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**redsoxfan325** I read that for any $\displaystyle a>0, x^a > \ln x$ eventually, though it is possible that $\displaystyle \ln x>x^a$ for finitely many $\displaystyle x$. I figured that $\displaystyle \ln x = x^a$ has zero or two solutions for any given $\displaystyle a>0$ EXCEPT one. Graphing the equations with different values of $\displaystyle a$ led me to conjecture that this occurs when $\displaystyle a=\frac{1}{e}$. Maple confirmed that the equation $\displaystyle \sqrt[e]{x} = \ln x$ has one solution: $\displaystyle x=e^e$. I was wondering two things though:

1) How do you analytically show that if $\displaystyle \sqrt[e]{x} = \ln x$, then $\displaystyle x=e^e$?

2) Is there anything else that's special about the function $\displaystyle \sqrt[e]{x}$?