Results 1 to 2 of 2

Thread: Tangent Curve to a Curve

  1. #1
    Super Member redsoxfan325's Avatar
    Joined
    Feb 2009
    From
    Swampscott, MA
    Posts
    943

    Tangent Curve to a Curve

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

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

    2) Is there anything else that's special about the function $\displaystyle \sqrt[e]{x}$?
    Follow Math Help Forum on Facebook and Google+

  2. #2
    MHF Contributor
    Joined
    Nov 2008
    From
    France
    Posts
    1,458
    Quote Originally Posted by redsoxfan325 View Post
    I read that for any $\displaystyle a>0, x^a > \ln x$ eventually, though it is possible that $\displaystyle \ln x>x^a$ for finitely many $\displaystyle x$. I figured that $\displaystyle \ln x = x^a$ has zero or two solutions for any given $\displaystyle a>0$ EXCEPT one. Graphing the equations with different values of $\displaystyle a$ led me to conjecture that this occurs when $\displaystyle a=\frac{1}{e}$. Maple confirmed that the equation $\displaystyle \sqrt[e]{x} = \ln x$ has one solution: $\displaystyle x=e^e$. I was wondering two things though:

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

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

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

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

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

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

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

    And strangely $\displaystyle f(e^e) = 0$
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. Tangent to Given Curve
    Posted in the Calculus Forum
    Replies: 3
    Last Post: Oct 8th 2010, 05:40 AM
  2. Replies: 6
    Last Post: Apr 7th 2010, 02:34 PM
  3. Tangent to a curve
    Posted in the Pre-Calculus Forum
    Replies: 7
    Last Post: Dec 27th 2009, 12:51 PM
  4. tangent to curve
    Posted in the Calculus Forum
    Replies: 7
    Last Post: Aug 6th 2009, 05:19 PM
  5. Tangent to a curve
    Posted in the Calculus Forum
    Replies: 1
    Last Post: Oct 31st 2008, 05:48 PM

Search Tags


/mathhelpforum @mathhelpforum