Question

I have done this problem many times and it is still wrong. I know where it's wrong, at least, but do not know the answer. I look here for some different guidance from what I have been getting, hoping it will help.

( x^(2/3) / 4y(-2) ) ^(1/2)

x^(2/3)(1/2) = x^(2/6) = x^(1/3)

4^(1/2) = 2

And this is allegedly where I am making a mistake:

y^-2 * 1/2 = y^(-2/2) = y^-1 = 1/y

So for my final answer I have

x^(1/3)
-------
2/y

But this is wrong, I guess. Any help is appreciated!

Answer 1

Is it 4(y)^-2 or (4y)^-2

Answer 2

Neither. There were no brackets or grouping symbols in the denominator. It was (is) literally:

(4y^-2)^1/2

Answer 3

Should I ask this question in a different section? It is pre-algebra as I understand it. Should it be in an even lower section?

Answer 4

You are not really wrong, but you have not simplified your result.

you got
\[ \frac{x^\frac{1}{3}}{\frac{2}{y}}\]

fractions with fractions as their top or bottom are ugly.
if you multiply top and bottom by y/2 you get
\[ \frac{x^\frac{1}{3} \cdot \frac{y}{2}}{\frac{2}{y}\cdot \frac{y}{2}}
= x^\frac{1}{3} \cdot \frac{y}{2}
=\frac{x^\frac{1}{3}y}{2}
\]