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Math Help - one more question on logs

  1. #1
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    one more question on logs

    Hello all,

    I have one more that I tried but I dont think i did it right.

    1. Coroners often use liver temperature and Newton's Law of Cooling to estimate time of death. Newton's Law of Cooling is given by the fomula:

    T- T(base R) = {T (base 0) - T(base R)}e^kt

    where T is the temperature of the body at the time t, T(base R) is the room temperature, T(base0) is the body temperature at the time of death, and k is the relative rate fo cooling. A recently deceased body is found to be 28C at 3:30 pm, and 25C at 4:30 pm. The air temperature in the room is 20C. If the body's temperature was 37C at the time of death, approximately when did the individual die? (Hint: recall that logex= lnx)



    Thanks,
    David
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  2. #2
    Senior Member Danneedshelp's Avatar
    Joined
    Apr 2009
    Posts
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    Hmmm...I have not done one of these in a long time.
    I am going to use the equation in this form,
    u(t)=T+(u_{0}-T)e^kt
    its the same equation, just rearranged.

    Let u_{0}=28C, T=20C, t=60mins (time elapse from 3:30 to 4:30) and u(t)=25.
    Then,
    25=20+(28-20)e^{k60} \Rightarrow 5=8e^{k60} \Rightarrow \ln(\frac{5}{8})=k60 \Rightarrow \frac{\ln(\frac{5}{8})}{60}=k so , k\approx-0.007833
    Now, let u_{0}=37C, T=20C, k=\frac{\ln(\frac{5}{8})}{60} and u(t)=25.
    Then,
    25=20+(37-20)e^{\frac{\ln(\frac{5}{8})}{60}t} \Rightarrow \ln(\frac{5}{17})=\frac{\ln(\frac{5}{8})}{60}t \Rightarrow t= \frac{\ln(\frac{5}{17})}{\frac{\ln(\frac{5}{8})}{6  0}}\approx156.23mins

    So, the person died at about 1:06 PM

    Thats my stab at it.
    Last edited by Danneedshelp; May 29th 2009 at 09:55 AM.
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