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Rules of logarithms

  1. May 18, 2013 #1
    Hi,

    So I'm doing boltzmann's entropy hypothesis.

    I have a basic question about the mathematics of logarithms.

    For [itex]\frac{ΔS}{K_B}=ln(W_f)-ln(W_i)[/itex], I do the correct maths and go [itex]\frac{ΔS}{K_B}=ln(\frac{W_f}{W_i})[/itex], and finally take the log of the equation to get:
    [tex]e^{\frac{ΔS}{K_B}}=\frac{W_f}{W_i}[/tex]
    This is correct, according to my worksheet.

    However, I was wondering why making [itex]ln(W_f)-ln(W_i)=ln(\frac{W_f}{W_i})[/itex] is necessary in order to get the correct equation. Why can't taking the log of [itex]ln(W_f)-ln(W_i)[/itex] work?

    If I do it, [itex]\frac{ΔS}{K_B}=ln(W_f)-ln(W_i)[/itex] is [itex]e^{\frac{ΔS}{K_B}}=W_f-W_i[/itex], which is incorrect.

    What is the mathematics behind this and what am I missing? Or is it just a general rule that you have to simplify the logs in order to proceed? It's bugging me.
     
  2. jcsd
  3. May 18, 2013 #2

    tiny-tim

    User Avatar
    Science Advisor
    Homework Helper

    hi theojohn4! :smile:

    eA+B isn't the same as eA + eB

    eAB isn't the same as eAeB

    similary, ln(A+B) isn't the same as lnA + lnB :wink:
     
  4. May 18, 2013 #3

    Mark44

    Staff: Mentor

    You're not taking the log of each side - you're exponentiating each side of the equation. That is, you are making each side the exponent of e. There's a big difference.
     
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