Prove that $e^x\geq x^{a}$

Question

Prove that $\forall a\in\mathbb{R}_{>0}:((\forall x\in\mathbb{R}_{>0}:e^x\geq x^{a})\iff a\leq e$).

I have looked up numerous theorems and MSE posts and strangely, didn't find anything similar or anything that could help me with this.

Answer 1

For the first part take log and plug $x=e$(as it is given for all $x$ in positive reals.) Next use the fact $x^{\frac{1}{x}}\leq e^{\frac{1}{e}}$ if $x$ is positive real. Then $e^{\frac{1}{e}}\leq e^{\frac{1}{a}}$ by the provided inequality on lhs part we are done, thus $x^a\leq e^x$