# How do you solve for x in -5 = log _2 x?

$x = \frac{1}{32}$
We start with $- 5 = {\log}_{2} x$. Now, we can rewrite a logarithm as an exponentiation equation, like this: color(red)(y)=color(blue)(b)^(color(green)(x) becomes ${\log}_{\textcolor{b l u e}{b}} \textcolor{red}{y} = \textcolor{g r e e n}{x}$ and vice versa. So, ${\log}_{\textcolor{b l u e}{2}} \textcolor{red}{x} = \textcolor{g r e e n}{- 5}$ becomes color(red)(x)=color(blue)(2)^(color(green)(-5).
And if we solve ${2}^{-} 5$ we get $x = \frac{1}{32}$ or $.03125$.