Math help

1. Feb 12, 2005

footprints

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

2. Feb 12, 2005

VietDao29

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

3. Feb 12, 2005

footprints

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

4. Feb 12, 2005

Nylex

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).

5. Feb 12, 2005

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.

6. Feb 12, 2005