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Logarithm problem that I'm stuck on

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  1. Jul 6, 2015 #1

    Rectifier

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    The problem statement
    ln(x) = 5 -x
    Solve for x.

    The attempt at a solution
    [tex]ln x = 5 - x \\ e^{ln x} = e^{5 - x} \\ e^{ln x} = \frac{e^5}{e^x} \\ x e^x = e^5[/tex]
    Here is the place where I get stuck.
     
  2. jcsd
  3. Jul 6, 2015 #2

    DrDu

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    There is no analytical solution. You have to solve it numerically, e.g. ##x_{n+1}=5-\ln(x_n)## starting with ##x_0=5##.
     
  4. Jul 6, 2015 #3

    Orodruin

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    In addition to what DrDu said already, the formal solution to the problem is given in terms of the Lambert W function. As indicated in the previous post, you will have to find the solutions to this numerically.
     
  5. Jul 6, 2015 #4
    ##5 = x + ln(x)##

    Consider:
    ##y = 5##
    and
    ##y = x + ln(x)##

    Take an educated guess at a solution.

    ##y_n = x_n + ln(x_n)##
    _
    Consider the slope of the tangent line to this function at ##x_n##.

    The slope of the tangent line of this function at ##x_n## is approximately equal to the slope of the secant line of this function between ##x_n## and ##x_{n+1}##.

    Let ##y_{n+1} = 5##.

    Find your new ## x_{n+1}##.

    Repeat iteratively.

    Within 3 iterations (by hand), I converged on the solution.
     
  6. Jul 7, 2015 #5

    HallsofIvy

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    The "Lambert W function" that Orodruin referred to is defined as the inverse function to [itex]f(x)= xe^x[/itex]. So, immediately, x= W(e^5).
     
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