Does $x^x$ ever overtake $e^{x^2}$?

Question

I feel like it should, but I cannot prove it. If it does, how can we find the precise value of $x$ for which $x^x$ overtakes the other?

Answer 1

No. $x\ln x < x^2\ \forall\ x > 0$, and $\exp(x \ln x) = x^x$.

Answer 2

Raise both sides to the power of $1/x$ $$x^x = e^{x^2} \implies x = e^x$$ Since it's well known that $e^x > x+1 >x$ we are done, as we see that we have no points in common and that $e^{x^2}> x^x$

Answer 3

(I assume this problem is restricted to real, positive values of $x$)

It doesn't overtake it - $e^{x^2}$ is always larger.

Proof: (for $x > 0$)

\begin{align} x &> \ln x \\ x^2 &> x \ln x \\ e^{x^2} &> x^x \end{align}