Solve the equation: $\;$ $3 + 2 \log_{x+1} 3 = 2 \log_3 \left(x+1\right)$
Given equation: $\;\;$ $3 + 2 \log_{x+1} 3 = 2 \log_3 \left(x+1\right)$
i.e. $\;$ $2 \log_3 \left(x + 1\right) - 2 \log_{x+1} 3 = 3$
i.e. $\;$ $\log_3 \left(x + 1\right) - \log_{x+1} 3 = \dfrac{3}{2}$
i.e. $\;$ $\log_3 \left(x+1\right) - \dfrac{1}{\log_3 \left(x + 1\right)} = \dfrac{3}{2}$ $\;\;\; \cdots \; (1)$
Let $\;$ $\log_3 \left(x + 1\right) = p$ $\;\;\; \cdots \; (2)$
Then in view of equation $(2)$, equation $(1)$ becomes
$p - \dfrac{1}{p} = \dfrac{3}{2}$
i.e. $\;$ $2p^2 - 3p - 2 = 0$
i.e. $\;$ $\left(2p + 1\right) \left(p - 2\right) = 0$
i.e. $\;$ $p = \dfrac{-1}{2}$ $\;$ or $\;$ $p = 2$
Substituting the value of $p$ in equation $(2)$ gives
when $\;$ $p = \dfrac{-1}{2}$, $\;$ then $\;$ $\log_3 \left(x + 1\right) = \dfrac{-1}{2}$
i.e. $\;$ $x + 1 = 3^{\frac{-1}{2}} = \dfrac{1}{\sqrt{3}} = \dfrac{\sqrt{3}}{3}$
$\implies$ $x = \dfrac{\sqrt{3}}{3} - 1 = \dfrac{\sqrt{3} - 3}{3}$
when $\;$ $p = 2$, $\;$ then $\;$ $\log_3 \left(x + 1\right) = 2$
i.e. $\;$ $x + 1 = 3^2 = 9$ $\implies$ $x = 8$
$\therefore \;$ The solution to the given equation is $\;\;$ $x = \left\{\dfrac{\sqrt{3} - 3}{3}, \; 8 \right\}$