# Thread: Solving the unknown base of exponents

1. ## Solving the unknown base of exponents

So i realised that exponents are the same thing as indices which I put down to the different countries that we all come from.

Anyway - I have this question which i didn't really know how to approach:

solve for 'x':

x^3/2 = 8^1/2 * 9^1/3 / 6^4/3

I did some working and got:
x^3/2 = 2^3/4 * 3^-1/4

I'm not sure if my working was right, but even so, I don't know how to proceed with this question.

2. Hello, Ogsta101!

Solve for $x\!:\;\;x^{\frac{3}{2}} \;=\;\frac{8^{\frac{1}{2}}\cdot9^{\frac{1}{3}}}{6^ {\frac{4}{3}}}$
Note that all the bases have factors of 2 or 3.

We have: . $x^{\frac{3}{2}} \;=\;\frac{\left(2^3\right)^{\frac{1}{2}}\cdot\lef t(3^2\right)^{\frac{1}{3}}} {(2\cdot3)^{\frac{4}{3}}} \;=\;\frac{2^{\frac{3}{2}}\cdot3^{\frac{2}{3}}}{2^ {\frac{4}{3}}\cdot3^{\frac{4}{3}}}\quad\Rightarrow \quad x^{\frac{3}{2}} \;=\;\frac{2^{\frac{1}{6}}}{3^{\frac{2}{3}}}$

Raise both sides to the power $\frac{2}{3}\!:\;\;\left(x^{\frac{3}{2}}\right)^{\f rac{2}{3}} \;=\;\left(\frac{2^{\frac{1}{6}}}{3^{\frac{2}{3}}} \right)^{\frac{2}{3}}$

Therefore: . $\boxed{x\;=\;\frac{2^{\frac{1}{9}}}{3^{\frac{4}{9} }} }$