TITLE: How to compare $2^{\pi}$ and $\pi^2$ using calculus
QUESTION [6 upvotes]: How to compare $2^{\pi}$ and $\pi^2$ using calculus
I guess $$f(x)=\frac{\ln x}{x}$$ wont help here since $2 \lt e \lt \pi$

REPLY [10 votes]: $2$ and $\pi$ are on differing sides of the maximum of $\frac{\log(x)}x$ at $x=e$, so it is hard to compare them. However, since $4=2^2$, we have $\frac{\log(2)}2=\frac{\log(4)}4$, and we can compare
$$
\frac{\log(\pi)}\pi\gt\frac{\log(4)}4=\frac{\log(2)}2
$$
because $\frac{\log(x)}x$ is monotonically decreasing for $x\ge e$. Therefore,
$$
\pi^2\gt2^\pi
$$

Further Result
$2$ and $4$ is not the only pair of unequal numbers we can compute for which $\frac{\log(x)}x=\frac{\log(y)}y$. The same is true for
$$
x=\left(1+\frac1t\right)^t\quad\text{and}\quad y=\left(1+\frac1t\right)^{t+1}
$$
where $t\gt0$. $t=1$ gives $2$ and $4$.