Answer
$-5$
Work Step by Step
Let $y=\log_2{\frac{1}{32}}$.
Use the definition of logarithm $\log_b {x} = y\longleftrightarrow b^{y} = x$, to write an exponential equation.
In this exercise, the base $b$ is $2$, $y$ is the exponent, and $x$ is $\frac{1}{32}$:
$2^{y} = \frac{1}{32}$
With $32=2^5$, the equation above is equivalent to
$2^{y} = \frac{1}{2^5}$
Use the rule $\frac{1}{a^m}=a^{-m}$ to obtain:
$2^{y} = 2^{-5}$
If two numbers having the same base are equal, that means that their exponents are also the same, so set the exponents equal to one another to solve for $y$:
$y = -5$
Therefore, $\log_2{\frac{1}{32}}=-5$.
