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