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E^ln(x)=x Is it true always?

  1. Jul 4, 2011 #1
    consider this f(x)=eln(sin(x)) f:R-->R.
    Can we write this function defination like this f(x)=x f:R-->R
    I think no because if we put x as any negative number in first function(function in first line) then there will no solution exist for this but if we put x as any negative number in second function then their will be a solution.
    So does it mean that eln(x)=x Is not true always.
    ln represent natural log wih base e. R represent set of all real numbers.
     
  2. jcsd
  3. Jul 4, 2011 #2

    S_Happens

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    It's only true for x greater than zero, so the positive real numbers onto the real numbers. In my limited experience I believe it is true for any log base.
     
  4. Jul 4, 2011 #3
    If you extend the logarithms to negative numbers (to http://en.wikipedia.org/wiki/Complex_logarithm" [Broken]), you do have
    [tex]e^{\ln x}=x[/tex]
     
    Last edited by a moderator: May 5, 2017
  5. Jul 4, 2011 #4
    thanks for confirming my thought.

    I have not read any such thing.
     
    Last edited by a moderator: May 5, 2017
  6. Jul 4, 2011 #5

    HallsofIvy

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    Then I presume you haven't dealt with functions of a complex variable.

    For real numbers,
    [tex]e^{ln(x)}= x[/tex]
    for any x such that ln(x) is defined- i.e. for positive real numbers.
     
  7. Jul 4, 2011 #6
    To avoid getting into complex numbers, can't you just use [itex]x = -e^{\ln(-x)} [/itex] for [itex] x<0 [/itex]?

    Or for any nonzero (real) x, [itex] x = \frac{x}{|x|}e^{ln|x|}[/itex]?
     
    Last edited: Jul 4, 2011
  8. Jul 4, 2011 #7
    I apologize if I have gone too far, but my point is that if you somehow extend the logarithms to allow negative numbers, [itex]e^{\ln x}=x[/itex] is true for all x.
     
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