Task 2. Find all values of $a$ for which the equation

$$
2 \lg (x+3)=\lg (a x)
$$

has a unique solution.
The first solution. Obviously, $x>-3, a x>0$, $x \neq 0$. From equation (1) we get:

$$
(x+3)^{2}=a x, a=\frac{(x+3)^{2}}{x}=\left(\sqrt{x}+\frac{3}{\sqrt{x}}\right)^{2}, \text { if } x>0
$$

Obviously,

$$
\left(\sqrt{x}+\frac{3}{\sqrt{x}}\right)=\left(\sqrt{x}-\frac{3}{\sqrt{x}}\right)^{2}+12
$$

From here it is clear that for positive $x$ the minimum value of the function

$$
a=\frac{(x+3)^{2}}{x}
$$

is 12. If $-3-3$. Now it is clear that for all $a>0$, the line $y=a x$ has a unique common point with the curve $y=(x+3)^{2}$ for $x>-3$ only when $y=a x$ is tangent to this curve. From this we get $a=12$.