Which one is bigger, $e ^ {3 \pi}$ or $ \space 3 ^ {e \pi}$?

Question

I have this equation, and I've tried checking which is bigger,

$ e ^ {3 \pi} \space ? \space 3 ^ {e \pi}$

What I've tried:

$ e ^ {3 \pi} \space ? \space 3 ^ {\pi e} / \sqrt[\pi]{} $

$ e ^ {3} \space ? \space 3 ^ {e}$

$ (e ^ {1\over e}) ^ {3e} \space ? \space (3 ^ {1\over 3}) ^{3e} / \sqrt[3e]{}$

$ e ^ {1\over e} \space ? \space 3 ^ {1\over 3} / \ln()$

$ {1\over e} \space ? \space {1\over 3} \ln(3) $

and now because ln(3) > 1, I assumed the right side was bigger, apparentaly it wasn't. For future references. How do I go about checking which one is bigger?

Answer 1

You have come up till comparing $e^\frac{1}{e}$ and $3^\frac{1}{3}$. Now, observe the function $x^\frac{1}{x}$. Prove that it takes the maximum value at e. This can be done by implicit differentiation.

Answer 2

Without finding max/minima:

We have $$e^{3\pi}>3^{e\pi}\impliedby e^3>3^e\impliedby\ln e^3>\ln 3^e\impliedby 3>e\ln 3$$ and since the Taylor series for $\ln x,\, x>0$ is $$\ln x=2\left[\frac{x-1}{x+1}+\frac13\left(\frac{x-1}{x+1}\right)^3+\frac15\left(\frac{x-1}{x+1}\right)^5+\cdots\right]$$ we plug in $x=3$ to get $$\ln3=2\left[\frac12+\frac13\left(\frac12\right)^3+\frac15\left(\frac12\right)^5+\cdots\right]=\sum_{k=0}^n\frac1{2k+1}\left(\frac14\right)^k$$ as $n\to\infty$. We can see that it converges to the value of $1.0986\cdots$ after trying for $n$ from $1$ to $12$.

Now $$e\ln3>2.72\times 1.1=2.992$$ so the result follows.

Answer 3

For positive $x,y$ we have $x^y>y^x\iff \ln (x^y)>\ln (y^x)\iff y\ln x>x\ln y$ because the $\ln$ function is strictly increasing on $\Bbb R^+.$

We have $y\ln x>x\ln y\iff \frac {\ln x}{x}>\frac {\ln y}{y}$ for positive $x,y.$

Let $f(z)=\frac {\ln z}{z}$ for $z>0.$ We have $f'(z)=\frac {1-\ln z}{z^2},$ which is negative for $z>e$ and is $0$ for $z=e.$ So $f(z)$ is strictly decreasing for $z\geq 1.$ Therefore $f(e)>f(3).$ We have $$f(e)>f(3)\implies\frac {\ln e}{e}>\frac {\ln 3}{3}\implies 3\ln e>e\ln 3\implies \ln (e^{3})>\ln (3^e)\implies$$ $$\implies e^3>3^e.$$ For positive $a,b,c$ we have $a>b \iff a^c>b^c,$ so $$e^3>3^e\implies (e^3)^{\pi}>(3^e)^{\pi}\implies e^{3\pi}>3^{e\pi}.$$