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Simple logarithm simplifications

  1. Jul 28, 2009 #1
    1. The problem statement, all variables and given/known data
    Hey all, I just need some help remembering my basic simplification rules.

    Can the expression eC*ln(x) be simplified anymore. It would certainly help out an integral I'm working on but I don't remember my simplification real well.

    2. Relevant equations

    3. The attempt at a solution
    I'm doubting there is a solution since eC*ln(x) = eclnx and I don't see how that helps much. But I thought I'd check with you guys and see if anyone had a better idea.
  2. jcsd
  3. Jul 28, 2009 #2


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    No, [itex]e^{C ln(x)}[/itex] is NOT equal to [itex]e^{C^{ln x}}[/itex]. I don't know where you got that. It is equal to [itex]e^{ln(x^C)}[/itex] (because [/itex]C ln(x)= ln(x^C)[/itex]).
    Last edited by a moderator: Jul 28, 2009
  4. Jul 28, 2009 #3
    ohhhhh, I see now, thanks!

    edit: also ecln(x) is equivalent to ecln(x)
    this is the same as saying that 23*2 = 232 = 64
    Last edited: Jul 28, 2009
  5. Jul 28, 2009 #4
    Are you sure? Because 23*2=26 and 232=29. What you said is equivalent to saying 4*6 = 46 or 64.

    As HallsofIvy said, ecln(x)=eln(xc). This can be simplified. Can you see how?
  6. Jul 28, 2009 #5


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    No, it is not. You are not distinguishing between

    What you are writing, eab, is equivalent to
    which is NOT the same as
    [tex]\left(e^a\right)^b= e^{ab}[/tex].
  7. Jul 28, 2009 #6
    ya, you're right, I meant to write what you have in the 2nd part there, I just wasn't liberal enough with parenthesis. Sorry, my bad.

    so what I should have written in the first place is ecln(x) = (ec)ln(x) , right?

    also, Kaimyn, eln(xc) simplifies to xc, if I remember my log stuff right (which obviously isn't a given) the e and the ln should essentially cancel each other out.

  8. Jul 28, 2009 #7


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    Yes, ec ln(x)= eln xc= xc.
  9. Jul 31, 2009 #8
    He thought:

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