help with the powers on my first post

**gillyman** · December 30th 2009, 10:44 AM

Reduce to the lowest power?

$/sqrt^6/frac{3}-{8}/frac/frac{1}-{4}$

**bigwave** · December 30th 2009, 10:50 AM

Originally Posted by gillyman

Reduce to the lowest power?

$/sqrt^6/frac{3}-{8}/frac/frac{1}-{4}$

could you rewrite this using latex

**gillyman** · December 30th 2009, 11:31 AM

Hope this is correct, its also my first time using LATEX
$(/sqrt)^{6}-/frac{3}{8}/frac{1}{4}$

**bigwave** · December 30th 2009, 11:53 AM

Originally Posted by gillyman

Hope this is correct, its also my first time using LATEX
$(/sqrt)^{6}-/frac{3}{8}/frac{1}{4}$

$\sqrt{6}-\frac{\frac{3}{8}}{\frac{1}{4}}$

or

$\frac{\sqrt{6}}{\frac{3}{8}} - \frac{8}{\frac{1}{4}}$

sorry i know this is a pain but...

**gillyman** · December 31st 2009, 03:30 AM

$\frac{(\sqrt)^6\frac{3}{8}}{\frac{1}{4}}$

I hope this is correct. I do need a crash course in LATEK.

**HallsofIvy** · December 31st 2009, 04:50 AM

Do you mean $\frac{\sqrt[6]{\frac{3}{8}}}{\frac{1}{4}}$ ?

That is $4(3^{1/6})2^{1/2})= 2^{5/2}3^{1/6}$ .

Thread: help with the powers on my first post