Answer
$\log_b \dfrac{r^{1/4}s^2}{t^{2/3}}$
Work Step by Step
Using the properties of logarithms, the given expression, $
\dfrac{1}{4}\log_b r+2\log_b s-\dfrac{2}{3}\log_b t
,$ is equivalent to
\begin{align*}\require{cancel}
&
\log_b r^{1/4}+\log_b s^2-\log_b t^{2/3}
&(\text{use }\log_b x^y=y\log_b x)
\\&=
\log_b r^{1/4}s^2-\log_b t^{2/3}
&(\text{use }\log_b (xy)=\log_b x+\log_b y)
\\\\&=
\log_b \dfrac{r^{1/4}s^2}{t^{2/3}}
&(\text{use }\log_b \dfrac{x}{y}=\log_b x-\log_b y)
.\end{align*}
Hence, the expression $
\dfrac{1}{4}\log_b r+2\log_b s-\dfrac{2}{3}\log_b t
$ is equivalent to $
\log_b \dfrac{r^{1/4}s^2}{t^{2/3}}
$.
