find x in the equation

$\log_3 x+\log_2 (2x+2)=4$

$\implies log_3 x+\log_2 (x+1)+\log_2 2=4$

$\implies \log_3 x+\log_2 (x+1)=3=\log_2 8$

$\implies \log_3 x=\log_2 8-\log_2 (x+1)=\log_2 (\frac{8}{x+1})=k$ (say).

Then $x=3^k, \dfrac{8}{x+1}=2^k$

$\implies 2^k+6^k=8=2^1+6^1$ .

Hence $k=1$ and $x=3^1=\boxed 3$ .

$\begin{aligned} \log_3 x + \log_2 (2x+2) & = 4 \\ \frac {\log x}{\log 3} + \frac {\log (2(x+1)}{\log 2} & = 4 \\ \frac {\log x}{\log 3} + \frac {\log 2 + \log (x+1)}{\log 2} & = 4 \\ \frac {\log \red x}{\log 3} + \frac {\log (\red x+1)}{\log 2} & = 3 & \small \red{\text{By observation }x=3} \\ \frac {\log \red 3}{\log 3} + \frac {\log (\red 3+1)}{\log 2} & = 3 \\ 1 + 2 & = 3 \end{aligned}$

Therefore $x = \boxed 3$ .