How do u solve these kinds of problems?1/3=x^(2/3)

1. May 29, 2005

gillgill

how do u solve these kinds of problems?

1/3=x^(2/3)

2. May 29, 2005

Gale

its called algebra. you isolate the variable. in this case you would raise both sides to the reciprocal exponent.

$$(\frac {1}{3})^{\frac{3}{2}} = x = \frac{1}{\sqrt{27}}$$

3. May 29, 2005

gillgill

okay..i see...thanks....

4. May 29, 2005

dextercioby

Gale,i think you missed a valid solution...

Daniel.

5. May 30, 2005

Ouabache

Perhaps Gale, you will see what Daniel means, if you break it
down into steps

$$(x^{\frac{2}{3}})^3 = ({\frac{1}{3}})^3$$

$$x^2 = \frac{1}{27}$$

which of course, has two solutions

6. May 30, 2005

Gale

yes, i was aware thanks... but hey, if its not enough that daniel made me feel stupid for forgetting the dumb negative answer, then i feel stupider now that someone actually thought they had to explain it to me... THANKS

7. May 30, 2005

whozum

Oh just in case you didnt quite catch it Gale the answers were

$$\frac{\sqrt{27}}{27}, \frac{-\sqrt{27}}{27}$$