# Math Help - Solve x^(1/2)+ln(x^(1/2))=u-ln(d)

1. ## Solve x^(1/2)+ln(x^(1/2))=u-ln(d)

I was in the middle of a finding a minimum, when I got stuck with this incredibly annoying equation. I need urgent help as I need to submit my paper by tomorrow:

x^(1/2)+ln(x^(1/2))=u-ln(d) [u and d are parameters).

Is there anything to be done with it to simplify the equation? In part two of the question, I am to assume that u=1 and d=2, but I still can't solve it.

Thanks!

If you let y= 1/x, then the equation becomes y+ ln(y)= 1- ln(2). Taking the exponential of both sides, $ye^y= e/2$. That can be solved using the Lambert W functions which is defined as the inverse of the function $f(x)= xe^{x}$: $W(ye^y)= y= W(e/2)$. Then $x= \frac{1}{y}= \frac{1}{W(e/2)}$.