Question

If $2^x=16$, find
(i) $x$
(ii) $2^{\frac{x}{2}}$
(iii) $2^{2x}$
(iv) $2^{x+2}$
(v) $\sqrt{2^{-x}}$

Answer 1

$2^x=16$

$=2^4$

$\Rightarrow x=4$

So,

$\left(i\right)$

$2^{\frac{x}{2}}=2^{\frac{4}{2}}$

$=2^2$

$=4$

$\left(ii\right)$

$2^{2x}=2^{2\times 4}$

$=2^8$

$=256$

$\left(iii\right)$

$2^{x+2}=2^{4+2}$

$=2^6$

$=64$

$\left(iv\right)$

$\sqrt{2^{-x}}=\sqrt{2^{-4}}$

$=2^{-4/2}$

$=2^{-2}$

$=\frac{1}{4}$