Almost as you wrote, you need to solve equation
$$2 e^x-4=\log \left(\frac{x+4}{2}\right)$$ for which you cannot expect explicit solutions.
Consider then that you look for the zeros of function
$$f(x)=2 e^x-4-\log \left(\frac{x+4}{2}\right)$$ which, because of the loagithm is defined for $-4 < x < \infty$.
Consider its derivatives
$$f'(x)=2 e^x-\frac{1}{x+4}\qquad \qquad f''(x)=2 e^x+\frac{1}{(x+4)^2} >0\qquad \forall x$$ Sooner or later, you will learn the the solution of $f'(x)=0$ is given in terms of Lambert function and the solution is given by
$$f'(x)=0\implies x=W\left(\frac{e^4}{2}\right)-4\approx -1.57782$$ For this value $f(x)\approx -3.77867 $ and the second derivative being always positive, $f(x)=0$ shows two roots.
By inspection, you could notice that there is one root between $0$ and $1$ since
$$f(0)=-2-\log (2)\approx -2.69315\qquad \text{and}\qquad f(1)=-4+2 e-\log \left(\frac{5}{2}\right)\approx 0.520273$$ and, as said in another answer, you can use Newton method starting, say, using $x_0=1$. The iterates would be
$$\left(
\begin{array}{cc}
n & x_n \\
0 & 1 \\
1 & 0.900646 \\
2 & 0.895101 \\
3 & 0.895084
\end{array}
\right)$$ which is the solution for six significant figures.
For the other root, which we can suppose to be close to $-4$, in a first approximation we could write
$$f(x)\approx 2 e^{-4}-4-\log \left(\frac{x+4}{2}\right)=0 \implies x=2 e^{\frac{2}{e^4}-4}-4\approx -3.96200 $$ which is probably more than sufficient for your needs.
Edit
For the positive root, we could also build the simplest Padé approximant of the function
$$f(x)=\frac{f(a)+ \left(f'(a)-\frac{f(a) f''(a)}{2 f'(a)}\right) (x-a)} {1-\frac{ f''(a)}{2 f'(a)}(x-a) }$$ and, using $a=1$, solve for $0$ the numerator. This would give the nice
$$x=\frac{-34+578 e-100 e^2+(150 e-9) \log
\left(\frac{5}{2}\right)}{6+158 e+100 e^2+(50 e+1) \log
\left(\frac{5}{2}\right)}\approx 0.895201$$
Best Answer
Saying $f(x) = f^{-1}(x)$ means that they intersect along the line $y=x$. We could equivalently ask for either
$$x = e^x-1$$
$$x = \ln(x+1)$$
Which both have solutions at $x=0$ by guessing. A little IVT magic proves that this is the only solution for either equation.