2. Solve the equation

$$
\log _{3 x+4}(2 x+1)^{2}+\log _{2 x+1}\left(6 x^{2}+11 x+4\right)=4
$$
Solution. For the bases of the logarithms to be positive, we need $3x + 4 > 0$ and $2x + 1 > 0$, from which it follows that $x > -\frac{1}{2}$. For such values of $x$, $2x + 1$ and $6x^2 + 11x + 4 = (2x + 1)(3x + 4)$ are also positive. Therefore, the equation is meaningful only for $x > -\frac{1}{2}$. Step by step, we have:

$$
\begin{aligned}
& \log_{3x+4}(2x+1)^2 + \log_{2x+1}(6x^2 + 11x + 4) = 4 \\
& 2 \log_{3x+4}(2x+1) + \log_{2x+1}(2x+1)(3x+4) = 4 \\
& \log_{3x+4}(2x+1) + \log_{2x+1}(3x+4) = 3 \\
& \log_{3x+4}(2x+1) + \frac{1}{\log_{3x+4}(2x+1)} = 3
\end{aligned}
$$

We introduce the substitution $\log_{3x+4}(2x+1) = t$ and obtain the equation $2t + \frac{1}{t} = 3$, whose solutions are 1 and $\frac{1}{2}$. Substituting these values of $t$ back into the substitution, we get solutions for $x$, namely $x_1 = -3$, $x_2 = -1$, and $x_3 = \frac{3}{4}$. Only for the third value of $x$ is the equation meaningful. Therefore, the only solution to the given equation is $x = \frac{3}{4}$.