Exponential Drug Decay and Dose Timing
A patient receives an initial dose that produces a peak concentration of 80 mg/L. The drug’s concentration decays exponentially with a half-life of 6 hours, so $C(t)=80(0.5)^{t/6}$, where t is hours after the peak. A nurse wants to know when the concentration falls below 10 mg/L, what the concentration is at 9 hours, and whether a second identical dose given at 8 hours would be safe if the safety limit is 120 mg/L immediately after any dose. Assume concentrations add directly and the second dose produces the same 80 mg/L peak contribution.
1. Solve for the first time after the initial peak when the concentration falls below 10 mg/L.
Read the worked answer
Set $C(t)<10$:
$$80(0.5)^{t/6}<10.$$
Divide by 80:
$$(0.5)^{t/6}<0.125.$$
Since $0.125=0.5^3$,
$$(0.5)^{t/6}<(0.5)^3.$$
Because the base is less than 1, the inequality reverses when comparing exponents:
$$\frac{t}{6}>3 \quad\Rightarrow\quad t>18.$$
The concentration is exactly 10 mg/L at 18 hours and falls below 10 mg/L immediately after 18 hours.
2. Find the concentration at 9 hours.
Read the worked answer
Substitute $t=9$:
$$C(9)=80(0.5)^{9/6}=80(0.5)^{1.5}.$$
Since $0.5^{1.5}=\frac{1}{2\sqrt{2}}\approx 0.3536$,
$$C(9)\approx 80(0.3536)=28.3 \text{ mg/L}.$$
3. Determine whether a second identical dose at 8 hours keeps the immediate post-dose concentration below the 120 mg/L safety limit.
Read the worked answer
At 8 hours, the remaining concentration from the first dose is
$$C(8)=80(0.5)^{8/6}=80(0.5)^{4/3}.$$
Now $0.5^{4/3}=2^{-4/3}\approx 0.3969$, so
$$C(8)\approx 80(0.3969)=31.75 \text{ mg/L}.$$
Adding the second dose’s 80 mg/L peak contribution gives
$$31.75+80=111.75 \text{ mg/L},$$
which is below 120 mg/L. The second dose at 8 hours is safe under these assumptions, with about 8.25 mg/L of margin.