How do i find the point of intersection of an exponential and its inverse Given that I have an exponential function I was to find its inverse. Afterwards get the points of intersection. My problem is I can't get the intersection. 
$$1. f(x) = 2e^x -4$$
$$2.  f^{-1}(x) = log\,e (\frac{x+4}{2})$$
Solving should be 
$2e^x = log\,e (\frac{x+4}{2})$
But I don't know how to get the points from equating the two
$(x,y)$
$(x,y)$
 A: 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 
$$x=\frac{-34+578 e-100 e^2+(150 e-9) \log
   \left(\frac{5}{2}\right)}{6+158 e+100 e^2+(50 e+1) \log
   \left(\frac{5}{2}\right)}\approx 0.895201$$
A: you can use a numerical method,e.g. the Newton-Raphson method, the solution is given by
$$x_1\approx -3.9619479677925718987$$
$$x_2\approx 0.89508432127523123175$$
