# Math Help - Prove witch is greater π^3 or 3^π using calculus? URGENT

1. ## Prove witch is greater π^3 or 3^π using calculus? URGENT

My professor gave us some homework. He asked us to prove witch is greater π^3 > 3^π or π^3 < 3^π using calculus? Typing it in the calculator tells me that 3^π is greater, but how do I prove this using calculus? All I need is some help or some one to point me in the right direction bcz I really don't understand how I prove this using calculus. What method should I use?

2. Originally Posted by mbailey14
My professor gave us some homework. He asked us to prove witch is greater π^3 > 3^π or π^3 < 3^π using calculus? Typing it in the calculator tells me that 3^π is greater, but how do I prove this using calculus? All I need is some help or some one to point me in the right direction bcz I really don't understand how I prove this using calculus. What method should I use?
Consider the function $y=\frac{lnx}{x}$.
$\frac{dy}{dx}=\frac{x.\frac{1}{x}-lnx*1}{x^2}$

$\frac{dy}{dx}=\frac{1-lnx}{x^2}$

$\frac{dy}{dx}<0$ in the interval $(e, \infty)$.

So $y=\frac{lnx}{x}$ is decreasing in the interval $(e, \infty)$.

$\pi>3$

$\implies \frac{ln3}{3}>\frac{ln\pi}{\pi}$

$\implies \pi ln3>3ln\pi$

$\implies ln3^{\pi}>ln\pi^3$

$\implies 3^{\pi}>\pi^3$

3. Originally Posted by alexmahone
Consider the function $y=\frac{lnx}{x}$ .
$\frac{dy}{dx}=\frac{x.\frac{1}{x}-lnx*1}{x^2}$

$\frac{dy}{dx}=\frac{1-lnx}{x^2}<0$ in the interval $(e, \infty)$
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