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

[gillgill]

how do u solve these kinds of problems?

1/3=x^(2/3)

 

[Gale]

how do u solve these kinds of problems?

1/3=x^(2/3)

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

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

 

[gillgill]

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

 

[dextercioby]

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

Daniel.

 

[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

 

[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

 

[whozum]

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

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