How do you solve -5 = log_2 x?

Oct 12, 2015

$x = \frac{1}{32}$

Explanation:

${\log}_{2} x = - 5$

Using the rule ${\log}_{a} M = N \Rightarrow M = {a}^{N}$, you can rewrite ${\log}_{2} x = - 5$ as $x = {2}^{-} 5$

Using the indices rule ${a}^{-} m = \frac{1}{a} ^ m$,

$x = \frac{1}{2} ^ 5$

$x = \frac{1}{32}$