# rational exponent of a fraction

• Jan 23rd 2009, 09:17 PM
newgirl
rational exponent of a fraction
Hello everyone. I've been browsing through the forums a bit, and it looks like this is probably going to be one of the easier questions on here. I haven't taken a math class in over 10 yrs and can't remember anything (Doh).. could someone help me with this problem? I've searched all over online, but can't find an example exactly like this one. I know the answer is supposed to be 16/9, but how do you get there?

$\displaystyle {\frac{-27}{64}}^{\frac{-2}{3}}$ (I should also add that there are parenthesis around $\displaystyle {\frac{-27}{64}}$)

(Bow)
• Jan 23rd 2009, 10:23 PM
o_O
A few properties we're going to use:
• $\displaystyle a^{-m} = \frac{1}{a^m}$
• $\displaystyle a^{\frac{m}{n}} = \sqrt[m]{a^n} = \left(\sqrt[m] {a}\right)^n$
• $\displaystyle \sqrt[n]{\frac{a}{b}} = \frac{\sqrt[n]{a}}{\sqrt[n]{b}}$

So we have:
$\displaystyle \left(\frac{-27}{64}\right)^{-\frac{2}{3}} = \frac{1}{ \left( \displaystyle \frac{-27}{64}\right)^{\frac{2}{3}} } = \frac{1}{\displaystyle \left(\sqrt[3]{\frac{-27}{64}}\right)^2} = \frac{1}{\displaystyle \frac{\left(\sqrt[3]{-27}\right)^2}{\left(\sqrt[3]{64}\right)^2}} = \frac{\left(\sqrt[3]{64}\right)^2}{\left(\sqrt[3]{-27}\right)^2} = \cdots$
• Jan 23rd 2009, 10:48 PM
newgirl
Thank you so much! That explains a lot. i wasn't sure what to do with the fraction, but now i get you have to invert the numerator and denominator

(Clapping)