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Homework Help: Solve e^(0.1x) = x

  1. May 19, 2010 #1
    Sorry, I'm not sure what is considered precalculus and what isn't, so hopefully this is in the right section.

    1. The problem statement, all variables and given/known data

    [tex]e^{0.1x} = x[/tex]

    2. Relevant equations

    3. The attempt at a solution
    [tex](1.105170918)^{x} = x[/tex]

    Unfortunately there is no context to the question, so I have no idea what approach should be taken. Can somebody explain what should be done?
  2. jcsd
  3. May 19, 2010 #2
    You can express the solution in terms of the Lambert-W function. Try and review this link and then study the examples in the Application section and see if you can then solve your problems.

  4. May 19, 2010 #3
    There is no analytic solution in terms of familiar mathematical functions, but there is a solution in terms of the Lambert W function. See


    Transform the equation to
    [tex](-.1 x) e^{-.1 x} = -.1[/tex]
    [tex]-.1 x = W(-.1)[/tex]
    [tex]x = -10 W(-.1) \approx 1.11833[/tex]

    [edit]Beaten to the punch![/edit]
    Last edited: May 19, 2010
  5. May 19, 2010 #4

    Char. Limit

    User Avatar
    Gold Member

    There are two solutions, not one:

    [tex]x=-10W(-.1) = 1.11833...[/tex]


    You can see that if you graph the two.
  6. May 20, 2010 #5
    I see no indication anywhere where only real values were requested and therefore there are actually an infinite number of solutions since the W function is infinitely-valued. For example, [itex] 44.491 - 73.0706 i[/itex] is an approximation to one of the complex values. Plot the real and imaginary components of the function and you'll see what I mean. :)

    But more importantly, does Stolen know how to arrive at that expression? Just divide by [itex]e^{ax}[/itex] then multiply by -a and get:


    which is now in a form that the W function can be taken on both sides (see reference);
    Last edited: May 20, 2010
  7. May 20, 2010 #6
    Well I can see how to get [tex]W(-0.1)[/tex] by approximating with Newton's method:

    [tex]w_{n+1} = w_{n} - \frac{w_{n}e^{w_{n}}-(-0.1)}{e^{w_{n}} + w_{n}e^{w_{n}}}[/tex]

    picking any number for [tex]w_{0}[/tex]

    How do I know that [tex]W_{-1}(-0.1)[/tex] exists if I don't graph it, as Char.Limit says? And how do I approximate [tex]W_{-1}(-0.1)[/tex]?
  8. May 20, 2010 #7

    D H

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    Staff Emeritus
    Science Advisor

    Use a fixed point iterator. You have [itex]x=e^{0.1x}[/itex], which is of exactly the right form needed for fixed point iteration, [itex]x=f(x)[/tex]. The only concern is stability, and for that you must have [itex]|f'(x_0)| < 1[/itex] and [itex]|f'(x_f)| < 1[/itex]. Here, x0 is the initial guess and xf is the final solution.
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