# confused... fractions and indicies

• Jan 8th 2009, 09:36 AM
coyoteflare
confused... fractions and indicies
Simplify

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

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...

sorry i cant get the hang of the [tex] thing , its suppose to be 16^half all over (divided by) 81^three quarters
• Jan 8th 2009, 09:42 AM
shinhidora
Quote:

Originally Posted by coyoteflare
Simplify

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

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...

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

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

Edit: Ow like this?

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

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

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

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

$\displaystyle \frac{16^{\frac{1}{2}}}{81^{\frac{3}{4}}}= \frac{4}{27}$
• Jan 8th 2009, 09:46 AM
coyoteflare
$\displaystyle \frac{16^{1/2}}{81^{3/4}}$

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
• Jan 8th 2009, 09:51 AM
shinhidora
Quote:

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

By the way, I edited my original post with the answers and stuff :o
• Jan 8th 2009, 09:54 AM
masters
Quote:

Originally Posted by shinhidora
I don't get it... you mean...

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

Edit: Ow like this?

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

That's what I understood, too. Using radicals it would be

$\displaystyle \frac{\sqrt{16}}{(\sqrt[4]{81})^3}$
• Jan 8th 2009, 09:58 AM
shinhidora
In general it's

$\displaystyle \sqrt[a]{n^{b}} = n^{\frac{b}{a}}$

This will make it more clear I guess ;)
• Jan 8th 2009, 10:02 AM
coyoteflare
without soundign too stupid
how is$\displaystyle \sqrt[4]{81^{3}} = 81^{\frac{3}{4}}$ = 27?

i do $\displaystyle \sqrt[4]{81}$ then cube the answer and it coems out massive!?
• Jan 8th 2009, 10:07 AM
masters
Quote:

Originally Posted by coyoteflare
how is$\displaystyle \sqrt[4]{81^{3}} = 81^{\frac{3}{4}}$ = 27?

i do $\displaystyle \sqrt[4]{81}$ then cube the answer and it coems out massive!?

Hello coyote,

The 4th root of 81 is 3.

3 cubed = 27. Not too massive.