For which values of the parameter $a$ does the equation $\lg (a x)=2 \lg (x+1)$ have exactly one solution?
Solution. Equation (1) makes sense if $a x>0$ and $x+1>0$. If $a>0$, this holds for positive numbers, if $a=0$, there is no such number, and if $a<0$, it holds for negative numbers. If $a>0$, the equation can be rewritten as $x^2 + (2-a)x + 1 = 0$. The discriminant of this quadratic equation is $\Delta = (2-a)^2 - 4 = a^2 - 4a$. This is non-negative if $a^2 - 4a \geq 0$, i.e., $a(a-4) \geq 0$, which means $a \leq 0$ or $a \geq 4$. Therefore, the equation has real roots if $a \leq 0$ or $a \geq 4$.

For $a>4$, $T_{a}$ is the set of positive numbers, and equation (1) has no roots. (Since $x_{1}+x_{2}=a-2>0$, both roots are positive, so in this case, equation (1) has two roots.)

For $a \leq 0$, $T_{a}$ is the set of negative numbers, and equation (1) has exactly one root. The well-known graphs immediately show that in the first case, the discriminant of equation (2) characterizes the behavior of equation (1) in the usual way, and if $a<0$, equation (1) has exactly one solution.