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Question about why ln(e^x) =/= x

  1. Mar 7, 2013 #1
    Why is ln(e^x) =/= x?

    The domain and range of the LHS are the same as the RHS, so I don't understand why this equation is false, where e^ln(x) = x, and the LHS and RHS of this does not have the same domain...

    I know that e^x and ln(x) are inverse functions, so please don't only tell me this. Why does e^ln(x) = x, while ln(e^x) =/= x?

    Like, I understand why sin^-1(sin(x)) =/= x, whereas sin(sin^-1(x)) = x, and this is because sin^-1(x) has a range of -pi/2 to pi/2
    Last edited: Mar 7, 2013
  2. jcsd
  3. Mar 7, 2013 #2
    They are equal.

    ln(e^x) = x*ln(e) = x

    What/who told you differently?
  4. Mar 7, 2013 #3
    I could have sworn wolfram alpha was telling me the equation is false, but I just tried again and it told me the equation is indeed true :blushing:
  5. Mar 7, 2013 #4


    Staff: Mentor

    It's the other way around that you run into domain considerations:

    $$ e^{\ln x} = x$$

    The left side is defined only for x > 0 (considering only the real-valued log function).
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