# Finding k in this logarithmic model

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.

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

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