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Simplifying logarithms

  1. Jun 26, 2006 #1
    my question is simplify:
    logpie(1-cosx)+logpie(1+cosx)-2logpie sinx (i dont know how to make the pie symbol)

    i thought it was going to be
    logpie-logpiecosx+logpie+logpiecosx-2logpiesinx
    =
    2logpie-2logpiesinx

    But, from my other post I was told you cannot do that with ln's.. is it the same for logs, that I cannot do that? If not, any suggestions on how i could simplify this?
     
  2. jcsd
  3. Jun 26, 2006 #2

    arildno

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    1. Open up your textbook where the rules for the arithmetic of logarithms stand.

    2. Post those rules in your next post.

    3. THINK about those very rules, and see how you may apply them!
     
  4. Jun 26, 2006 #3
    log(a+b) does not equal log(a)+log(b)
    rather,
    log(ab)=log(a)+log(b)
    be careful
     
  5. Jun 26, 2006 #4
    OK
    So laws of logarithms:
    1) loga1=0
    2)loga(xy)=logax+logay
    3) loga (1/x)=-logax
    4) loga(x/y)=logax-logay
    5)loga(x^y)=ylogax
    6)logax=logbx/logba


    so
    logpie(1-cosx)+logpie(1+cosx)-2logpie sinx

    using 2)loga(xy)=logax+logay
    logpie(1-cosx)(1+cosx)-2logpiesinx
    =-2logpiecosx-2logpiesinx
    then from 4) loga(x/y)=logax-logay
    logpie=-2cosx/-2sinx
    so logpie=cosx/sinx

    That seems better to me, or did I make a mistake somewhere?
     
  6. Jun 26, 2006 #5

    arildno

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    Great, star321!
    Now, for your mistakes:
    logpie(1-cosx)(1+cosx)-2logpiesinx
    =-2logpiecosx-2logpiesinx

    This is wrong!
    You have:
    [tex]\log_{\pi}((1-\cos(x))(1+\cos(x)))=\log_{\pi}(1-\cos^{2}(x))[/tex]
    This is NOT equal to [itex]2\log_{\pi}(\cos(x))[/itex]!!

    However, we DO have the trigonometric identity [itex]1-\cos^{2}(x)=\sin^{2}(x)[/itex]

    Therefore, you have:
    [tex]\log_{\pi}((1-\cos(x))(1+\cos(x)))=2\log_{\pi}(\sin(x))[/tex]

    Your application of the fraction rule is also faulty.
     
    Last edited: Jun 26, 2006
  7. Jun 26, 2006 #6
    just out of curiosity... why did you put pie instead of pi?
     
  8. Jun 26, 2006 #7
    oh is it sposed to be pi?? hehe.. i didnt know.
     
  9. Jun 26, 2006 #8

    TD

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    Yes, pie is something you eat. You don't eat pi :smile:
    You could try eating pi pies though :wink:
     
  10. Jun 26, 2006 #9

    HallsofIvy

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    It seems better?? What do you think "[itex]log_\pi[/itex]" MEANS??
    [itex]log_\pi[/itex] is a function. The function by itself is meaningless:
    [itex]log_\pi[/itex] of what?
     
  11. Jun 26, 2006 #10
    Could you? Is it really possible to eat Pi(Pies)?

    At some point your pie will need to be divided into such a small fraction that there is no way to seperate the individual pie mixture ( the collection of molecules that can be said to make up a pie ) into that small a part of the whole without it ceasing to be a true pie....

    So as you approach the 50th decimal expansion of Pi(Pies) you would be in the realm of protons, and as such, it would be impossible to distinguish a pie from a mellon....

    Is this correct?
     
  12. Jun 26, 2006 #11

    arildno

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    Nor would it be distinguishable from cows' dung. :smile:
     
  13. Jun 26, 2006 #12
    if you are in a non-Euclidean space-time where pi=3. You can eat 3 pies!
     
  14. Jun 28, 2006 #13
    How long would it take you to eat them?
     
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