Second Dose Timing on the Ward
A patient is given a single 80 mg dose of a drug that clears from the bloodstream so that the amount remaining after t hours is M(t) = 80(0.5)^(t/4). Hospital policy allows a second 80 mg dose only once the residual amount has fallen to 15 mg, because the total drug in the bloodstream must never exceed the 95 mg safety threshold. A nurse proposes giving the second dose at t = 8 h, arguing that after two half-lives 'most of the drug is gone.' A pharmacy student is asked to check that proposal, to find the earliest safe time for the second dose to the nearest 0.1 h, and to rewrite the model as M(t) = 80e^(kt) because the hospital spreadsheet accepts only continuous-growth formulas with base e.
1. Check the nurse's proposed time of 8 h against the safety threshold, and comment on the reasoning 'most of the drug is gone.'
Read the worked answer
Evaluate the model at t = 8:
M(8) = 80(0.5)^(8/4) = 80(0.5)² = 80(0.25) = 20 mg
Adding a full 80 mg dose on top of 20 mg gives 100 mg in the bloodstream, which exceeds the 95 mg threshold by 5 mg. The proposal is unsafe and must be rejected.
The reasoning is also faulty in kind: two half-lives remove half, then half of the remainder, leaving one quarter of the original amount, not 'most' of it gone. Twenty milligrams is 25% of the dose, and because the second dose is large relative to the residual, a small error in timing translates into a direct overshoot of the limit.
2. Find the earliest time, to the nearest 0.1 h, at which the second dose may be given, and verify that the resulting total is within the threshold.
Read the worked answer
Require M(t) = 15:
80(0.5)^(t/4) = 15 (0.5)^(t/4) = 0.1875
Take logarithms of both sides (any base works; base 2 is cleanest): (t/4) log(0.5) = log(0.1875) t = 4 log(0.1875)/log(0.5)
t = 4 × (-1.6740)/(-0.6931) = 4 × 2.4150 ≈ 9.66 h
Equivalently t = 4 log₂(80/15) = 4 log₂(5.333) ≈ 9.66 h. Rounded to the nearest 0.1 h, the earliest time is 9.7 h.
Check: M(9.66) ≈ 15.0 mg, and 15.0 + 80 = 95.0 mg, exactly at the threshold and therefore permitted (the rule forbids exceeding 95 mg, not reaching it). Give at 9.7 h instead and the residual is about 14.9 mg, total 94.9 mg, still safe. Note the answer must be rounded up in practice, never down, or the threshold is breached.
3. Rewrite the model in the form M(t) = 80e^(kt) and explain how the two forms are related.
Read the worked answer
Write 80(0.5)^(t/4) = 80e^(kt). Matching the exponential parts:
e^(kt) = (0.5)^(t/4) = e^((t/4)ln 0.5)
so k = (ln 0.5)/4 = -0.6931/4 ≈ -0.1733, giving M(t) = 80e^(-0.1733t).
Check the half-life: M(4) = 80e^(-0.1733 × 4) = 80e^(-0.6931) = 80(0.5) = 40 mg, as required.
The two expressions are the same function written in different bases; no rounding of the original data is involved, only of k. The student should warn the ward that k must be stored to at least four decimal places, because a spreadsheet entry such as k = -0.17 would shift the half-life to about 4.08 h and make the 95 mg limit appear safe when it is not.