How are these two equivalent?

**Joker37** · November 24th 2010, 01:12 PM

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

Any help showing detailed working out would be appreciated!

**Plato** · November 24th 2010, 01:17 PM

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

**Quacky** · November 24th 2010, 01:20 PM

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

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

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

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

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

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

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

**Joker37** · November 24th 2010, 01:27 PM

Originally Posted by Plato

Work with this.
$2^{-\frac{{ 4}} {3}}$

Wait a minute. How did you get $2^{-\frac{{ 4}} {3}}$ ?

**Plato** · November 24th 2010, 01:28 PM

Originally Posted by Joker37

Wait a minute. How did you get $2^{-\frac{{ 4}} {3}}$ ?

I understand the way exponents work.

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