# Thread: How are these two equivalent?

1. ## How are these two equivalent?

Show that $\displaystyle \frac{2^\frac{2}{3}}{4}=\sqrt[3]{\frac{1}{16}}$.

Any help showing detailed working out would be appreciated!

2. Work with this.
$\displaystyle \sqrt[3]{{\frac{1} {{16}}}} = 2^{-\frac{{ 4}} {3}}$

3. ## Start by writing everything as a power of 2

$\displaystyle \frac{2^\frac{2}{3}}{4}=\sqrt[3]{\frac{1}{16}}$

$\displaystyle \displaystyle\frac{2^\frac{2}{3}}{4}$

$\displaystyle =\displaystyle\frac{2^\frac{2}{3}}{2^2}$

$\displaystyle =\displaystyle(2^\frac{2}{3})(2^{-2})$

Using the law of indices, $\displaystyle (x^a)(x^b)=(x^{a+b})$

Apply that here, then tidy up to give you the answer you seek.

Edit: Unfortunately, I am far slower than Plato, as usual. =.=

4. Originally Posted by Plato
Work with this.
$\displaystyle 2^{-\frac{{ 4}} {3}}$
Wait a minute. How did you get $\displaystyle 2^{-\frac{{ 4}} {3}}$?

5. Originally Posted by Joker37
Wait a minute. How did you get $\displaystyle 2^{-\frac{{ 4}} {3}}$?
I understand the way exponents work.