If $m<n$, how do the two quantities compare: $m^n$ and $n^m$?

Example: $2^5>5^2$, $3^7>7^3$, ... and so on. Is it true in general as in is there a sound proof?

  • $\begingroup$ You need to take into account whether you will accept numbers less than $1$. $\endgroup$ – fleablood Feb 15 '18 at 0:09
  • $\begingroup$ @Clayton That was before the edit. Before the edit, the post said $m > n$, not $m < n$. Check the edit history. I assumed it was a typo and pointed it out. $\endgroup$ – Arnav Borborah Feb 15 '18 at 0:13

As you've written it, it isn't true in general. You can take $m=2<4=n$ as a counterexample since $2^4=4^2$. However, if $3\leq m<n$, then we have $m^n>n^m$.

To see this, take the logarithm of both sides and reduce, so we have $$\frac{\log m}{m}>\frac{\log n}{n}.$$This inequality is true since $f(x)=\frac{\log x}{x}$ has a negative derivative for $x>e$ (which implies $f(x)$ is decreasing).

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