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Ln x/y = ln x - ln y From Integral 1/x dx only ?

  1. Jan 13, 2012 #1

    morrobay

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    Is the derivation of this relationship ln x/y = ln x - ln y
    exclusively from, and originating from the evaluation of the Integral from 1 --> x 1/t dt = ln x ?
    To say another way can ln x/y = ln x - ln y always be considered to apply to this integral ?
     
  2. jcsd
  3. Jan 13, 2012 #2

    Mentallic

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    Not really, there are other ways to show it's true, such as using basic exponential formulae:

    [tex]e^{x-y}=\frac{e^x}{e^y}[/tex]
     
  4. Jan 14, 2012 #3

    AlephZero

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    Yes you can show it direct from the integral.
    [tex]\int_1^x\frac{dt}t - \int_1^y\frac{dt}t = \int_y^x\frac{dt}t[/tex]
    Then substitute u = t/y.
     
  5. Jan 14, 2012 #4

    morrobay

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    Thanks to both of you . Im not questioning the validity of the correspondence to
    the integral: 1 to x 1/x dx = ln x - ln 1
    Rather confirmation that this correspondence is one to one .
    That ln x is derived from the above integral alone.
     
    Last edited: Jan 15, 2012
  6. Jan 14, 2012 #5

    Char. Limit

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    I'm really not sure what you mean. Do you mean that ln(x) is the only function f(x) that satisfies the property f(x/y) = f(x) - f(y)? Or do you mean that the integral is the only way of proving this property? Or perhaps something else entirely?
     
  7. Jan 15, 2012 #6

    Office_Shredder

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    I think he means don't use the fact that you know ln(x) is an inverse of ex
     
  8. Jan 15, 2012 #7

    morrobay

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    Im trying to show that this property : ln x/y = ln x - ln y
    is derived from the integral 1 to x 1/x dx

    see this http://www.eoht.info/page/S+=+k+ln+W
    I want to show that the natural logarithm in the statistical mechanics formulation
    of entropy change , delta S = k ln W2/W1
    is based on the natural logarithm for thermodynamic isothermal gas expansion:
    delta S = nR integral V1 to V2 dv/V = ln V2/V1
    That is based on the integral 1 to x 1/x dx = ln x - ln 1
     
    Last edited: Jan 15, 2012
  9. Jan 15, 2012 #8
    You can prove ln(x/y) = ln(x)-ln(y), its similar to proving ln(xy)=ln(x)+ln(y), but you would use a different substitution. Start with the integral between 1 and x/y, and then split it into two integrals one between x and 1 AND the other between x/y and x. Then for the second integral use the substitution t=u/x,
     
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