# simplifying with exponents

• Sep 15th 2009, 09:05 PM
snypeshow
simplifying with exponents
I just cant seem to get it. i know its simple, but i keep geting lost with the exponents, could anyone help?

http://img3.imageshack.us/img3/8582/259003.gif
• Sep 15th 2009, 09:29 PM
pickslides
Quote:

Originally Posted by snypeshow
I just cant seem to get it. i know its simple, but i keep geting lost with the exponents, could anyone help?

http://img3.imageshack.us/img3/8582/259003.gif

Use $\displaystyle (a^m)^n = a^{mn}$ and $\displaystyle a^m\div a^n = a^{m-n}$

Difference of two squares will also come in handy.
• Sep 16th 2009, 01:26 PM
snypeshow
so for the top part, (a-b) = m and (a+b) would be n, so you would get(a-b)(a+b)/x^b^2

which would give you $\displaystyle x^(a^2 - b^2)$

so would the answer be x^a^2 because there would be 2 x^b^2, both in the divisor and divident, so it would cancel it out?
• Sep 16th 2009, 01:56 PM
pickslides
I think I am following what you are saying, not 100% sure it is correct.

Spoiler:
$\displaystyle \frac{(x^{(a-b)})^{(a+b)}}{x^{b^2}} = \frac{x^{(a-b)(a+b)}}{x^{b^2}}= \frac{x^{(a^2-b^2)}}{x^{b^2}} = x^{(a^2-b^2)-b^2} = x^{a^2-2b^2}$
• Sep 16th 2009, 06:34 PM
snypeshow
$\displaystyle x^{(a^2-b^2)-b^2}$

why the $\displaystyle -b^2$ at the end?
• Sep 16th 2009, 07:07 PM
pickslides
$\displaystyle a^m\div a^n = a^{m-n}$
• Sep 16th 2009, 07:15 PM
snypeshow
oh yeah i completely forgot about that, i though they'd just cancel each other out

thanks alot!!!