Answer
$\log_b \dfrac{s^3}{t}$
Work Step by Step
Using the properties of logarithms, the given expression, $
3\log_bs-\log_bt
,$ is equivalent to
\begin{align*}\require{cancel}
&
\log_b s^3-\log_bt
&(\text{use }\log_b x^y=y\log_b x)
\\\\&=
\log_b \dfrac{s^3}{t}
&(\text{use }\log_b \dfrac{x}{y}=\log_b x-\log_b y)
.\end{align*}
Hence, the expression $
3\log_bs-\log_bt
$ is equivalent to $
\log_b \dfrac{s^3}{t}
$.
