Answer
$\frac{1}{2}log_{b}7+\frac{1}{2}log_{b}x$
Work Step by Step
The product property of logarithms tells us that $log_{b}xy=log_{b}x+log_{b}y$ (where x, y, and, b are positive real numbers and $b\ne1$).
Therefore, $log_{b}\sqrt(7x)= log_{b}7^{\frac{1}{2}}x^{\frac{1}{2}}= log_{b}7^{\frac{1}{2}}+log_{b}x^{\frac{1}{2}}$.
The power property of logarithms tells us that $log_{b}x^{r}=r log_{b}x$ (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$.
