Answer
$\log_b \left(m^2\sqrt{n} \right)$
Work Step by Step
Using the properties of logarithms, the given expression, $
2\log_b m+\dfrac{1}{2}\log_b n
,$ is equivalent to
\begin{array}{l}\require{cancel}
\log_b m^2+\log_b n^{1/2}
\\\\=
\log_b m^2+\log_b \sqrt{n}
\\\\=
\log_b \left(m^2\sqrt{n} \right)
.\end{array}