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I'm currently working on algebraic manipulation, changing algebraic fractions into a chosen alternate form but I've hit a brick wall.

I'm trying to get: $$\frac{2(3^x - 2^x)}{3^{x+1} - 2^{x+1}}$$

to the alternate form:

$$\frac{2}{\frac{1}{(3/2)^x-1}+3}$$

Any help will be seriously appreciated!

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Well, to start with, you should be able to see that $$\frac{2(3^x-2^x)}{3^{1+x}-2^{1+x}}=\cfrac2{\frac{3^{1+x}-2^{1+x}}{3^x-2^x}},$$ so the only thing left to do is show that $$\frac{3^{1+x}-2^{1+x}}{3^x-2^x}=\cfrac1{\left(\frac32\right)^x-1}+3.$$

The key here is to use your exponent rules.

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Starting with $$\frac{2(3^x-2^x)}{3^{1+x}-2^{1+x}},$$ divide top and bottom by the parenthesisized part of the top, and split off $3$ and $2$ from the $x+1$ powers, and also rewrite $-2\cdot2^x=-3\cdot 3^3+2^x:$ $$\frac{2}{\frac{3\cdot 3^x-3\cdot 2^x+2^x}{3^x-2^x}}.$$ The denominator is now $$3+\frac{2^x}{3^x-2^x}=3+\frac{1}{(3/2)^x-1}.$$

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  • $\begingroup$ cheers buddy! I feel as if I'm missing a trick here, but where did 3*2^x come from? I just assumed: 3^(1+x)-2^(1+x) = 3(3^x) - 2(2^x) $\endgroup$ – therunningman Nov 28 '13 at 4:12
  • $\begingroup$ The term $3(3^x)-2(2^x)$ was rewritten as $3(3^x)-3(2^x)+2^x$ so that the first two terms would be 3 times the denominator of the thing under the 2 ... that's where the 3 comes from in the denominator. What's left is the $2^x/(3^x-2^x)$, and the last step is to divide top and bottom of this term by $2^x.$ $\endgroup$ – coffeemath Nov 28 '13 at 4:24
  • $\begingroup$ @therunningman In my last response I forgot to tag you, so you may not have noticed it. Let me know if the explanation is clear now. $\endgroup$ – coffeemath Nov 28 '13 at 5:11

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