2. Solve the equation

$$
\log _{(3 x+7)}\left(9+12 x+4 x^{2}\right)+\log _{(2 x+3)}\left(6 x^{2}+23 x+21\right)=4
$$
Solution. Since

$$
9+12 x+4 x^{2}=(3+2 x)^{2}, \quad 6 x^{2}+23 x+21=(2 x+3)(3 x+7)
$$

the given equation reduces to the equation

$$
2 \log _{(3 x+7)}(2 x+3)+1+\log _{(2 x+3)}(3 x+7)=4
$$

i.e.

$$
2\left[\log _{(3 x+7)}(2 x+3)\right]^{2}-3 \log _{(3 x+7)}(2 x+3)+1=0
$$

from which we obtain

$$
\log _{(3 x+7)}(2 x+3)=1, \quad \log _{(3 x+7)}(2 x+3)=\frac{1}{2}
$$

Solving these equations, we get $x_{1}=-4, x_{2}=-2$ and $x_{3}=-\frac{1}{4}$. It is easily verified that the solution to the equation is only $x_{3}=-\frac{1}{4}$.