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Lambert function

  1. Dec 14, 2012 #1
    Dear Forum,

    I am a researcher in the field of microeconomics and I came across this equation which I would like to solve for [itex]k[/itex].
    [itex]\Omega = \rho^k (1-k\cdot \ln \rho) [/itex]

    It looks a little bit like the Lambert function. But I am stuck here.
    Do you have an idea how I could proceed?

    Kind regards,
  2. jcsd
  3. Dec 14, 2012 #2


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    First, of course, [itex]k ln(\rho)= ln(\rho^k)[/itex] so I would start by letting [itex]x= \rho^k[/itex]. Then your equation becomes [itex]\Omega= x(1- ln(x))[/itex]. Now take the exponential of both sides: [itex]e^{\Omega}= e^x(e^{1- ln(x)})= e^x(e)/x[/itex] and then [itex]\frac{e^x}{x}= e^{\Omega- 1}[/itex] or [itex]xe^{-x}= e^{1- \Omega}[/itex].

    Now let y=- x so that [itex]-ye^y= e^{1- \Omega}[/itex] or [itex]ye^y= -e^{1- \Omega}[/itex]. You can apply Lambert's function to both sides of that to find y, then go back to find [itex]\rho[/itex].
  4. Dec 14, 2012 #3
    I can perfectly follow you, thank you for you quick reply.
    But I am not sure if the exponential of [itex]x(1-\ln x)[/itex] equals [itex]e^x(e^{1-\ln x})[/itex]?

    Kind regards,
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