Example 7 In the equation $\frac{\lg 2x}{\lg (x+a)}=2$ concerning $x$, for what real number $a$ does the equation have no solution? One solution? Two solutions?
Given that the equation is equivalent to
\[
\left\{\begin{array}{l}
2 x>0 \\ 
x+a>0, \text { and } x+a \neq 1 \\ 
2 x=(x+a)^{2}
\end{array}\right.
\]
From (3) we get
\[
\begin{array}{c}
x^{2}+2(a-1) x+a^{2}=0 \\
\Delta=4(a-1)^{2}-4 a^{2}=-4(2 a-1) .
\end{array}
\]
(1) $-2 a+1>0 \Rightarrow a<\frac{1}{2}$, i.e., $x_{1}$ satisfies (1)(2). Also, $x_{1}+a=1-\sqrt{1-2 a}, \Rightarrow x_{1}+a \leqslant 0 \Leftrightarrow a \leqslant 0$, i.e., when $a \leqslant 0$, $x_{1}$ does not satisfy (2). When $0<a<\frac{1}{2}$, $x_{1}+a=1-\sqrt{1-2 a} \in (0,1)$, i.e., $x_{1}$ satisfies (1)(2). From (3), $x_{2}+a=1+\sqrt{1-2 a} \geqslant 1$, i.e., $x_{2}$ does not satisfy (2). When $a>\frac{1}{2}$, (4) has no real roots, i.e., the original equation has no solution. In summary, when $a \geqslant \frac{1}{2}$, the original equation has no solution, and when $0<a<\frac{1}{2}$, the original equation has a unique solution $x_{1}=1-\sqrt{1-2 a}$.

Given the system of equations
\[
\left\{\begin{array}{l}
2 x>0 \\ 
x+a>0, \text { and } x+a \neq 1 \\ 
2 x=(x+a)^{2}
\end{array}\right.
\]
When $x+a=1$, from (3) we get $x=\frac{1}{2}$. When $x=\frac{1}{2}$, $x+a=1, a=\frac{1}{2}$. Since $x+a \neq 1$, $a \neq \frac{1}{2}$. From (1)(3), $x+a>0$ is naturally satisfied. From (3), we get $x^{2}+2(a-1) x+a^{2}=0$. When the discriminant of its roots $\Delta=0$, $a=\frac{1}{2}$. For (3), let $f(x)=\sqrt{2 x}, g(x)=x+a$ (with parameter, simple form), where $a \neq \frac{1}{2}, x \neq \frac{1}{2}$.
The solution to the original equation is the x-coordinate of the intersection points of the functions $f(x)$ and $g(x)$ in the same coordinate system. As shown in Figure 5-1,
(1) When $a \leqslant 0$, the equation has one and only one solution;
(2) When $0<a<\frac{1}{2}$, the equation has two and only two solutions;
(3) When $a \geqslant \frac{1}{2}$, the solution set of the equation is empty.