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Logarithmic lawsapplication

  1. Jan 12, 2014 #1
    I'm using this computer program:
    It tells me that:
    log(x+1/3)=log(3x+1)
    Why is that so? I plug in values: they give different answers.....!
     
  2. jcsd
  3. Jan 12, 2014 #2

    Mentallic

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    [tex]\log(3x+1)=\log(3(x+1/3))=\log(3)+\log(x+1/3)[/tex]

    Since [itex]\log(3)\neq 0[/itex] for any base of log, something more elusive must be happening.
     
  4. Jan 12, 2014 #3
    The reason I am asking this is that I am trying to calculate the integral of 1/(3x+1).
    I found it equals (1/3)(integral of (1/(x+1/3))=1/3ln((x+1/3)). But, the computer says it is wrong!
     
  5. Jan 12, 2014 #4

    Mentallic

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    [tex]\frac{d(\log{x})}{dx}=\frac{1}{x}[/tex]

    only applies when the base of log is e (usually denoted ln(x)). In general,

    [tex]\frac{d(\log_{a}{x})}{dx}=\frac{1}{x\ln{a}}[/tex] since [tex]\log_{a}{x}=\frac{\ln{x}}{\ln{a}}[/tex] hence [tex]\frac{d(\log_{a}{x})}{dx} = \frac{d(\frac{\ln{x}}{\ln{a}})}{dx}=\frac{1}{\ln{a}}\frac{d(\ln{x})}{dx}=\frac{1}{\ln{a}}\cdot\frac{1}{x}[/tex]

    Maybe your computer is using a base other than e?
     
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