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

• Sep 25th 2010, 08:19 PM
loststudent
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!
• Sep 25th 2010, 08:23 PM
Educated
If you've read the forum rules, you'll see that we cannot help you on questions that count for your grade.

Forum Rules
Rule #6
• Sep 25th 2010, 08:26 PM
loststudent
Thanks but it doesn't. I am actually auditing this class but got permission to submit the problem sets nevertheless. While I don't get a grade, I get useful feedback. The reason I am auditing this class and not taking it for credit is precisely because I don't have enough mathematical background.
• Sep 25th 2010, 09:04 PM
HallsofIvy
If you would show some working, it would at least help us to see what you should know or what kind of course this is. The solution to this problem cannot be written in terms of "elementary functions".

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 $x= \frac{1}{y}= \frac{1}{W(e/2)}$.
• Sep 25th 2010, 09:07 PM
loststudent
Thanks! much appreciated!