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Basic Clarification on Laws of Logs

  1. Oct 5, 2014 #1
    Hi, I know the basic laws of logs but just wanted some clarification on a few things; your help would be much appreciated :) (warning, these questions may seem stupid!)

    I know that [itex] \log{xy} = \log{x} + \log{y} [/itex]
    but does [itex] 4\log{xy} = 4\log{x} + 4\log{y} [/itex] OR [itex] = 4\log{x} + \log{y} [/itex] ?

    Again, I know that [itex] \log{xy} = \log{x} + \log{y} [/itex]
    but does [itex]- \log{xy} = - ( \log{x} + \log{y}) [/itex] OR [itex]= - \log{x} + \log{y} [/itex]

    Again, I know these questions may seem stupid, my head tells me that it is the first case for both but I do not know for certain.

    Thanks :)
  2. jcsd
  3. Oct 5, 2014 #2


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    This won't be hard to figure out for yourself if you know how to approach it. Just test some values.

    Let's say that

    $$\log{x}=2, \log{y}=3$$

    So then it's clear that


    but is this equal to $$4\log{x}+\log{y}$$ or $$4(\log{x}+\log{y})$$?
  4. Oct 5, 2014 #3
    Of course :oops: . I was so caught up in trying to look at it algebraically that I did not even think to try and test it out with numbers!

    My heads not in the game this morning, so much stuff to do and so little time!

    Thanks :D
  5. Oct 5, 2014 #4


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    Your idea of putting parenthesis around the log is a good one. You can apply the rules inside the parenthesis and then use the law of distribution of multiplication over addition. That will help you in many situations, not just with logarithms.
  6. Oct 8, 2014 #5


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    What FactChecker said: 4 log(xy)= 4(log(x)+ log(y))= 4 log(x)+ 4 log(y).
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