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

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

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Answer 1 · 2015-10-30T04:18:24

Use your knowledge of indices

Ans: $8^{\frac{5}{3}}$ $8^(5/3)$ or $\sqrt[3]{8^{5}}$ $root(3)(8^5)$ or $32$ $32$

Explanation:

$8^{\frac{2}{3}}$ $8^(2/3)$ can be written as $\sqrt[3]{8^{2}}$ $root(3)(8^2)$ .

If you know that $8^{2}$ $8^2$ is $64$ $64$ , and that cube root of $64$ $64$ ( $\sqrt[3]{64})$ $root(3)64)$ is $4$ $4$ , then this it is easy to solve the problem in this way.

You could also write $8^{\frac{2}{3}}$ $8^(2/3)$ as $\sqrt[3]{8 \cdot 8}$ $root(3)(8*8)$ .

Take the cube root of each $8$ $8$ separately and multiply them

$8^{\frac{2}{3}} \cdot 8$ $8^(2/3)*8$

( $\sqrt[3]{8} \cdot \sqrt[3]{8}$ $root(3)8*root(3)8$ )*8

$(2 \cdot 2) \cdot 8$ $(2*2)*8$

$4 \cdot 8$ $4*8$

$32$ $32$

Another way to solve this:

When two numbers are being multiplied ( $8^{\frac{2}{3}}$ $8^(2/3)$ and $8$ $8$ ) and their bases ( $8$ $8$ ) are the same, the product can simply be written as the base ( $8$ $8$ ) to the power of the addition of the powers ( $\frac{2}{3} + 1$ $2/3+1$ )

$8^{\frac{2}{3}} \cdot 8^{1}$ $8^(2/3)*8^1$

$8^{\frac{2}{3} + 1}$ $8^(2/3+1)$

$8^{\frac{5}{3}}$ $8^(5/3)$

$8^{\frac{5}{3}}$ $8^(5/3)$ can be written as $\sqrt[3]{8^{5}}$ $root(3)(8^5)$

$\sqrt[3]{8^{5}}$ $root(3)(8^5)$

$\sqrt[3]{8 \cdot 8 \cdot 8 \cdot 8 \cdot 8}$ $root(3)(8*8*8*8*8)$

$2 \cdot 2 \cdot 2 \cdot 2 \cdot 2$ $2*2*2*2*2$

$32$ $32$

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

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