Show that a positive constant $t$ can satisfy $e^x>x^t$ for any $x>0$ if and only if $t<e$.
I tried this: since $x>0$, the inequality is equivalent to $x>t \log x$. If $0<x \le 1$ it is $t \log x \le 0$ and so the inequality is true because $x>0$. If $x > 1$ it is $x> t \log x \iff \frac{x}{\log x}>t$.
Studying the function $f:(1,\infty) \to \mathbb{R}$ defined by $f(x)=\frac{x}{\log x}$, I found that $f$ has a minimum in $x=e$ and it is $f(e)=e$.
So the inequality $e^x>x^t$ is always verified for $0<x \le 1$ and, as $t$ varies in $(0,\infty)$, it is verified for any $x>1$ only if $t$ doesn't get larger than the minimum of $f$, that is if $t<e$. Hence the inequality is satisfied for any $x>0$ if and only if $t<e$.
Is this a correct way to prove this statement?
