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I've been asked by a friend to help him solve this equation, but since we couldn't find the right answer, I thought about posting it here. Firstly, I thought about derivating both sides and get:

$a=be^{bx}$

and from here we could find the answer pretty quickly, but then I noticed that what I did was incorrect because I can't derivate both sides - the functions aren't equal.

ekosman
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  • Firstly, you can't differentiate both sides of an equation. – DHMO Apr 16 '17 at 11:57
  • Secondly, it is a transcendental equation, which means that you cannot solve it analytically in elementary functions. – DHMO Apr 16 '17 at 11:58
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    You can either use Lambert W function for an exact form of the solution or Newton's method for an approximate solution. – DHMO Apr 16 '17 at 11:58
  • $f(x)=g(x)$ for some number $x$ does not imply that $f'(x)=g'(x)$. For example, for $x=1$ we have $x^2=x$ but not $2x=1$.
    You can differentiate respect to some variable $x$ both sides of an equality when the equality holds for every value of $x$ (or, at least, for every value in some neighbourhood, and the equality of the derivatives will hold in this neighbourhood).
    – ajotatxe Apr 16 '17 at 12:09
  • :-( Fairly certain there is a good duplicate out there, but I can't find it at the moment. – Simply Beautiful Art Apr 16 '17 at 12:18

1 Answers1

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By the Lambert W function, we have,

$$ax=e^{bx}$$

$$xe^{-bx}=\frac1a$$

$$-bxe^{-bx}=-\frac ba$$

$$-bx=W\left(-\frac ba\right)$$

$$x=-\frac1bW\left(-\frac ba\right)$$

If $0<b/a<e^{-1}$, then there will be two real solutions, denoted by the two branches of the Lambert W function.

If $b/a<0$, then there is only one real solution.

If $b=0$, then the solution is $x=\frac1a$.

If $b/a=e^{-1}$, then $x=\frac1b$.

If $b/a>e^{-1}$, then there are no real solutions.


See here and here for some basic approximations.

I also made a quick approximation calculator-ish thing for the Lambert W function:

Desmos.com

$f(x)$ works best for large inputs and $g(x)$ works best for inputs close to zero.