# Negative fraction exponents

• January 7th 2011, 12:58 PM
BIOS
Negative fraction exponents
Just wondering if this is the case:

$\displaystyle 3^\frac{m}{n}=\frac{m}{3^n}$

The exponent in the first term should be negative but i don't know how to do that in latex sorry!

Thanks

BIOS
• January 7th 2011, 01:00 PM
snowtea
$3^{-\frac{m}{n}} = \frac{1}{3^{\frac{m}{n}}}$

click on the formula to see the latex.
• January 7th 2011, 01:11 PM
Quote:

Originally Posted by BIOS
Just wondering if this is the case:

$\displaystyle 3^\frac{m}{n}=\frac{m}{3^n}$

The exponent in the first term should be negative but i don't know how to do that in latex sorry!

Thanks

BIOS

$\displaystyle\frac{m}{3^n}=m\left(\frac{1}{3^n}\ri ght)=m\left(\frac{3^0}{3^n}\right)=m\left(3^{0-n}\right)=m3^{-n}$
• January 7th 2011, 02:02 PM
BIOS
Thanks alot for the replies and the clarification.

So is this the case:

$3^{-\frac{m}{n}} = \frac{1}{3^{\frac{m}{n}}}=\frac{1}{\sqrt [n]{3^m}}$
• January 7th 2011, 02:03 PM
snowtea
Quote:

Originally Posted by BIOS
Thanks alot for the replies and the clarification.

So is this the case:

$3^{-\frac{m}{n}} = \frac{1}{3^{\frac{m}{n}}}=\frac{1}{\sqrt [n]{3^m}}$

Yup.
• January 7th 2011, 02:07 PM
BIOS
Sweet. Thanks again for the help. I have most of the rules down now. Just need to work on my logarithms!