Answer
Graphically: $x=2.264$
Algebraically: $x=2.264$
Work Step by Step
Let $y_1=\ln (x+1)$ and $y_2=2-\ln (x)$.
The graph is as shown using a graphing utility.
From the graph, the intersection point is at $(2.264,1.183)$.
Thus, graphically, the solution is $x=2.264$.
Solving algebraically:
$$\ln (x+1)+\ln (x)=2-\ln (x)+\ln (x)$$ $$\ln (x+1)x=2$$ $$\ln (x^2+x)=2$$
Applying log rules which is $\ln(x)=y$ is equivalent to $x=e^y$:
$$x^2+x=e^2$$ $$x^2+x-e^2=0$$ $$x=\frac{-1\pm\sqrt{1^2-4(1)(-e^2)}}{2(1)}=\frac{-1\pm\sqrt{1+4e^2}}{2}$$ $$x_1=\frac{-1+\sqrt{1+4e^2}}{2}$$ $$x_2=\frac{-1-\sqrt{1+4e^2}}{2}$$
Checking $x_1$:
$$\ln \left(\frac{-1+\sqrt{1+4e^2}}{2}+1\right)=2-\ln \frac{-1+\sqrt{1+4e^2}}{2}$$ $$1.183=1.183~True$$
Checking $x_2$:
$$\ln \left(\frac{-1-\sqrt{1+4e^2}}{2}+1\right)=2-\ln \frac{-1-\sqrt{1+4e^2}}{2}$$ $$undefined=undefined\Rightarrow Not~valid$$
Simplifying $x_1$:
$$x_1=\frac{-1+\sqrt{1+4e^2}}{2}=2.264$$
Thus, algebraically, the solution is $x=\frac{-1+\sqrt{1+4e^2}}{2}=2.264$.
