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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 $x$?
On
It's not possible to solve this algebraically.
Look at Lambert W function. In your case, the solution is $x=\frac{\ln(5)}{W(\ln(5))}$.
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 $x\approx2.129372$.