Solving the Lambert Function: Advice from Forum

[Sammuueel]

Dear Forum,

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

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,
Samuel

 

[HallsofIvy]

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

Now let y=- x so that -ye^y= e^{1- \Omega} or ye^y= -e^{1- \Omega}. You can apply Lambert's function to both sides of that to find y, then go back to find \rho.

 

[Sammuueel]

I can perfectly follow you, thank you for you quick reply.
But I am not sure if the exponential of x(1-\ln x) equals e^x(e^{1-\ln x})?

Kind regards,
Samuel