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Logarithm equation with different base

  1. Nov 9, 2012 #1
    1. The problem statement, all variables and given/known data
    Find x that satisfies:

    log11 (x3 + x2 - 20x) = log5 (x3 + x2 - 20x)


    2. Relevant equations
    logarithm


    3. The attempt at a solution
    x3 + x2 - 20x = 1

    x3 + x2 - 20x - 1 = 0

    Then stuck...
     
  2. jcsd
  3. Nov 9, 2012 #2
    Edit: Disregard that. I was incorrect.
     
    Last edited: Nov 9, 2012
  4. Nov 9, 2012 #3

    SammyS

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    Actually folks, this is correct:
    x3 + x2 - 20x = 1 .​
    The only way for log5(u) = log11(u) is for u = 1 .

    That cubic function does not factor nicely.
     
  5. Nov 9, 2012 #4
    Since RHS and LHS have different base, the only possible case for RHS = LHS is when the argument of the log = 1

    For your hint:
    11k = 5k is only correct for k = 0. It means that the argument of the log = 0, which is not correct since log 0 is undefined.

    Or maybe I get it wrong?

    Thanks
     
  6. Nov 9, 2012 #5

    ehild

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    You are not lucky with your book of problems. That equation has only complex roots. Check it at wolframalpha.com.
    http://www.wolframalpha.com/input/?i=x^3+%2B+x^2+-+20x+-+1+%3D+0

    ehild
     
  7. Nov 9, 2012 #6
    Hm...using my calculator, I got three real roots. They are:

    x = - 4.9776
    x = 4.02750
    x = - 0.04988

    And those numbers are really similar to the x-intercept of root plot (diagram) of wolframalpha. But I don't understand why there are also three alternatives complex solutions.

    OK, the point is this equation can't be solved manually.

    Thanks a lot for the help :smile:
     
  8. Nov 9, 2012 #7

    ehild

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    Well, even wolframalpha can be wrong.

    ehild
     
  9. Nov 9, 2012 #8

    SammyS

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    The fact that the roots given by WolframAlpha are complex, is likely an artifact of whatever numerical method is being used there. The imaginary part of the roots listed there is extremely small relative to the real part, with the real part matching the roots as you have listed them and the imaginary part on the order of 10-15 .
     
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