What is the 4th root of 80?

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What is the 4th root of 80?

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Answer 1 · 2015-09-19T22:47:03

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

Explanation:

If $a, b, c > 0$ $a, b, c > 0$ then ${(a b)}^{c} = a^{c} b^{c}$ $(ab)^c = a^cb^c$

$\sqrt[4]{x} = x^{\frac{1}{4}}$ $root(4)(x) = x^(1/4)$ , so $\sqrt[4]{a b} = \sqrt[4]{a} \sqrt[4]{b}$ $root(4)(ab) = root(4)(a)root(4)(b)$

If $a, b, c > 0$ $a, b, c > 0$ then ${(a^{b})}^{c} = a^{b c}$ $(a^b)^c = a^(bc)$

So: $\sqrt[4]{a^{4}} = {(a^{4})}^{\frac{1}{4}} = a^{4 \cdot \frac{1}{4}} = a^{1} = a$ $root(4)(a^4) = (a^4)^(1/4) = a^(4*1/4) = a^1 = a$

So:

$\sqrt[4]{80} = \sqrt[4]{2^{4} \cdot 5} = \sqrt[4]{2^{4}} \sqrt[4]{5} = 2 \sqrt[4]{5}$ $root(4)(80) = root(4)(2^4*5) = root(4)(2^4)root(4)(5) = 2root(4)(5)$

Answer 2 · 2015-09-19T22:53:25

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

Explanation:

The expression $\sqrt[4]{80}$ $root(4)80$ can be simplified if it is possible to factor out the fourth power of some integer from $80$ $80$ .

Through prime factorization, we find that $\sqrt[4]{80} = \sqrt[4]{2 \cdot 2 \cdot 2 \cdot 2 \cdot 5} = \sqrt[4]{2^{4} \cdot 5} = \sqrt[4]{2^{4}} \sqrt[4]{5} = 2 \sqrt[4]{5}$ $root(4)80 = root(4)(2*2*2*2*5) = root(4)(2^4*5) = root(4)(2^4)root(4)(5) = 2root(4)5$

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What is the 4th root of 80?

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