Answer
$\frac{1}{2}log_{b}7+\frac{1}{2}log_{b}x$
Work Step by Step
We know that $log_{b}xy=log_{b}x+log_{b}y$ (where $x$, $y$, and $b$ are positive real numbers and $b\ne1$).
Therefore, $log_{b}\sqrt 7x=log_{b}(7^{\frac{1}{2}}x^{\frac{1}{2}})=log_{b}7^{\frac{1}{2}}+log_{b}x^{\frac{1}{2}}$.
We know that $log_{b}x^{r}=rlog_{b}x$ (where $x$ and $b$ are positive real numbers, $b\ne1$, and $r$ is a real number).
Therefore, $log_{b}7^{\frac{1}{2}}+log_{b}x^{\frac{1}{2}}=\frac{1}{2}log_{b}7+\frac{1}{2}log_{b}x$.