## Simplification of complex fractions.

Quadratic equations and inequalities, variation equations, function notation, systems of equations, etc.
Ian
Posts: 4
Joined: Tue Feb 25, 2014 8:02 am
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### Simplification of complex fractions.

This is my attempt at simplification below:

$\frac{(8^x)(8^{3x})}{(2^{3x})(4^{x+2})}$
= $\frac{8^{4x}}{(2^{3x})(2^{2x})(2^4)}$
= $\frac{2^{12x}}{(2^{3x})(2^{2x})(2^4)}$
= $(2^{12x})(2^{-3x})(2^{-2x})(2^{-4})$
= $(2^{12x-3x-2x})(2^{-4})$
= $(2^{7x})(\frac{1}{16})$

The revision handbook gives the following as options:
1. $(2^{7x})(16)$
2. $\frac{2^{7x}}{16}$
3. $2^{17x+4}$
4. $\frac{64^{4x}}{8^{4x+2}}$
5. none of the above.

I'm second-guessing my calculation. Please could someone review my method?

buddy
Posts: 197
Joined: Sun Feb 22, 2009 10:05 pm
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### Re: Simplification of complex fractions.

Ian wrote:This is my attempt at simplification below:

$\frac{(8^x)(8^{3x})}{(2^{3x})(4^{x+2})}$
= $\frac{8^{4x}}{(2^{3x})(2^{2x})(2^4)}$
= $\frac{2^{12x}}{(2^{3x})(2^{2x})(2^4)}$
= $(2^{12x})(2^{-3x})(2^{-2x})(2^{-4})$
= $(2^{12x-3x-2x})(2^{-4})$
= $(2^{7x})(\frac{1}{16})$

The revision handbook gives the following as options:
1. $(2^{7x})(16)$
2. $\frac{2^{7x}}{16}$
3. $2^{17x+4}$
4. $\frac{64^{4x}}{8^{4x+2}}$
5. none of the above.

I'm second-guessing my calculation. Please could someone review my method?

yrs is the same as #2

Ian
Posts: 4
Joined: Tue Feb 25, 2014 8:02 am
Contact:

### Re: Simplification of complex fractions.

Voilà! I completely (ashamedly) missed that. Thanks.