How do you determine which is greater ${(8^{2})}^{2} (\Leftrightarrow) {(8^{2})}^{3}$ $(8^2)^2 (<=>) (8^2)^3$ ?

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Answer 1 · 2018-03-02T12:05:19

${(8^{2})}^{3}$ $(8^2)^3$

Let $x = {(8^{2})}^{2}$ $x=(8^2)^2$ .

We have:

${(8^{2})}^{3}$ $(8^2)^3$

$= {(8^{2})}^{2} \cdot 8^{2}$ $=(8^2)^2*8^2$

$= x \cdot 8^{2}$ $=x*8^2$

$= 64 x$ $=64x$

Since $64 > 1$ $64>1$ , we have $64 x > x$ $64x>x$ , or ${(8^{2})}^{3} > {(8^{2})}^{2}$ $(8^2)^3>(8^2)^2$ .

How do you determine which is greater (8^2)^2 (<=>) (8^2)^3(82)2(⇔)(82)3(8^2)^2 (<=>) (8^2)^3?