# A problem with fractional indices

• Sep 9th 2007, 10:18 AM
Coach
A problem with fractional indices
I have tried everything I know, but I can't solve this. Can you please show me the right direction?

$\displaystyle \frac{16^\frac{1}{3}*4^\frac{1}{3}}{8}$

Thank you!
• Sep 9th 2007, 10:27 AM
Jhevon
Quote:

Originally Posted by Coach
I have tried everything I know, but I can't solve this. Can you please show me the right direction?

$\displaystyle \frac{16^\frac{1}{3}*4^\frac{1}{3}}{8}$

change everything to base 2

$\displaystyle \frac {16^{\frac {1}{3}} \cdot 4^{\frac {1}{3}}}{8} = \frac { \left( 2^{4} \right)^{\frac {1}{3}} \cdot \left( 2^{2} \right)^{\frac {1}{3}}}{2^3}$

Now continue
• Sep 9th 2007, 10:46 AM
Coach
Thank you so much!

But I would still have another question

$\displaystyle \frac{x^{2n+1}*x^\frac{1}{2}}{\sqrt{x^{3n}}}$

Thank you!
• Sep 9th 2007, 11:03 AM
Jhevon
Quote:

Originally Posted by Coach
Thank you so much!

But I would still have another question

$\displaystyle \frac{x^{2n+1}*x^\frac{1}{2}}{\sqrt{x^{3n}}}$

change the square root in the denominator to a power (you know how to do that, right? see my post here)

so we get $\displaystyle \frac{x^{2n+1}*x^\frac{1}{2}}{\sqrt{x^{3n}}} = \frac{x^{2n+1}*x^\frac{1}{2}}{x^\frac {3n}{2}}$

Now continue
• Sep 9th 2007, 11:24 AM
Coach
ok, so is it like

$\displaystyle x^{2n+1+\frac{1}{2}}-^{\frac{3n}{2}}$ ?
• Sep 9th 2007, 12:07 PM
Jhevon
Quote:

Originally Posted by Coach
ok, so is it like

$\displaystyle x^{2n+1+\frac{1}{2}}-^{\frac{3n}{2}}$ ?

yes