What is the minimum value of $a$ such that $x^a \geq \ln(x)$ for all $x > 0$?

This is probably just elementary, but I don't know how to do it. I would like to find the minimum value of $a$ such that $x^a \geq \ln(x)$ for all $x > 0$. Numerically, I have found that this minimum value lies between 0.365 and 0.37 (i.e., $x^{0.37} > \ln(x)$ for all $x > 0$, but $x^{0.365}$ is not). Is there any analytical way to find out exactly this minimum value?

EDIT: Based on the received answers, I finally came up with my own one as follows.

Consider the function $f(x) = x^a - \ln(x).$ This function is convex in $x$, and hence, achieves the unique minimum as $x^*$ such that $f'(x^*) = 0.$ Solving that equation yields $$f_{\mathrm{min}} = \min\limits_{x>0} f(x) = \frac{\ln(a)+1}{a}.$$

Now, by letting $f_{min} = 0$, we get the desired value $a^* = 1/e.$

Thank everyone for the answers!

$a = \max_{x>0} \frac{\ln(\ln(x))}{\ln(x)} = 1/e$.

Consider the minimum of the function $f(x)=x^{1/e}-\ln(x)$.

EDIT: As a second step, verify that $x^a < \ln(x)$ at $x=(1/a)^{1/a}$, for any $0<a<1/e$.