7.307 For what values of a does the equation $2 \lg (x+3)=\lg (a x)$ have a unique solution?
Solution. The original equation is equivalent to the system:

$$
\left\{\begin{array} { c } 
{ \operatorname { l g } ( x + 3 ) ^ { 2 } = \operatorname { l g } ( a x ) } \\
{ x + 3 > 0 }
\end{array} \Leftrightarrow \left\{\begin{array} { c } 
{ ( x + 3 ) ^ { 2 } = a x , } \\
{ x + 3 > 0 }
\end{array} \Leftrightarrow \left\{\begin{array}{c}
x^{2}+(6-a) x+9=0 \\
x>-3
\end{array}\right.\right.\right.
$$

Let $f(x)=x^{2}+(6-a) x+9$. For the last system to have a unique solution, it is necessary and sufficient that the parabola $f(x)$ has one of the forms shown in Figures 4-6 $\left(x_{0}=\frac{a-6}{2}\right)$.

![](https://cdn.mathpix.com/cropped/2024_05_21_f024bf2ff7725246f3bfg-099.jpg?height=237&width=332&top_left_y=1193&top_left_x=176)

Fig. 4

![](https://cdn.mathpix.com/cropped/2024_05_21_f024bf2ff7725246f3bfg-099.jpg?height=237&width=342&top_left_y=1193&top_left_x=522)

Fig. 5

![](https://cdn.mathpix.com/cropped/2024_05_21_f024bf2ff7725246f3bfg-099.jpg?height=220&width=336&top_left_y=1183&top_left_x=876)

Fig. 6

1) The situation in Fig. 4 is ensured by the following condition:

$$
\left\{\begin{array} { l } 
{ f(-3) - 3 }
\end{array} \Leftrightarrow \left\{\begin{array}{l}
a=0 \\
\frac{a-6}{2}>-3
\end{array} \Leftrightarrow \varnothing\right.\right.
$$

3) The situation in Fig. 6 is ensured by the following conditions:

$$
\left\{\begin{array} { l } 
{ x _ { 0 } > - 3 , } \\
{ D = 0 }
\end{array} \Leftrightarrow \left\{\begin{array}{l}
a>0, \\
(6-a)^{2}-36=0
\end{array} \Leftrightarrow a=12\right.\right.
$$

Answer: $a \in(-\infty ; 0) \cup\{12\}$.

## Solve the equations (7.308-7.333):