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I'm confused. Why is it that for a problem in the form of: $(2^{x+1})(2^{x-1})$ we get $2^{2x}$ instead of $4^{2x}$; Shouldn't we multiply the $2$s by each other..?

Similarly for a problem like: $(2^{x+1})(4^{x-1})$, why do we get $2^{3x-1}$ rather than $8^{2x}$?

And for $(3^{2x}/3^{x-1})$ we get $3^{x+1}$...shouldn't the $3$s cancel?

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Some parentheses would help a lot... lack of them may even be why you're confused. Please see here for how to typeset common math expressions with MathJax, and see here for how to use Markdown formatting. – Zev Chonoles Jul 3 '13 at 2:40
Lack of parentheses isn't why I'm confused – jaykirby Jul 3 '13 at 2:46
I tried to typeset your problem, could you check to make sure I didn't change the content? – Tyler Jul 3 '13 at 3:01
Appreciate it, content is same. – jaykirby Jul 3 '13 at 3:28
up vote 5 down vote accepted

Recall that $$\large a^b\cdot a^c = a^{b+ c}$$

$$\large \dfrac{a^b}{a^c} = a^{b - c}$$ $$\large \left(a^b\right)^c = a^{bc}$$

So, using these "laws of exponents", we have: $$\begin{align} (1)\quad \large 2^{(x + 1)}\cdot 2^{(x - 1)} & = \large 2^{(x + 1) + (x - 1)}\\ \\ & = \large 2^{2x}\end{align}$$ $$ $$ $$\begin{align} (2)\quad \large 2^{x+1} \cdot 4^{x-1} & = \large 2^{x + 1} \cdot \left(2^2\right)^{(x-1)}\\ \\ &= \large \large 2^{x + 1} \cdot 2^{2(x-1)} \\ \\ & = \large 2^{(x+1) + 2(x-1)} \\ \\ & = \large 2^{3x - 1}\end{align}$$

Now, for the last question, I'll let you try to work that one out.

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got it so we don't multiply the bases then if the bases are the same.. – jaykirby Jul 3 '13 at 3:29
In general, we don't multiply bases...$2^a\cdot 3^b \neq 6^{a+b} $ – amWhy Jul 3 '13 at 3:33
What about for something like (3^x/2^y)^ we get 9^2x/4^2y? – jaykirby Jul 3 '13 at 3:38
$\left(\dfrac{3^x}{2^y}\right)^2 = \dfrac{3^x\cdot 3^x}{2^y\cdot 2^y} = \dfrac {3^{2x}}{2^{2y}} = \dfrac{9^x}{4^y}$ – amWhy Jul 3 '13 at 3:41
Nice Amy. I am glad I could catch you here before you leave for bed. – Babak S. Jul 3 '13 at 4:09

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