# Thread: Another problem with fractional indices

1. ## Another problem with fractional indices

I feel so stupid for having top ask for help again, but I have been trying to figure this out for several hours now

$\displaystyle 2x^{\frac{1}{3}}={\frac{81}{8}}^{-1}$

I multiplied both sides by two to eliminate the fraction

$\displaystyle 4x^{\frac{1}{3}}=27x^{-1}$

Then I multiplied both sides by $\displaystyle x^{\frac{1}{3}}$

Result

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

And I'm stuck.

Would you say I have any hope of getting into Advanced Mathematics, if I can't even solve a problem like this?

Thank you!

2. Originally Posted by Coach
I feel so stupid for having top ask for help again, but I have been trying to figure this out for several hours now

$\displaystyle 2x^{\frac{1}{3}}={\frac{81}{8}}^{-1}$
I looked at the raw Latex and it seems that the entire fraction on the right-hand side is supposed to be to the power of $\displaystyle ^{-1}$

$\displaystyle 2x^{\frac{1}{3}}=( {\frac{81}{8}} ) ^{-1}$

$\displaystyle 2x^{\frac{1}{3}}= \frac{8}{81}$

$\displaystyle x^{\frac{1}{3}}= \frac{4}{81}$

$\displaystyle x= ( {\frac{4}{81}} ) ^\frac{3}{1}$

$\displaystyle x = \frac{64}{531441}$

3. Thank you so much, but now I feel even more stupid, because I made a typo in the original question
the correct one is supposed to be

$\displaystyle 2x^{\frac{1}{3}}={\frac{81}{8}}x^{-1}$

4. Originally Posted by Coach
Thank you so much, but now I feel even more stupid, because I made a typo in the original question
the correct one is supposed to be
$\displaystyle$
$\displaystyle 2x^{\frac{1}{3}}={\frac{81}{8}}x^{-1}$
$\displaystyle 2x^{\frac{1}{3}}={\frac{81}{8}}x^{-1}$

$\displaystyle x^{\frac{1}{3}}={\frac{81}{16}}x^{-1}$

$\displaystyle ( x^{\frac{1}{3}} )( x ) = {\frac{81}{16}}$

$\displaystyle ( x^{\frac{4}{3}} ) = {\frac{81}{16}}$

$\displaystyle ( x ) = ( {\frac{81}{16}} )^{\frac{3}{4}}$

$\displaystyle ( x ) = ( {\frac{27}{8}} )$

5. Thank you so much!

6. Originally Posted by Coach
Thank you so much!
You're welcome and I look forward to helping you again