Is $x^x=y$ solvable for $x$?

As Aryabhata mentions this is another application for the Lambert W function. The solution to your problem is presented in the wikipedia article. Using elementary substitutions you have

$$x=\frac{\ln(y)}{W(\ln y)}$$

If you are interested in the asymptotic growth of $x$ relative to $y$, note that for every $z$: $W(z) = \ln{z} - \ln\ln{z} + o(1)$. Hence:

$$x=\frac{\ln(y)}{\ln{\ln y} - \ln\ln{\ln y} + o(1)} = \Theta\left( \frac{\ln y}{\ln \ln y}\right)$$

You should try WolframAlpha for similar problems. WolframAlpha would solve y=x^x for y=5 as shown here (using Lamber W Function as suggested before).