# Algebraic Manipulation question - trying to get alternate form

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!

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 $$\frac{3^{1+x}-2^{1+x}}{3^x-2^x}=\cfrac1{\left(\frac32\right)^x-1}+3.$$
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 $$3+\frac{2^x}{3^x-2^x}=3+\frac{1}{(3/2)^x-1}.$$
• 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 $2^x.$ – coffeemath Nov 28 '13 at 4:24