Answer
$\log_b \dfrac{3x}{y^2}$
Work Step by Step
Using the properties of logarithms, the given expression, $
\log_b3+\log_bx-2\log_by
$, is equivalent to
\begin{align*}\require{cancel}
&
\log_b3+\log_bx-\log_by^2
&(\text{use }\log_b x^y=y\log_b x)
\\\\&=
\log_b3x-\log_by^2
&(\text{use }\log_b (xy)=\log_b x+\log_b y)
\\\\&=
\log_b \dfrac{3x}{y^2}
&(\text{use }\log_b \dfrac{x}{y}=\log_b x-\log_b y)
.\end{align*}
Hence, the expression $
\log_b3+\log_bx-2\log_by
$ is equivalent to $
\log_b \dfrac{3x}{y^2}
$.
