What is the inverse of $f(x) = a⋅e^{bx} + cx + d$ Does an inverse function for 
$f(x) = a⋅e^{bx} + cx + d$
exist where a, b, c, d are constants?  If so, what is it?
I've tried lots of methods, but they've all failed.
What I ended up doing to approximate an inverse:
For very small (very negative) values of y, I ended up using the following function:
$x = \frac{ln(\frac{y-cx-d}{a})}{b}$
to assist in a recursive approximation by which, after seeding x with a guess, I feed the function back on itself a few times.  For greater values of y, I used a similar method/function inspired by binary search algorithms. My implementations converged quickly for my purposes.
Thanks everyone who commented.
 A: Consider $$y = a⋅e^{bx} + cx + d$$ and make a change of variable $e^{bx}=z$. This then leads to $$y=a.z+\frac cb .\log(z)+d$$Any equation which can write $A+B x+C\log(Dx+E)=0$ has solution(s) in terms of Lambert function. For the particular case in the post, we should then obtain $$z=\frac{c }{a b}W(t)$$ where $$t=\frac{a b }{c}e^{\frac{b (y-d)}{c}}$$ and, back to $x$ (taking into account the fact that $t=W(t).e^{W(t)}$), $$x= \frac{y-d}{c}-\frac{1}{b}W(t)$$ The advantage of this is that there are very good approximations of Lambert function.
For illustration purposes, let us use $a=1,b=2,c=3,d=10^6$ and let us try to solve the equation for $y=2.10^6$. So, replacing, we have $$x=\frac{1000000}{3}-\frac{1}{2} W\left(\frac{2 e^{2000000/3}}{3}\right)$$ Now, let us use for the approximation $$W(t)\approx L_1-L_2+\frac{L_2}{L_1}+\cdots$$ where $L_1=\log(t)$,$L_2=\log(L_1)$. This then leads to $x=6.907744917$ which is the exact solution for ten significant figures.
A: It is quite possible for a function to have an inverse but with no closed formula for the inverse.
In this case, $f(x)$ is $1-1$ if and only if $f'(x)\neq 0$ for all $x$ in the (interior of the) domain, which means if $abe^{bx}\neq -b$ has no solution, so if $\frac{-d}{ab}$ is negative.
However, combinations of $x$ and $e^x$ generally make it impossible to find a formula for the inverse function. 
We can easily find a way to compute the inverse when $f$ is increasing or decreasing - just apply a binary-search algorithm to get a sequence of approximations to the inverse.
