1. ## fractional indices

can anyone help me with this question
simplify

9^-1/2 x 8^2/3 ?

2. Originally Posted by daniel123
9^-1/2 x 8^2/3 ?
$\displaystyle 9^{\frac{-1}{2}}\times 8^{\frac{2}{3}}$

$\displaystyle \frac{1}{9^{\frac{1}{2}}}\times 8^{\frac{2}{3}}$

$\displaystyle \frac{1}{\sqrt{9}}\times (\sqrt[3]{8})^2$

Can you finish it?

3. $\displaystyle 9^{-1/2} \cdot 8^{2/3}$

First, if you have a negative exponent, you can make it positive by putting it in the denominator of a fraction:
$\displaystyle = \frac{1}{9^{1/2}} \cdot 8^{2/3}$

An exponent of 1/2 is the same as square root, so the square root of 9 is 3:
$\displaystyle = \frac{1}{3} \cdot 8^{2/3}$

The 8 to the 2/3 power can be rewritten using the power-of-a-power property:
$\displaystyle = \frac{1}{3} \cdot (8^{1/3})^2$

An exponent of 1/3 is the same as cube root, so the cube root of 8 is 2:
$\displaystyle = \frac{1}{3} \cdot 2^2$

Simplify:
$\displaystyle = \frac{4}{3}$

EDIT: Sorry pickslides!

4. ok thankyou very much. I was going to leave it as square root of 9^-1 x cube root of 8^2