solve in R the equation

If $x > 2$ then $\log(x)/\log(2) > 1$ and $\log(x+1)/\log(3) > 1,$ so the left side is too big.

If $x < 2$ then $\log(x)/\log(2) < 1$ and $\log(x+1)/\log(3) < 1,$ so the left side is too small.

If $x=2$ then the equation is clearly true. Hence the only solution is $\fbox{2}.$

By inspection, $x=2$ .

$\begin{aligned} \frac {\log x}{\log 2} + \frac {\log(x+1)}{\log 3} & = 2 & \small \blue{\text{Multiply both sides by }\log 2 \log 3} \\ \log x \log 3 + \log (x+1) \log 2 & = 2\log 2\log 3 & \small \blue{\text{Putting }x=2} \\ \log 2 \log 3 + \log 3 \log 2 & = 2 \log 2 \log 3 \end{aligned}$

Therefore, $x= \boxed 2$ .