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Is this equation possible to solve exactly?

  1. Jun 27, 2012 #1
    [itex]ln(x) = x - 2[/itex]

    Any ideas?
     
  2. jcsd
  3. Jun 27, 2012 #2
    I plugged it into wolfram alpha and it gave me two solutions in the form of values in the product log function: a function I had never heard of but seems to not be solvable exactly.
     
  4. Jun 27, 2012 #3

    phyzguy

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    What do you mean by "solve exactly"? It has two well-defined solutions, and you could calculate their values to as many decimal places as you want, but they are irrational numbers, so you could never write them down exactly, just like you could never write down pi or sqrt(2) exactly. Is it possible to solve x^2 = 2 exactly?
     
  5. Jun 27, 2012 #4
    You can write down pi and sqrt(2) exactly with no problems whatsoever. Here, I'll show you:

    pi
    sqrt(2)

    However, if you attempted to solve an equation who's answer was pi, but you had no way to analytically prove that it was pi, you would be unable to give an exact answer to the problem.

    I want to know if there is an analytic method to solve an equation like this, just like there are analytic methods to exactly solve many other equations with irrational answers.

    Back to the original question, do you have any ideas?

    If you think the answer to this problem truly is irrational and simply cannot be evaluated exactly (with the use of e, pi, sqrts, and other such symbols/constants), then please let me know the reasoning behind your hypothesis.

    e-
    Let me reword my question.
    Is it possible to analytically solve for x in the equation above? If so, how could it be done?
     
  6. Jun 27, 2012 #5

    phyzguy

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    I can also write down the solution to your equation. Here, I'll show you:

    -ProductLog(-1/e^2)

    Does that answer your question?
     
  7. Jun 27, 2012 #6
    Oh thank you so much :) I'm happy I finally know how to analytically solve an equation like this when I come across one. The method you taught me is so simple.
     
  8. Jun 27, 2012 #7
    You could use Newton's method to find an approximation.
     
  9. Jun 27, 2012 #8
    You know thats the first time I've ever seen a good legitimate use for newton's method. I guess there was a reason I learned it after all.
     
  10. Jul 4, 2012 #9

    Ray Vickson

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    You can solve it in terms of the Lambert W function. Here is what Maple gives:

    Solution 1:
    x = exp(-LambertW(-exp(-2))-2)

    Solution 2:
    x = exp(-LambertW(-1,-exp(-2))-2).

    RGV
     
  11. Jul 5, 2012 #10

    HallsofIvy

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    The "Lambert W function" is another name for the "product log" that phyzguy mentioned.
     
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