Question as per the title.
Show that $e^x > x^t$ where $x > 0$ and $t < e$
It was suggested that to arrive at the proof, the maclaurin series be used.
Question as per the title.
Show that $e^x > x^t$ where $x > 0$ and $t < e$
It was suggested that to arrive at the proof, the maclaurin series be used.
Write $x^t= e^{t \log x}$. Then $e^x > x^t$ iff $t < x/\log x$. Now, for $x>0$, the minimum value of $x/\log x$ is $e$, attained at $x=e$.