10 - x + 10 - x} = (2)^{2 \cdot (10 - x)} = 4^{10 - x}}\tag{1.b}$$ $$\text{Yes, that is that is allowed.}$$

$$\underbrace{ \large \frac{4^x}{2^y} = 2^{x - y}}\;\;\; ?\tag{2 ?}$$ $$\text{NOT correct. See below:}$$

$$(2)\quad \large \frac{4^x}{2^y} \;=\; 4^x \cdot 2^{-y} \;= \;(2^2)^x \cdot 2^{-y} \;=\; 2^{(2x)} \cdot 2^{(-y)} \;=\;2^{(2x-y)}$$

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@Kaish: Does this answer your question? Is it all clear now? – amWhy Dec 24 '12 at 22:21

First statement is correct, but $4^x/2^y = 2^{2x-y}$.

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Is that correct?

Yes!

Notice how you converted $2^2$ to $4$ in your first statement. You can do the vice-versa as well. Let's make a little alteration:$${4^x \over 2^y} \Leftrightarrow {2^{{\color{#00}{2x}}} \over 2^y} \Leftrightarrow {2^{2x - y}}$$That's it!

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