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During my research I ran into the following general type of equation:

$$\exp(ax+b)=\frac{cx+d}{hx+f}. $$

Does anyone have an idea how to go about solving this equation?

Thanks in advance.

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  • $\begingroup$ Welcome to Math SE ! $\endgroup$ – Claude Leibovici Nov 22 '14 at 8:54
  • $\begingroup$ It's possible to reduce the parameters to $e^x = a + \frac{b}{x}$ or $W(x)=ax+b$, though this doesn't seem to help much. $\endgroup$ – sdcvvc Feb 15 '15 at 8:49
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I think that, for the most general case, there is no closed form to the solution of the equation $$e^{a x+b}=\frac{c x+d}{h x+f}$$ For the particular case where $c=0$, you would have $$x=\frac{W\left(\frac{a d e^{\frac{a f}{h}-b}}{h}\right)}{a}-\frac{f}{h}$$ For the particular case where $h=0$, you would have $$x=-\frac{W\left(-\frac{a f e^{b-\frac{a d}{c}}}{c}\right)}{a}-\frac{d}{c}$$ where effectively appear Lambert function. Even the case $c=h$ does not seem to have a solution.

For your more general case $(c \neq 0,h\neq 0)$, I wonder if numerical method should not be the only practical way.

If there is a specific case you would like to be solved, post the values of the six parameters and I should try to show you.

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  • $\begingroup$ To reduce confusion, I changed $ex+f$ in the original question to $hx+f$. Please change your answer correspondingly. Thanks a lot! $\endgroup$ – mike Nov 22 '14 at 9:08
  • $\begingroup$ There are 3 e's in your special solution for c=0 case. Two of the e's are for the parameter e. The third e is for the exponential. $\endgroup$ – mike Nov 22 '14 at 9:10
  • $\begingroup$ @mike. Your idea was very good, indeed ! Thanks. $\endgroup$ – Claude Leibovici Nov 22 '14 at 9:13
  • $\begingroup$ Dear mike, I still try to find a solution in general case. Thank you so much for your comments. $\endgroup$ – John Nov 24 '14 at 4:38
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Given: $$\exp(ax+b)=\frac{cx+d}{hx+f}. $$ substituting $ax+b=z$ we obtain: $$e^z=\frac{cz-bc+da}{hz-bh+fa} $$

setting: $$A=c/h$$ $$t=(bc-da)/c$$ $$s=(bh-fa)/h$$ the equation becomes: $$e^z=A\frac{z-t}{z-s} $$

This equation can be solved by means of a generalization of Lambert W function that has been described in:

"On the generalization of the Lambert W function with applications in theoretical physics", István Mező, Árpád Baricz , http://arxiv.org/abs/1408.3999

The solution (obtained by Lagrange inversion) can be written as a series involving Laguerre polynomials. In particular the authors found that above seen equation can be formally solved by Lagrange series (related to Lambert W function):

$$z(A,t,s)=t- (t-s) \sum_{n=1} \frac{L_n' (n(t-s))}{n} e^{-nt} A^n$$

where:

$L_n(x)$ is n-th Laguerre polynomial and $L_n'(x)$ its first derivative

Replacing $t$, $s$, $A$, $z$, the formal solution of $\exp(ax+b)=\frac{cx+d}{hx+f}$ has been obtained.

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