First-order elimination in one line
Most drugs at normal concentrations are eliminated by first-order kinetics: the body removes the same fraction per hour, not the same milligrams. After an intravenous bolus in a one-compartment model, the concentration falls exponentially: C(t) = C0 × e^(−k×t). k, in 1/h, is the elimination rate constant. The half-life is the time in which the concentration halves, and it is the same at any concentration.
Step 1: k from two concentrations
Divide the two levels and take the natural log: =LN(C2/C3)/(B3-B2) = LN(16 ÷ 6) ÷ (10 − 2) = 0.1226 per hour. Use LN, not LOG, and the time between the two samples, not the time since the dose. Every hour the drug loses about 11.5% of what is left (1 − e^(−0.1226)).
Step 2: the half-life
=LN(2)/B5 = 0.693 ÷ 0.1226 = 5.65 hours. The relation works both ways: k = 0.693 ÷ t½. A quick sanity check: the concentration fell from 16 to 6 mg/L in 8 hours, a bit more than one halving (16 to 8 mg/L) and less than two (16 to 4 mg/L), so a half-life between 4 and 8 hours is plausible.
Worked example: Drug Q, two samples after an IV bolus
Rows 2 and 3 are the two measured levels, rows 5 to 10 the results and rows 12 to 17 the concentration after whole half-lives. Drug Q is a placeholder.
| A | B | C | D | |
|---|---|---|---|---|
| 1 | Drug Q, IV bolus | Time (h) | Conc. (mg/L) | |
| 2 | Sample 1 | 2 | 16.0 | |
| 3 | Sample 2 | 10 | 6.0 | |
| 4 | Result | Value | ||
| 5 | Elimination rate constant k (1/h) | 0.1226 | ||
| 6 | Half-life t½ (h) | 5.65 | ||
| 7 | Back-extrapolated C0 (mg/L) | 20.45 | ||
| 8 | Reference concentration (mg/L) | 2.0 | ||
| 9 | Time after dose to reach it (h) | 19.0 | ||
| 10 | Share eliminated after 24 h | 94.7% | ||
| 11 | Half-lives | Time (h) | Conc. (mg/L) | Remaining |
| 12 | 0 | 0.0 | 20.45 | 100.0% |
| 13 | 1 | 5.7 | 10.22 | 50.0% |
| 14 | 2 | 11.3 | 5.11 | 25.0% |
| 15 | 3 | 17.0 | 2.56 | 12.5% |
| 16 | 4 | 22.6 | 1.28 | 6.3% |
| 17 | 5 | 28.3 | 0.64 | 3.1% |
- B5
=LN(C2/C3)/(B3-B2)Elimination rate constant k - B6
=LN(2)/B5Half-life - B7
=C2*EXP(B5*B2)Back-extrapolated C0 - B9
=LN(B7/B8)/B5Time to reach the reference level - C13
=$B$7*EXP(-$B$5*B13)Concentration after one half-life
Practice only: a placeholder drug with illustrative values. This is not clinical guidance or a dosing recommendation. Real dosing follows local protocols, the product information and the review of a qualified professional.
Step 3: back to C0 and forward to any level
Run the curve backwards to the time of the dose: C0 = C1 × e^(k×t1) = 16 × e^(0.1226 × 2) = 20.45 mg/L. C0 is used to estimate the volume of distribution, dose ÷ C0. Running it forwards, the time to fall to any level C is LN(C0 ÷ C) ÷ k: the concentration reaches 2 mg/L after 19.0 hours, and 94.7% of the drug is eliminated after 24 hours.
The half-life rule of thumb
The lower table steps through whole half-lives: 50% left after one (5.7 h), 25% after two (11.3 h), 12.5% after three (17.0 h), 6.3% after four and 3.1% after five (28.3 h). That is why a drug is often considered practically eliminated after four to five half-lives, and why repeated doses take about the same four to five half-lives to reach steady state.
Limits of the two-point method
Two samples give a quick estimate, but every measurement has noise. Pharmacokinetic studies fit many samples with a regression of LN(concentration) on time, the slope being −k, which the LINEST guide shows how to do in Excel. Take both samples in the terminal elimination phase: right after a dose, distribution into the tissues makes the concentration fall faster, and k would come out too high. Drugs with saturable elimination follow zero-order kinetics at high levels, and the half-life is then not constant.
Common mistakes
- Using LOG instead of LN. The formula needs the natural logarithm. With LOG (base 10), k comes out 2.3 times too small.
- Dividing by the time since the dose. k uses the time between the two samples, t2 − t1.
- Sampling during the distribution phase. Early samples fall faster than elimination alone explains. Use samples from the terminal phase.
- Mixing units. If time is in hours, k is per hour and the half-life is in hours. Minutes and hours in one formula give nonsense.
- Assuming the drug is gone after one or two half-lives. After two half-lives 25% is still there. It takes about five to get down to roughly 3%.
Frequently asked questions
How do you calculate the elimination rate constant from two concentrations?
Divide the earlier concentration by the later one, take the natural log and divide by the time between the samples: k = ln(C1 / C2) / (t2 - t1). The unit is 1 per time unit, for example per hour.
What is the formula for half-life from k?
Half-life = ln(2) / k, about 0.693 / k. A drug with k = 0.1226 per hour has a half-life of 5.65 hours.
How many half-lives until a drug is eliminated?
After five half-lives about 3% of the starting concentration remains, so four to five half-lives is the usual rule of thumb for practical elimination and for reaching steady state with repeated doses.
Does the half-life depend on the dose?
Not for first-order elimination: the same fraction is removed per hour at any concentration. For drugs with saturable, zero-order elimination at high levels, the apparent half-life gets longer as the dose increases.