Answer
$\approx-1.4125$
Work Step by Step
Using $\log_a b=\dfrac{\log b}{\log a}$ or the Change-Of-Base Theorem, the given expression, $
\log_{0.32} 5
,$ evaluates to
\begin{array}{l}\require{cancel}
\dfrac{\log 5}{\log 0.32}
\\\\
\approx-1.4125
.\end{array}