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How to solve x^x = 13 using logarithms

  1. May 24, 2006 #1
    I was trying to solve a problem using logarithm. It says x^x =13 I tried to solve it using logarithm but I couldn't. I used a graphing calculator solver and I found it is 2.6410619... Is it possible to solve using logarithm? If not is there another way to solve besides trying and checking?
     
  2. jcsd
  3. May 24, 2006 #2
    I'm not aware of any way to solve such a problem other than numerically (IE using some numeric approximation). Ah I see VietDao has a solution.
     
    Last edited by a moderator: May 24, 2006
  4. May 24, 2006 #3

    VietDao29

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    Homework Helper

    You can solve it by Lambert W-function, i.e the inverse function of f(x) = xex.
    [tex]x ^ x = 13[/tex]
    [tex]\Leftrightarrow \ln (x ^ x) = \ln (13)[/tex]
    [tex]\Leftrightarrow x \ln x = \ln (13)[/tex]
    [tex]\Leftrightarrow (\ln x ) e ^ {\ln x} = \ln (13)[/tex]
    [tex]\Leftrightarrow \ln x = W( \ln (13))[/tex]
    [tex]\Leftrightarrow x = e ^ {W( \ln (13))}[/tex]
    ------------------
    Or you can get an approximation of it by Newton's Method.
    First, we'll try to rearrange the equation to give the form of f(x) = 0
    So
    [tex]\x ^ x = 13[/tex]
    [tex]\Leftrightarrow \x ^ x - 13 = 0[/tex]
    Let f(x) = xx - 13
    Then we differentiate f(x) with respect to x to give:
    f'(x) = xx(ln(x) + 1)
    And we choose an arbitrary x0, say x0 = 2 (we should graph it first to chose x0 precisely, we choose x0 since there's one solution near 2).
    And we apply the formula:
    [tex]x_{n + 1} = x_n - \frac{f(x_n)}{f'(x_n)}[/tex], and let n increase without bound to obtain the desired solution, i.e:
    [tex]x = \lim_{n \rightarrow \infty} x_n[/tex].
    The root of that equation is about 2.641061916.
    :)
     
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