How do you simplify $8^{\frac{2}{3}}$ $8^(2/3)$ ?

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How do you simplify $8^{\frac{2}{3}}$ $8^(2/3)$ ?

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Answer 1 · 2016-07-24T20:03:06

$= 4$ $=4$

Explanation:

$8^{\frac{2}{3}}$ $8^(2/3)$ can be written as
$\sqrt[3]{8^{2}}$ $root3(8^2$
$= \sqrt[3]{64}$ $=root3(64)$
$= \sqrt[3]{(4) (4) (4)}$ $=root3((4)(4)(4)$
$= 4$ $=4$

Answer 2 · 2017-06-30T07:07:59

$4$ $4$

Explanation:

One of the laws of indices states: $x^{\frac{m}{n}} = \sqrt[n]{(x^{m})} = {(\sqrt[n]{x})}^{m}$ $x^(m/n) = rootn((x^m)) = (rootn(x))^m$

Therefore $8^{\frac{2}{3}}$ $8^(2/3)$ can be written as $\sqrt[3]{(8^{2})} or {(\sqrt[3]{8})}^{2}$ $root3((8^2)) or (root3(8))^2$

I prefer to use the second form because it uses smaller numbers - the root is found first and then that is squared.

$\sqrt[3]{8} = 2 and 2^{2} = 4$ $root3(8) = 2 and 2^2 =4$

So: ${(\sqrt[3]{8})}^{2} = 2^{2} = 4$ $(color(blue)(root3(8)))^2 = color(blue)(2)^2 = 4$

Consider a question such as $32^{\frac{3}{5}}$ $32^(3/5)$

$\sqrt[5]{(32^{3})}$ $root5(color(blue)((32^3)))$ would mean finding $32^{3}$ $32^3$ first ... ouch!

${(\sqrt[5]{32})}^{3}$ $(color(blue)(root5 (32)))^3$ would mean finding $\sqrt[5]{32}$ $root5(32)$ first. That is $2$ $color(blue)(2)$

${(\sqrt[5]{32})}^{3} = 2^{3} = 8$ $(color(blue)(root5 (32)))^3 = color(blue)(2)^3 = 8$

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How do you simplify $8^{\frac{2}{3}}$ $8^(2/3)$ ?

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