Answer
-.125
Work Step by Step
We are given $log_{b}2=.43$ and $log_{b}3=.68$.
We know that $log_{b}\frac{x}{y}=log_{b}x-log_{b}y$ (where $x$, $y$, and $b$ are positive real numbers and $b\ne1$).
Therefore, $log_{b}\sqrt\frac{2}{3}=log_{b}\frac{2^{\frac{1}{2}}}{3^{\frac{1}{2}}}=log_{b}2^{\frac{1}{2}}-log_{b}3^{\frac{1}{2}}$.
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, $log_{b}2^{\frac{1}{2}}-log_{b}3^{\frac{1}{2}}=\frac{1}{2}log_{b}2-\frac{1}{2}log_{b}3=\frac{1}{2}(log_{b}2-log_{b}3)=\frac{1}{2}(.43-.68)=.5\times-.25=-.125$.
