3. Solve the equation

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

where $3 x+7$ and $2 x+3$ are positive real numbers and different from 1.
Solution. From $3 x+7>0, 2 x+3>0, 3 x+7 \neq 1, 2 x+3 \neq 1$ it follows that the solution to the equation can be a real number $x \in\left(-\frac{3}{2},-1\right) \cup(-1,+\infty)$. Since

\[
\begin{aligned}
& 9+12 x+4 x^{2}=(2 x+3)^{2} \\
& 6 x^{2}+23 x+21=(2 x+3)(3 x+7)
\end{aligned}
\]

the given equation can be written in the form

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

from which we get

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

i.e., $x_{1}=-4, x_{2}=-2$ and $x_{3}=-\frac{1}{4}$. Since $x \in\left(-\frac{3}{2},-1\right) \cup(-1,+\infty)$, it follows that the solution to the equation is $x_{3}=-\frac{1}{4}$.