Lambert W function with rational polynomial

1. Jul 9, 2014

nlooije

Hi all,

During my research i ran into the following general type of equation: $\exp(ax+b)=\frac{cx+d}{ex+f}$
does anyone have an idea how to go about solving this equation?

2. Jul 9, 2014

Ledsnyder

It doesnt show the steps but I got this from Wolfram

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3. Jul 9, 2014

micromass

Staff Emeritus
I think the idea is to solve for $x$ instead of $f$.

4. Jul 13, 2014

HallsofIvy

Staff Emeritus
Let $u= \frac{cx+ d}{ex+ f}$, the fraction on the right. Then, solving for $x$, $x= \frac{d- fu}{eu- c}= -\frac{f}{e}u+ \frac{fc}{e}$.

So the equation is, so far,
$$e^{ax+ b}= e^{-\frac{af}{e}u+ \frac{afc}{e}+ b}= u$$
$$e^{-\frac{af}{e}u}e^{\frac{afc+ bd}{d}}= u$$
$$ue^{\frac{af}{e}u}= e^{\frac{afc+ bd}{d}}$$

Let $v= \frac{af}{e}u$. Then $u= \frac{e}{af}v$ and we have
$$\frac{e}{af}ve^v= e^{\frac{afc+ bd}{d}}$$
$$ve^v= \frac{af(af+ bd)}{de}$$

$$v= W(\frac{af(af+ bd)}{de}$$

Now work back through the substitutions to find x.

5. Oct 25, 2014

giorgiomugnaini

I think this calculation is wrong.

6. Apr 20, 2015

7. Oct 9, 2015

yosdam

May I suggest that HallsofIvy changes his name to Half-fly or Highdive, or something?

8. Oct 9, 2015