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Log of 12*e3c

  1. Dec 26, 2009 #1
    If you have to take the log of 12*e3c
    Can I do it like this: 12*ln(e3c)?
    So ln e fall away, so the result is: 12*3c = 36c
    Or do I have to do it like this: ln(12*e3c?

    I'm not sure of it.
     
  2. jcsd
  3. Dec 26, 2009 #2

    Mentallic

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    Re: Log

    Use the rule that [itex]log(ab)=log(a)+log(b)[/itex] and I'm sure things should fall into place :smile:
     
  4. Dec 26, 2009 #3
    Re: Log

    Thanks
     
  5. Dec 26, 2009 #4

    Mentallic

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    Re: Log

    No problem :smile:
     
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