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Solving a transcendental equation

  1. Apr 15, 2012 #1
    Suppose you have the equation

    [tex]x^2*(log x)^{1/3}=-C [/tex]

    for very small C. I have a book claims that asymptotically, for very small C, the solution is:

    [tex]x^2=C*\frac{2}{[logC]^{1/3} } [/tex]

    I'm not quite sure how to show this. If the 2 wasn't there, it looks like what they did was:

    [tex]x^2=-\frac{C}{(log x)^{1/3}} [/tex]

    and just substituted -C for x on the RHS:

    [tex]x^2=\frac{C}{(log C)^{1/3}} [/tex]

    I was thinking maybe you could substitute -C for x2 instead to get:


    [tex]x^2=\frac{2^{1/3}C}{(log C)^{1/3}} [/tex]
     
  2. jcsd
  3. Apr 23, 2012 #2
    I agree, there is a typo in the formula of the book : the power 1/3 is missing in the coefficient 2^(1/3)
    An even better approximate can be derived (in attachment)
     

    Attached Files:

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