4・241 Given $2 \lg (x-2 y)=\lg x+\lg y$, try to find $x: y$.

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[Solution] The given equation can be transformed into $(x-2 y)^{2}=x y$, that is,
$$
\begin{array}{ll}
\text { i.e., } & x^{2}-5 x y+4 y^{2}=0, \\
\text { or } & (x-y)(x-4 y)=0 .
\end{array}
$$

From $x-y=0$, we get $\frac{x}{y}=1$. Since $x$ and $y$ should both be positive, in this case $x-2 y<0$, which contradicts the given equation. From $x-4 y=0$, we get $\frac{x}{y}=4$, so only $\frac{x}{y}=4$ is the solution.