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
$(x,y)$
$(x,y)$
Best Answer
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 $$x=\frac{-34+578 e-100 e^2+(150 e-9) \log \left(\frac{5}{2}\right)}{6+158 e+100 e^2+(50 e+1) \log \left(\frac{5}{2}\right)}\approx 0.895201$$