# Math Help - one more question on logs

1. ## 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

2. 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.