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What is the solution of x^2=ln(x)

  1. Apr 2, 2013 #1
    I was just wondering whether anyone could tell me what's the solution of x^2=ln(x).

    This isn't a homework question, I need to know the solution of this equation however for some of the maths that I do in my free time.

    Also, I'd be interested to see how the solution is obtained.

    Thanks.
     
    Last edited: Apr 2, 2013
  2. jcsd
  3. Apr 2, 2013 #2

    chiro

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    Science Advisor

    Hey karpmage.

    In this case there is no solution.

    Consider the following argument.

    If x is a real number then x >= 1 since ln(1) = 0.

    However the derivative of x^2 is 2x and this is less than the derivative of ln(x) which is 1/x.

    2x > 1/x for all x >= 1 and since 1^2 > ln(1) = 0, you will never have a point of intersection for the relationship x^2 = ln(x).

    So as long as x is a real number, no solution exists.
     
  4. Apr 2, 2013 #3

    DrClaude

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    Staff: Mentor

    When you have a simple equation like that, you can also plot both sides and realize that the two curves can never cross.
     
  5. Apr 2, 2013 #4
    I should've mentioned this, I'm sorry, I'm looking for a complex solution to this problem. I was already aware that there wasn't a real solution for x.
     
  6. Apr 2, 2013 #5

    Curious3141

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    In which case you should define what ln x means, since the complex logarithm is multi-valued. Let's say you mean the principal value.

    Let x = a + bi. Then perform the necessary operations on each side of the equation (easier when x is expressed in polar (exponential) form for the RHS), equate real and imaginary parts to get simultaneous equations. Plug that into a numerical solver and see if you get any real solutions for a and b.
     
  7. Apr 2, 2013 #6

    DrClaude

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  8. Apr 2, 2013 #7

    Curious3141

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  9. Apr 2, 2013 #8

    DrClaude

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    Yeah, I'm getting lazy :blushing:
     
  10. Apr 2, 2013 #9

    Curious3141

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    Mathspeak for "lazy" is "elegant". :rofl:
     
  11. Apr 2, 2013 #10
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