Is it possible in general to solve equations of the form $Ae^{Bx}+Cx+D=0$? The most I have been able to do is rearrange the equation as $(Bx+\frac{BD}{C})=-\frac{AB}{C}e^{-\frac{BD}{C}}e^{(Bx+\frac{BD}{C})}$, which is in the form of $x=Ae^x$ for $(Bx+\frac{BD}{C})$ which looks solvable. However I don't know how. Any help appreciated.
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1You'll need the Lambert W function to express the solution. – Daniel R Aug 30 '13 at 08:15
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Solution in terms of Lambert W function (for approximate solutions use a Taylor expansion of the exponential), see: https://math.stackexchange.com/a/3658405/532409 – Quillo May 05 '20 at 10:15
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Using Lambert W function, $x=a\mathrm e^x$ can be rewritten as $(-x)\mathrm e^{-x}=-a$, hence $x=-W(-a)$.
Note that this is not much more than a rewriting of the equation, hence the fact that Lambert W function was given a definition and that $x$ can be expressed through it does not lead us closer to a "solution", if you ask me.
Did
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@Did Well, at least I have a Taylor series to play with (therefore approximate solutions). – resgh Aug 30 '13 at 08:27
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