# How do you find the square root of 5 to the 4th power?

It is $\sqrt{{5}^{4}} = {5}^{2} = 25$

Sep 14, 2015

The question is slightly ambiguous, but both ${\left(\sqrt{5}\right)}^{4}$ and $\sqrt{{5}^{4}}$ are equal to $25$

#### Explanation:

${\left({a}^{b}\right)}^{c} = {a}^{b c}$ for $a , b , c \ge 0$

So:
${\left(\sqrt{5}\right)}^{4} = {\left({\left(\sqrt{5}\right)}^{2}\right)}^{2} = {5}^{2} = 25$

Also:
$\sqrt{{5}^{4}} = \sqrt{{\left({5}^{2}\right)}^{2}} = {5}^{2} = 25$

Alternatively using fractional exponents:

$\sqrt{a} = {a}^{\frac{1}{2}}$

So:
${\left(\sqrt{5}\right)}^{4} = {\left({5}^{\frac{1}{2}}\right)}^{4} = {5}^{\frac{1}{2} \cdot 4} = {5}^{2} = 25$

and:
$\sqrt{{5}^{4}} = {\left({5}^{4}\right)}^{\frac{1}{2}} = {5}^{4 \cdot \frac{1}{2}} = {5}^{2} = 25$