How do you solve  (1/2)^x =16^2?

Jul 16, 2015

$x = - 8$

Explanation:

You could use logarithms, but since $\frac{1}{2}$ and $16$ are both powers of 2, it's easier if you use that fact.

We have $\frac{1}{2} = {2}^{- 1}$ and $16 = {2}^{4}$. Therefore the equation ${\left(\frac{1}{2}\right)}^{x} = {16}^{2}$ can be written as ${\left({2}^{- 1}\right)}^{x} = {\left({2}^{4}\right)}^{2}$, or ${2}^{- x} = {2}^{8}$. This implies that $- x = 8$ so that $x = - 8$ (after all, ${2}^{z}$ is a one-to-one function of the variable $z$ (just to pick a different letter than $x$)).

You can check the answer like this:

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