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Simplify the logarithim expression

  1. May 30, 2010 #1
    not actually homework, but things I need to know:

    1: simplify the expression [tex] e^{a ln b} [/tex] (write it in a way that dosen't involve logarithims)

    i want to prove the identities:
    lnab = lna + lnb
    lna^b = blna
    [tex] \frac{d}{dx}lnx = \frac{1}{x} [/tex]

    i also want to derive the useful assumption
    [tex] ln(1+x) ~ aprrox = x [/tex]

    thank you in advance

    my working
    1: [tex] e^{a ln b} = e^a * e^{ln b} = e^a * b [/tex]
    that's as far as i got, i think that's correct but i'm not sure, because i used a way that INVOLVES logarithims

    2:proving lnab = lna + lnb
    not quite sure how to prove this..

    3:proving lna^b = blna
    again not sure how to prove this, i've just remembered the rule and assumed it was right

    4:proving d/dx lnx = 1/x
    should i be using the taylor series to prove these?
    i haven't officially been taught the taylor series yet, but i understand how it works
  2. jcsd
  3. May 30, 2010 #2


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    Re: logarithims !!!

    That's not correct. You're thinking of the rule ea+b=eaeb, but you're not adding a and ln b.
  4. May 30, 2010 #3
    Re: logarithims !!!

    is it not,

    b^a then?

    i know that is one of the rules, [tex] e^{a ln b} = b^a [/tex]
    i can't prove that it's correct but i remember that,
    so the first one is just b^a?
  5. May 30, 2010 #4
    Re: logarithims !!!

    Yes, because you already know that [tex] x^{yz} [/tex] can also be expressed as [tex] (x^y)^z [/tex] so you simply apply that to your problem and simplify the logarithm. The solutions to problems 2 and 3 can also be derived from these laws of exponents.
  6. May 30, 2010 #5
    Re: logarithims !!!

    can someone show me the proof for 3 please?
    only just managed to figure out a way for problem 2



    a*b = e^x *e^y
    a*b = e^(x+y)
    ln(a*b) = ln(e^x + e^y)
    ln(a*b) = x + y
    ln(a*b) = ln(a) + ln(y)
  7. May 31, 2010 #6


    Staff: Mentor

    Re: logarithims !!!

    For #3, let y = ln a.

    y = ln a <==> a = ey
    Then ab = (ey)b = eby

    Now, rewrite this exponential equation as a log equation, and the result you're looking for will fall out.
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