1. Limited time only! Sign up for a free 30min personal tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

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

    User Avatar
    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.
Share this great discussion with others via Reddit, Google+, Twitter, or Facebook