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Ln(x) + x = c

  1. Feb 19, 2007 #1


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    i heard there is some way of solving for x, but i can't for the life of me remember how. please help!

    ln(x) + x = 10

  2. jcsd
  3. Feb 19, 2007 #2
    Yeah, I think it should be something like this:
    solution set for x: [e,9]
  4. Feb 19, 2007 #3


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    Graph intersection: y=ln x and y=10-x.
  5. Feb 19, 2007 #4

    Gib Z

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    >.< I don't think thats it mate.

    Theres no way to solve for x, you can only approximate it.
  6. Feb 19, 2007 #5
    newton's method works quite nicely here

    we will need, the function:

    f (x) = ln(x) + x -10 = 0

    and its first derivative:

    f'(x) = 1/x + 1

    an initial guess:

    x0 = 1

    and we can begin:

    x1 = x0 - f(x0)/f'(x0) = 5.5

    x2 = x1 - f(x1)/f'(x1) = 7.865213153

    x3 = 7.929390693

    x4 = 7.929420095

    x5 = x4 as far as my calculator is concerned

    try it out
  7. Feb 19, 2007 #6
    Just curious how you would use Newton's Method, Gib Z, since you do not use a calculator.
    Last edited: Feb 19, 2007
  8. Feb 19, 2007 #7


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    I wonder if you are not thinking of the "Lambert W function". W(x) is defined as the inverse of the function f(x)= xex.

    If ln(x)+ x= 10, then, taking the exponential of each side, eln(x)+ x= xex= e10.

    x= W(e10).

    Of course, the only way to evaluate that is to do some kind of numerical approximation as others have said,
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