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Lambert W function with rational polynomial

  1. Jul 9, 2014 #1
    Hi all,

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

    thx in advance
     
  2. jcsd
  3. Jul 9, 2014 #2
    It doesnt show the steps but I got this from Wolfram
     

    Attached Files:

  4. Jul 9, 2014 #3

    micromass

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    I think the idea is to solve for ##x## instead of ##f##.
     
  5. Jul 13, 2014 #4

    HallsofIvy

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

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

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

    [tex]v= W(\frac{af(af+ bd)}{de}[/tex]

    Now work back through the substitutions to find x.
     
  6. Oct 25, 2014 #5
    I think this calculation is wrong.
     
  7. Apr 20, 2015 #6
  8. Oct 9, 2015 #7
    May I suggest that HallsofIvy changes his name to Half-fly or Highdive, or something?
     
  9. Oct 9, 2015 #8
    Added, Oct 2015: The cwr website has been taken down. The main facts are in an arXiv preprint with Istvan Mezo
     
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