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

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    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?
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    Quote Originally Posted by mbailey14 View Post
    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
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    Quote Originally Posted by alexmahone View Post
    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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