Logarithms

[Vempy]

Question:
Coroners estimate the time of death from a body temperature using the simple rule that a body cools about 1degree in the first hour after death and about 1/2degree for each additional hour. The temperature is measured using a small probe inserted into the liver, which holds body heat well. Assuming air temperature of 20degrees and a living body of 37degrees, the temperature is given by T(t)=20+17e^-kt, where t=0 is the instant of death.
a) for what value of k will the body cool by 1degree in the first hour?
b) using the value of k found in a), after how many hours will the temperature of the body be decreasing at a rate of 1/2degree per hour?
c) using the value of k found in a), show that, after 24h, the coroner's rule gives approximately the same temperature as the formula.


Okay, so for the first part I got:
t=1
36=20+17e^-k
16=17e^-k
16/17=e^-k
log(16/17)=-kloge
log(16/17)/loge=-k

But then for the next part, b, you have to take the derivative right, using that value as the k? I think so. But we never did derivatives with log functions, only ln functions, so I don't know what to do. But I might not have done part a) right either.

Could someone please help.

 

[skeeter]

Question:
Coroners estimate the time of death from a body temperature using the simple rule that a body cools about 1degree in the first hour after death and about 1/2degree for each additional hour. The temperature is measured using a small probe inserted into the liver, which holds body heat well. Assuming air temperature of 20degrees and a living body of 37degrees, the temperature is given by T(t)=20+17e^-kt, where t=0 is the instant of death.
a) for what value of k will the body cool by 1degree in the first hour?
b) using the value of k found in a), after how many hours will the temperature of the body be decreasing at a rate of 1/2degree per hour?
c) using the value of k found in a), show that, after 24h, the coroner's rule gives approximately the same temperature as the formula.


Okay, so for the first part I got:
t=1
36=20+17e^-k
16=17e^-k
16/17=e^-k
log(16/17)=-kloge why log? why not ln?
log(16/17)/loge=-k
since ln(e) = 1, k = ln(17/16)

But then for the next part, b, you have to take the derivative right, using that value as the k? I think so. But we never did derivatives with log functions, only ln functions, so I don't know what to do. But I might not have done part a) right either.

d/dt[T(t)=20+17e^-kt]
dT/dt = -17k*e^-kt
now set dT/dt = -1/2 and solve for t

 

[Vempy]

Yeah, ln makes more sense, doesn't it. Hahaha.

Thank you so much!