Answer
-1.29
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, $log_{b}\frac{4}{32}=log_{b}4-log_{b}32$.
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, $log_{b}4-log_{b}32=log_{b}2^{2}-log_{b}2^{5}=2log_{b}2-5log_{b}2=-3 log_{b}2=-3\times.43=-1.29$.