How to compare $2^{\pi}$ and $\pi^2$ using calculus
I guess $$f(x)=\frac{\ln x}{x}$$ wont help here since $2 \lt e \lt \pi$
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$2$ 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 $4$.
robjohn
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1Marvelous approach, but i think unfortunately cannot be used to compare $(1.9)^{\pi}$ and $\pi^{1.9}$ – Umesh shankar Jul 21 '18 at 17:13
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@Umeshshankar: actually, for $x\le e$, $\frac{\log(x)}x$ is monotonically increasing, so $\frac{\log(1.9)}{1.9}\lt\frac{\log(2)}2\lt\frac{\log(\pi)}\pi$. So... – robjohn Jul 21 '18 at 17:15
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Yes perfect, i think now we can compare $a^{\pi}$ and $\pi^a$ for any $a$ – Umesh shankar Jul 21 '18 at 17:19
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1@Umeshshankar: Not so fast! it doesn't work for $2.1^\pi$ and $\pi^{2.1}$. – robjohn Jul 21 '18 at 17:21
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Yes yes true about to post my edit, thanks – Umesh shankar Jul 21 '18 at 17:22
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Plus 1 for the extended post ..Great analysis – Umesh shankar Jul 21 '18 at 17:50