30. $\left\{\begin{array}{l}\log _{a} x+\log _{a} y=2 \\ \log _{b} x-\log _{b} y=4\end{array}\right.$
30. Potentiating, we get $x y=a^{2}, \frac{x}{y}=b^{4}, x= \pm a b^{2}, y= \pm \frac{a}{b^{2}}$. By the condition $x>0, y>0, a>0, b>0$. Answer: $x=a b^{2}, y=\frac{a}{b^{2}}$.