Two exponential terms equation solution Let $A_i$ and $B_i$ denote constants, I know this equation
$$A_1 \exp(B_1x) + A_2x + 1 = 0$$
can be solved using Lambert W function. But can I get a general solution of this equation?
$$A_1 \exp(B_1x) + A_2 \exp(B_2x) + A_3x + 1 = 0.$$
I searched a lot but did not get an answer. Thanks a lot for any help.
 A: I am sorry to disappoint you, but one of the things we have to consider when using the Lambert W function is that it abhors addition.
I am very good at manipulating the Lambert W function into providing me answers, but this only works with exponential, log, multiplication, and higher exponentiation.
Addition is out of the tool box when trying to use the Lambert W function.
Let me show you why.
To solve something with the Lambert W function requires changing all addition into multiplication.  I will attempt to do so for your question.
$$A_1e^{B_1x}+A_2e^{B_2x}+A_3x+1=0$$$$e^{A_1e^{B_1x}+A_2e^{B_2x}+A_3x+1}=e^0=1$$$$e^{A_1e^{B_1x}}e^{A_2e^{B_2x}}e^{A_3x}e^1=1$$
So we always want to remove addition by turning it into multiplication as I had done.
The second situation in which I remove addition comes about from $\log$ properties, but I will leave that up to you to decipher.
As for the given problem, you will note that I can divide both sides by $e$.$$e^{A_1e^{B_1x}}e^{A_2e^{B_2x}}e^{A_3x}=e^{-1}$$
You don't move the $e$'s with the variable you are trying to solve for onto the other side, at least not in this situation.  You only move variables to the other side when you have something in the form $f(x)^{g(x)}$.  In that situation, you want to separate the base and exponent and then divide to make the variable on the same side.
But there are too many different terms being multiplied here for us to root both sides or any manipulation as such.
Also, you may consider the exponent being too high, that is we have a $e^{e^x}$, but this problem can be fixed with the following substitution.
$$x=-W(u)$$$$e^x=e^{-W(u)}=(e^{W(u)})^{-1}=[\frac{u}{W(u)}]^{-1}=\frac{W(u)}u$$
Putting this into your problem produces the following (skipping the above work):
$$e^{A_1[\frac{W(u)}u]^{B_1}}e^{A_2[\frac{W(u)}u]^{B_2}}[\frac{W(u)}u]^{A_3}=e^{-1}$$
So the problem with this is most certainly your $B_1,B_2$, which cannot both be removed via substitution.
Even if they are removed, the problem would next lie in your $A_1,A_2,A_3$, which once again cannot be removed via substitution.
And lastly, if you manage all of that, then it no longer becomes a problem because we could have solved it easily in the beginning because it simplifies to a solvable equation.
Which means it is no longer a unique situation we are trying to solve, sadly.
