Solve, please, ${\left(\frac{1}{2}\right)}^{x}={16}^{2}$.
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Terry Ray
We have $\frac{1}{2}={2}^{-1}$ and $16={2}^{4}$. Thus, the equation should be written as ${\left({2}^{-1}\right)}^{x}={\left({2}^{4}\right)}^{2}$
${2}^{-x}={2}^{8}$, what implies that -x=8, so x=-8, and we have:
${\left(\frac{1}{2}\right)}^{-8}={2}^{8}=256={16}^{2}$
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sirpsta3u
Here you are:
${\left(\frac{1}{2}\right)}^{-8}={2}^{8}=256={16}^{2}$