Is $(e^\pi-\pi^e)$ positive or negative? why?

Question

Jack D'Aurizio · Answer 1 · 2015-02-07T20:55:52.663

1

If $a>b\geq e$, then $a^b<b^a$. It follows from the fact that the function $$ f(x)=\frac{\log x}{x} $$ is decreasing over $I=(e,+\infty)$, since: $$ f'(x) = \frac{1-\log x}{x^2} $$ is negative over $I$.

edited Feb 07 '15 at 20:55

answered Feb 07 '15 at 20:36

Jack D'Aurizio

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thanks. was helpful for me – Alireza Feb 07 '15 at 20:47
Shouldn't it be $a > b \geq e$ in order to apply this to the actual question? – rubik Feb 07 '15 at 20:52
@rubik: yes, you're right. Now fixed. – Jack D'Aurizio Feb 07 '15 at 20:56

Is $(e^\pi-\pi^e)$ positive or negative? why?

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