confused... fractions and indicies

**coyoteflare** · January 8th 2009, 09:36 AM

Simplify

$16^1/2/81^3/4$

i got the answer again but im not sure about the bottom half since i just cube rooted the 81 and wasnt sure what to do with the 4 part of it.... dont know if this is coincidently the right answer that i got...

please explain the method, thanks!

sorry i cant get the hang of the [tex] thing , its suppose to be 16^half all over (divided by) 81^three quarters

**shinhidora** · January 8th 2009, 09:42 AM

Originally Posted by coyoteflare

Simplify

$16^1/2/81^3/4$

i got the answer again but im not sure about the bottom half since i just cube rooted the 81 and wasnt sure what to do with the 4 part of it.... dont know if this is coincidently the right answer that i got...

please explain the method, thanks!

I don't get it... you mean...

$16$ divided by $2$ divided by $18^{3}$ divided by $4$ ?

Edit: Ow like this?

$\sqrt{16} = 16^{\frac{1}{2}}$

$16^{\frac{1}{2}} = 4$

$\sqrt[4]{81^{3}} = 81^{\frac{3}{4}}$

$81^{\frac{3}{4}} = 27$

$\frac{16^{\frac{1}{2}}}{81^{\frac{3}{4}}}= \frac{4}{27}$

**coyoteflare** · January 8th 2009, 09:46 AM

$\frac{16^{1/2}}{81^{3/4}}$

16 to the power of a half divided by 81 to the power of three quarters

i cant get the Latex to do it tooconfusing for me

**shinhidora** · January 8th 2009, 09:51 AM

Originally Posted by coyoteflare

[tex]{\frac{16^{\frac{1}{2}}}{81^{\frac{3}{4}}}

16 to the power of a half divided by 81 to the power of three quarters

i cant get the Latex to do it tooconfusing for me

It isn't that hard

In short, for your problem

\frac{a}{b} = $\frac{a}{b}$
a^{b} = $a^{b}$

By the way, I edited my original post with the answers and stuff

**masters** · January 8th 2009, 09:54 AM

Originally Posted by shinhidora

I don't get it... you mean...

$16$ divided by $2$ divided by $18^{3}$ divided by $4$ ?

Edit: Ow like this?

$\frac{16^{\frac{1}{2}}}{81^{\frac{3}{4}}}$