1. Limited time only! Sign up for a free 30min personal tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

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


    User Avatar
    Staff Emeritus
    Science Advisor

    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,
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Similar Discussions: Lambert function
  1. Lambert W Function (Replies: 2)

  2. Functions ? (Replies: 18)