Example 3. If $x>0, y>0$, and $21 \lg(x-2 y)=\lg x+\lg y$, then $x: y$ is ( )
(A) 4. (B) 1 .
(C) 1 or 4 .
(D) $\frac{1}{4}$.
Many students often transform the given condition
$21 g(x-2 y)=\lg x+\lg y$ into
$\lg (x-2 y)^{2}=\lg x y$. However, this transformation brings changes. Before the transformation, the conditions required $x-2 y>0, x>0, y>0$, while after the transformation, they became $x-2 y \neq 0, x y>0$, which weakens the conditions. If this change is not noticed, the solution might proceed as follows:
$$
\begin{array}{c}
\because \quad \lg (x-2 y)^{2}=1 \lg x y, \\
\therefore \quad(x-2 y)^{2}=x y . \\
\therefore \quad x^{2}-4 x y+4 y^{2}=x y \\
x^{2}-5 x y+4 y^{2}=0, \\
(x-y)(x-4 y)=0 . \\
x_{1}=y, \quad x_{2}=4 y .
\end{array}
$$

Thus, $x: y=1$ or $x: y=4$.
Therefore, the answer is (C).
In fact, if $x: y=1$, then $x=y$. Thus,
$$
x-2 y=y-2 y=-y<0$, which does not meet the original conditions.
Hence, the correct answer should be (A).