# What is the 4th root of 80?

Sep 19, 2015

$\sqrt[4]{80} = 2 \sqrt[4]{5}$

#### Explanation:

If $a , b , c > 0$ then ${\left(a b\right)}^{c} = {a}^{c} {b}^{c}$

$\sqrt[4]{x} = {x}^{\frac{1}{4}}$, so $\sqrt[4]{a b} = \sqrt[4]{a} \sqrt[4]{b}$

If $a , b , c > 0$ then ${\left({a}^{b}\right)}^{c} = {a}^{b c}$

So: $\sqrt[4]{{a}^{4}} = {\left({a}^{4}\right)}^{\frac{1}{4}} = {a}^{4 \cdot \frac{1}{4}} = {a}^{1} = a$

So:

$\sqrt[4]{80} = \sqrt[4]{{2}^{4} \cdot 5} = \sqrt[4]{{2}^{4}} \sqrt[4]{5} = 2 \sqrt[4]{5}$

Sep 19, 2015

$\sqrt[4]{80} = 2 \sqrt[4]{5}$
The expression $\sqrt[4]{80}$ can be simplified if it is possible to factor out the fourth power of some integer from $80$.
Through prime factorization, we find that $\sqrt[4]{80} = \sqrt[4]{2 \cdot 2 \cdot 2 \cdot 2 \cdot 5} = \sqrt[4]{{2}^{4} \cdot 5} = \sqrt[4]{{2}^{4}} \sqrt[4]{5} = 2 \sqrt[4]{5}$