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