Victor Wall

2021-12-12

Solve, please, $\left(\frac{1}{2}\right)}^{x}={16}^{2$ .

Terry Ray

Beginner2021-12-13Added 50 answers

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:

$\left(\frac{1}{2}\right)}^{-8}={2}^{8}=256={16}^{2$

sirpsta3u

Beginner2021-12-14Added 42 answers

Here you are:

$\left(\frac{1}{2}\right)}^{-8}={2}^{8}=256={16}^{2$

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