4B. 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 domain of solutions of the equation, it must hold that $2 x+1>0, 3 x+4>0, 6 x^{2}+11 x+4>0$ and $2 x+1, 3 x+4 \neq 1$. This yields the interval $\left(-\frac{1}{2}, \infty\right)$. We need to note that $6 x^{2}+11 x+4=(2 x+1)(3 x+4)$, and then the equation takes the form

\[
\begin{aligned}
& 2 \log _{3 x+4}(2 x+1)+\log _{2 x+1}(2 x+1)(3 x+4)=4 \\
& 2 \log _{3 x+4}(2 x+1)+\log _{2 x+1}(2 x+1)+\log _{2 x+1}(3 x+4)=4 \\
& 2 \log _{3 x+4}(2 x+1)+\log _{2 x+1}(3 x+4)=3 \\
& 2 \log _{3 x+4}(2 x+1)+\frac{1}{\log _{3 x+4}(2 x+1)}=3
\end{aligned}
\]

We introduce the substitution $\log _{3 x+4}(2 x+1)=t$. We obtain the equation $2 u+\frac{1}{u}=3$ which is equivalent to the quadratic equation $2 u^{2}-3 u+1=0$ whose solutions are $u_{1}=1$ and $u_{2}=\frac{1}{2}$. For the first solution from the substitution, we get $3 x+4=2 x+1$, or $x=-3$. However, $x=-3 \notin\left(-\frac{1}{2}, \infty\right)$, so it is not a solution. For $u=\frac{1}{2}$, we get the quadratic equation $3 x+4=(2 x+1)^{2} \Leftrightarrow 4 x^{2}+4 x-3=0$. Its solutions are $x_{1}=-1$ and $x_{2}=\frac{3}{4}$, of which only the latter is in the domain. Therefore, the only solution is $x=\frac{3}{4}$.

## 4th Year