Exponents, check.

mi986 · #1 $Old$ October 27th, 2007, 05:59 AM

Hey, could someone please check if I simplified this problem correctly?

$\frac {(2^{(a+b)}x^{2b})^{(a-b)}}{(\frac {2^ax}{x^{-b}})^a*(\frac {x^{(a-b)}}{2^b})^b}$

First I got rid of the exponents by multiplying.

= $\frac {2^{(a^2-b^2)}x^{(2ab-2b^2)}}{(\frac {x^{(ab-b^2)}}{2^{b^2}})*(\frac {2^{a^2}x^a}{x^{-ab}})}$

Then I simplified and combined the bottom fractions.

= $\frac {2^{(a^2-b^2)}x^{(2ab-2b^2)}}{x^{(2ab-b^2+a)}2^{(a^2-b^2)}}$

And then repeated that step and got this answer.

$= \frac {1}{x^{b^2+a}}$

Thanks in advance.

Mi986

qyzren · #2 $Old$ October 27th, 2007, 06:18 AM

yes that is correct.

Soroban · #3 $Old$ October 27th, 2007, 10:00 AM

Hello, mi986!

I can't follow your second line . . .

Quote:

$\displaystyle{\frac{\left(2^{a+b}\cdot x^{2b}\right)^{a-b}}{\left(\frac {2^ax}{x^{-b}}\right)^a*(\frac {x^{a-b}}{2^b})^b}}$

We have: . $\frac{2^{a^2-b^2}\cdot x^{2ab -b^2}}{\frac{2^{a^2}\cdot x^a}{x^{-ab}} \cdot \frac{x^{ab-b^2}}{2^{b^2}}} \;= \;\frac{2^{a^2-b^2}\cdot x^{2ab-b^2}}{2^{a^2-b^2}\cdot x^{a + 2ab -b^2}} \;=\;\frac{1}{x^a}$

mi986 · #4 $Old$ October 27th, 2007, 04:18 PM

Did you mean this part?
${x^{2b}}^{(a-b)}$
Don't I multiply the 2b by the a and then to the -b? So it becomes ${x^{2ab-2b^2}}$