28. $\left\{\begin{aligned} \log _{y} x-\log _{x} y & =\frac{8}{3} \\ x y & =16 .\end{aligned}\right.$

2 Order 5039
28. We should switch to logarithms with base $x(x>0, x \neq 1)$. We get

$$
\log _{x}^{2} y+\frac{8}{3} \log _{x} y-1=0 ; \quad \log _{x} y=\frac{1}{3} ; \log _{x} y=-3
$$

Considering these values together with the second equation, we find

$$
x_{1}=8 ; \quad y_{1}=2 ; \quad x_{2}=\frac{1}{4} ; \quad y_{2}=64
$$