For positive real numbers $x$ and $y$, the following holds:

$$
2 \log (x-2 y)=\log x+\log y
$$

What is $\frac{x}{y} ?$
III/1. For $\log (x-2 y)$ to be defined, it must hold that $x>2 y$. The equation can be written as $\log (x-2 y)^{2}=\log (x y)$, from which it follows that $(x-2 y)^{2}=x y$ or $(x-4 y)(x-y)=0$. Due to the condition that $x$ and $y$ must satisfy, the solution $x=y$ is not possible, so $x=4 y$ or $\frac{x}{y}=4$.

Condition $x>2 y$: $\mathbf{1}$ point. Transformation to $\log (x-2 y)^{2}=\log (x y)$: $\mathbf{2}$ points. Conclusion $(x-2 y)^{2}=x y$: $\mathbf{2}$ points. Equation $(x-4 y)(x-y)=0$ (or the solutions $x=y, x=4 y$): $\mathbf{1}$ point. Conclusion $x=4 y$ and the answer $\frac{x}{y}=4$: $\mathbf{1}$ point.