Answer
$\dfrac{1}{2}\log_6 p +\dfrac{1}{2}\log_6 q-\dfrac{1}{2}\log_6 7$
Work Step by Step
Using the properties of logarithms, the given expression, $
\log_6\sqrt{\dfrac{pq}{7}}
$, is equivalent to
\begin{align*}
&
\log_6\left(\dfrac{pq}{7}\right)^{1/2}
\\\\&=
\dfrac{1}{2}\log_6 \dfrac{pq}{7}
&(\text{use }\log_b x^y=y\log_b x)
\\\\&=
\dfrac{1}{2}\left(\log_6 (pq)-\log_6 7\right)
&(\text{use }\log_b \dfrac{x}{y}=\log_b x-\log_b y)
\\\\&=
\dfrac{1}{2}\left(\log_6 p +\log_6 q-\log_6 7\right)
&(\text{use }\log_b (xy)=\log_b x+\log_b y)
\\\\&=
\dfrac{1}{2}(\log_6 p) +\dfrac{1}{2}(\log_6 q)-\dfrac{1}{2}(\log_6 7)
&(\text{use Distributive Property}
\\\\&=
\dfrac{1}{2}\log_6 p +\dfrac{1}{2}\log_6 q-\dfrac{1}{2}\log_6 7
.\end{align*}
Hence, the expression $
\log_6\sqrt{\dfrac{pq}{7}}
$ is equivalent to $
\dfrac{1}{2}\log_6 p +\dfrac{1}{2}\log_6 q-\dfrac{1}{2}\log_6 7
$.
