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Simple Logarithm

  1. Dec 5, 2008 #1
    1. The problem statement, all variables and given/known data
    It was found that the percentage of the carbon-14,P, contained in the bones of an animal n years after it has died is given by P = 2[tex]^{-kn}[/tex] , where k is a positive constant. The percentage of carbon-14 contained in the bones after the animal has been dead for 5668 years was 50% . How long, to the nearest year, was the animal dead if the percentage of the carbon-14 found in the bones was 76%?

    I tried however my K is negative.


    2. Relevant equations



    3. The attempt at a solution
     
  2. jcsd
  3. Dec 5, 2008 #2

    HallsofIvy

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    You tried? Good! Now, what did you try? Show exactly how you got K negative and we may be able to help you. (Perhaps you used P= 50 instead of 0.50?)
     
  4. Dec 5, 2008 #3
    why is it 0.5 and not 50? i thought they were asking for the percentage and not the fraction?

    The k i got is log , base 2, 50 over -5668
     
  5. Dec 5, 2008 #4
    You use 0.5 because 50%=50/100=5/10=1/2=2-1
    [tex]2^{5668k}=2^{-1}[/tex]
    Now you need to find k and then substitute in 2-kn=76/100 and find n
     
  6. Dec 5, 2008 #5

    HallsofIvy

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    Where did you get the idea that they were asking for a percentage? The problem specifically said " How long, to the nearest year". "How long" is NOT a percentage!
     
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