Engineering
Exponential half-life model: remaining quantity and threshold time
Calculate an ideal exponential quantity and the time to a positive mathematical threshold.
A single-exponential decay example
Every half-life T_half reduces remaining quantity by 50%: N(t) = N0 * (0.5 ^ (t / T_half)) (equivalently N0 * EXP(-LN(2) * t / T_half)). Starting at N0 = 800 mg with a 5-hour half-life leaves 400 mg at 5h, 200 mg at 10h, and 100 mg at 15h (3 half-lives).
Inverting with natural logarithms LN solves the two reverse lab questions: (1) Threshold Time to drop to Target C2: t = T_half * LN(Target / N0) / LN(0.5); (2) Empirical Half-Life from two readings (N0 -> Nt over time t): T_half = t * LN(0.5) / LN(Nt / N0)!
| Half-Life Equation (N0 = 800 mg, T_half = 5 hrs) | Formula | Result |
|---|---|---|
| Remaining after t = 15 hrs | =800 * (0.5 ^ (15 / 5)) | 100.0 mg (12.5% left) |
| Hours to clear down to 25 mg | =5 * LN(25 / 800) / LN(0.5) | 25.0 hours (5 half-lives) |
| Decay Constant lambda | =LN(2) / 5 | 0.1386 hr^-1 |
Remaining quantity N(t) and threshold threshold-time formulas
Works seamlessly for fractional elapsed times t (e.g., t = 2.5 hours -> 0.5^0.5 = 70.71% remaining).
=ROUND(A2 * (0.5 ^ (C2 / B2)), 4)Solve elapsed time required to decay down to Target D2
Uses logarithm base-change LN(Target / N0) / LN(0.5) * HalfLife.
=ROUND(B2 * (LN(D2 / A2) / LN(0.5)), 4)Additional worked calculation
Using the initial example inputs:
=ROUND(C2 * (LN(0.5) / LN(100 / A2)), 4)Inputs and interpretation
This is an ideal single-exponential mathematical model with positive half-life, nonnegative initial quantity and elapsed time. The logarithmic time formula requires 0 < target <= initial quantity. Exact zero is approached asymptotically, not reached at a finite time. Do not use this example to choose medication dosing, clearance or radiation-safety thresholds.
Try the example
Use the original values or change them to test your understanding.