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Math Help - Tangent Curve to a Curve

  1. #1
    Super Member redsoxfan325's Avatar
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    Tangent Curve to a Curve

    I read that for any a>0, x^a > \ln x eventually, though it is possible that \ln x>x^a for finitely many x. I figured that \ln x = x^a has zero or two solutions for any given a>0 EXCEPT one. Graphing the equations with different values of a led me to conjecture that this occurs when a=\frac{1}{e}. Maple confirmed that the equation \sqrt[e]{x} = \ln x has one solution: x=e^e. I was wondering two things though:

    1) How do you analytically show that if \sqrt[e]{x} = \ln x, then x=e^e?

    2) Is there anything else that's special about the function \sqrt[e]{x}?
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  2. #2
    MHF Contributor
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    Quote Originally Posted by redsoxfan325 View Post
    I read that for any a>0, x^a > \ln x eventually, though it is possible that \ln x>x^a for finitely many x. I figured that \ln x = x^a has zero or two solutions for any given a>0 EXCEPT one. Graphing the equations with different values of a led me to conjecture that this occurs when a=\frac{1}{e}. Maple confirmed that the equation \sqrt[e]{x} = \ln x has one solution: x=e^e. I was wondering two things though:

    1) How do you analytically show that if \sqrt[e]{x} = \ln x, then x=e^e?

    2) Is there anything else that's special about the function \sqrt[e]{x}?
    Hi

    Let f(x) = \sqrt[e]{x} - \ln x = x^{\frac{1}{e}} - \ln x

    f'(x) = \frac{1}{e}\:x^{\frac{1}{e}-1} - \frac{1}{x}

    f'(x) = \frac{1}{ex}\:\left(x^{\frac{1}{e}} - e\right)

    f'(x) = 0 for x^{\frac{1}{e}} = e or x = e^e

    f is decreasing over )0 ; e^e] and increasing over [e^e ; +\infty(

    And strangely f(e^e) = 0
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