TITLE: How can I solve $x^x = 5$ for $x$?
QUESTION [5 upvotes]: Possible Duplicate:
Is $x^x=y$ solvable for $x$? 

I've been playing with this equation for a while now and can't figure out how to isolate $x$.
I've gotten to $x \ln x = \ln 5$, which seems like it would be easier to work with, but I can't figure out where to go from there.
Is it possible to solve this algebraically? If not, how can I find the value of $x$?

REPLY [5 votes]: We can find the result using the Lambert W function.
Let's define $y\,e^y=t$.  Then   $y=W(t)$   where $W(t)$  is the Lambert W function.
$$x=e^y\Rightarrow x^x= (e^y)^{e^y}=e^{y\,e^y}=5$$
$$\log e^{y\,e^y}=\log 5$$
$$y\,e^y\log e=y\,e^y=\log 5$$
Thus, $t=\log 5$,  and  from my first definition $y=W(\log 5)$, so $x=e^y=e^{W(\log 5)}$.
It can be expressed in another way too.
$$xy=y\,e^y=t=\log 5$$
$$x=\frac{\log 5}{y}=\frac{\log 5}{W(\log 5)}$$
I asked Wolfram Alpha what the numerical value is, and it said $x\approx2.129372$.