Lambert W function with rational polynomial

[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?

thx in advance

 

[Ledsnyder]

It doesn't show the steps but I got this from Wolfram

 

[micromass]

It doesn't show the steps but I got this from Wolfram

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

 

[HallsofIvy]

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.

 

[giorgiomugnaini]

x= \frac{d- fu}{eu- c}= -\frac{f}{e}u+ \frac{fc}{e}
.

I think this calculation is wrong.

 

[Keady]

I have a similar problem exp(2 x)= (x+y)/(x-y) and solve for x
Generalized Lambert functions, discussed at Lambert W on wikipedia or various papers on arXiv might help, e.g.
arXiv:1408.3999v1.pdf

My notes are currently at
http://www.cwr.uwa.edu.au/~keady/Seiches/rLambert/lambertWWave.pdf

 

[yosdam]

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.

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.

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

 

[Keady]

I have a similar problem exp(2 x)= (x+y)/(x-y) and solve for x
Generalized Lambert functions, discussed at Lambert W on wikipedia or various papers on arXiv might help, e.g.
arXiv:1408.3999v1.pdf

My notes are currently at
http://www.cwr.uwa.edu.au/~keady/Seiches/rLambert/lambertWWave.pdf

Added, Oct 2015: The cwr website has been taken down. The main facts are in an arXiv preprint with Istvan Mezo