How will you determine which number is bigger among $m^n$ and $n^m$ without using a calculator, $n$ and $m$ being real?

Question

We can use a calculator to calculate such problems but when $m$ and $n$ is a big value then we cannot calculate with a calculator..

Is there any way to solve such problems by hand?

Thank you.

Answer 1

They are $(m^{1/m})^{mn}$ and $(n^{1/n})^{mn}$.
It turns out that $x^{1/x}$ is an increasing function if $x<e$, and decreasing if $x>e$.
So if $e<m<n$ then $m^{1/m}>n^{1/n}$ and so $m^n>n^m$.

Answer 2

$m^n<n^m \Leftrightarrow \ln m^n< \ln n^m\Leftrightarrow n\ln m<m\ln n \Leftrightarrow \dfrac{\ln m}{m}<\dfrac{\ln n}{n}$

It is $\bigg(\dfrac{\ln x}{x}\bigg)'=\dfrac{1-\ln x}{x^2}\Rightarrow \dfrac{\ln x}{x}$ is increasing in $(0,e)$ and decreasing in $(e,\infty)$ ....