Closed form solution to exponential equation Is there a closed form solution to the following parametric equation?
$$e^{bx+c}(ax-1)=ax+1$$
where $x$ is the main variable and $a\in\mathbb{R^+}$ and $b,c\in\mathbb{R}$.
 A: In the same spirit as Jon's answer (and under the same conditions), you could approximate the exponential term by its simplest Padé approximant
$$e^{bx}=\frac{1+\frac{b }{2}x}{1-\frac{b }{2}x}=\frac{2+b x}{2-b x}$$ and solve for $x$ equation $$e^c \,\frac{2+b x}{2-b x}=\frac{a x+1}{ax-1}$$ which is a quadratic equation. $$a b \left(e^c+1\right) x^2+ (2 a-b)\left(e^c-1\right) x-2 (e^c+1)=0$$
Edit
If you have a look here (bottom of page 6), rewriting the equation as $$e^{bx}\,\,\frac{x- \frac 1a }{x+ \frac 1a} =e^{-c}$$ you will see that the solution can express in terms of the generalized  Lambert function.
A: You can hope for a numerical solution or just an approximate one. If you are happy with the latter, you can rewrite your equation as
$$
   e^{bx}=\frac{ax+1}{ax-1}e^{-c}\qquad {\bf (1)}
$$
and do a Taylor expansion like
$$
  1+bx=e^{-c}(-1-2ax)+O(x^2).
$$
Then,
$$
  x\approx -\frac{e^{-c}+1}{2ae^{-c}+b}.
$$
This is so provided $|bx|\ll 1$ and $|ax|\ll 1$.
Another approximate approach is by iteration. You take $x_0=1$ as your first iterate and from eq.(1) you will get ($a>1$)
$$
   e^{bx_1}=\frac{a+1}{a-1}e^{-c}.
$$
Then,
$$
    x_1=\frac{1}{b}\ln\left(\frac{a+1}{a-1}\right)-\frac{c}{b}.
$$
The best approach will depend on the parameters entering into the equation.
