How do I solve for y in the equation 3y^{\frac{2}{3}}=x?

[footprints]

3y^{\frac{2}{3}}=x
How do I make y the subject?
y^{\frac{2}{3}}=\frac{x}{3}
Then what?

 

[VietDao29]

Hi,
y >= 0
Then
y = \sqrt[\frac{2}{3}]{\frac{x}{3}} = \sqrt{(\frac{x}{3})^{3}}
Hope it help,
Viet Dao,

 

[footprints]

y = \sqrt[\frac{2}{3}]{\frac{x}{3}} = \sqrt{(\frac{x}{3})^{3}}

How did you go from \sqrt[\frac{2}{3}]{\frac{x}{3}} to \sqrt{(\frac{x}{3})^{3}}

 

[Nylex]

How did you go from \sqrt[\frac{2}{3}]{\frac{x}{3}} to \sqrt{(\frac{x}{3})^{3}}

Using the law of indicies that says a^mn = (a^m)^n.

y^(2/3) = x/3

Cube both sides: y^2 = (x/3)^3

Now square root both sides: y = (x/3)^3/2 = [(x/3)^3]^1/2, which is what you have (I can't use LaTeX properly, oops).

 

[dextercioby]

That root in the LHS is another way of writing
(\frac{x}{3})^{\frac{3}{2}}=[(\frac{x}{3})^{3}]^{\frac{1}{2}}=\sqrt{(\frac{x}{3})^{3}}

Daniel.

 

[footprints]

Using the law of indicies that says a^mn = (a^m)^n.

y^(2/3) = x/3

Cube both sides: y^2 = (x/3)^3

Now square root both sides: y = (x/3)^3/2 = [(x/3)^3]^1/2, which is what you have (I can't use LaTeX properly, oops).

Oh ya! Forgot about that.
Thanks for the help guys!