Problem 6: The amount of a certain medicine in the bloodstream decays exponentially with a half-life of 5 hours. In order to keep a patient safe during a one-hour procedure, there needs to be at least 50 mg of medicine per kg of body weight. How much medicine should be administered to a 60kg patient at the start of the procedure?

(from MIT 18.03 OCW)

Let $x(t)$ be the amount of the medicine in mg present in the bloodstream at time $t$ in hours.

Then we have the model: $x(t) = x_0e^{-kt}$

I wonder how in the solution they found $k = \frac{\ln2}{5}$.

How I tried: We know that after $t=5$ hours, the initial amount is halved, so $x(5) = (1/2)x_0$.

Then we use that information to solve for $k$:

$$ (1/2)x_0=x_0e^{-k(5)} $$

And then:

$$ ln(1/2) = -5k $$

Etc. so that $k=-\frac{\ln(1/2)}{5}$.

However in the given solution they leave out the minus sign when finding k (which I can understand) and more importantly, they find $k = \frac{\ln2}{5}$, where I have $\ln(1/2)$ in the numerator.

  • $\begingroup$ Note that $\ln a =-\ln a^{-1}$, so both answers are same. $\endgroup$
    – Anurag A
    Jul 31 '19 at 0:41

$$ - \frac{\ln(1/2)}{5} = - \frac{\ln 2^{-1}}{5} = \frac{\ln (2^{-1})^{-1}}{5} = \frac{\ln 2}{5}. $$


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