A step-by-step guide to the half-life formula, the ln(2) relationship, and how to use it to predict medication decay over time.
If you want to predict how much of a drug remains in the body after a given number of days, the half-life formula is the single equation you need. It is a first-order exponential decay model that converts a half-life value (usually reported in hours or days) into a precise remaining-amount at any future time.
This guide explains the formula itself, where the natural-logarithm term comes from, how to rearrange the equation to solve for different unknowns, and how to apply it to real-world injection schedules. As with every article on DoseCurve, this is educational background reading, not a recommendation to use any specific compound or dose.
The amount remaining after time t is:
remaining = initial × (1 / 2)^(t / half_life)
Where:
The exponent t / half_life counts how many half-lives have elapsed. After one half-life the factor is (1/2)^1 = 0.5, after two it is (1/2)^2 = 0.25, and so on.
Pharmacokinetic textbooks usually write the same relationship with the base-e exponential, because it makes calculus and multi-dose superposition much cleaner:
remaining = initial × exp(−k × t)
The elimination rate constant k is derived from half-life as:
k = ln(2) / half_life
ln(2) is approximately 0.693147. The two forms are mathematically identical because exp(−ln(2) × t / half_life) = (1/2)^(t / half_life). The ln(2) form is what DoseCurve uses internally when it builds the cumulative curve from a series of injections.
A half-life is defined as the time at which exactly half the amount remains. Setting remaining = initial / 2 and solving for the rate constant gives:
initial / 2 = initial × exp(−k × half_life)
1 / 2 = exp(−k × half_life)
ln(1 / 2) = −k × half_life
−ln(2) = −k × half_life
k = ln(2) / half_life
That is the origin of the 0.693 factor you will see quoted in every pharmacology textbook.
You inject 200 mg of a compound with an 8-day half-life. How much remains after 16 days?
Using the power form:
remaining = 200 × (1 / 2)^(16 / 8)
remaining = 200 × (1 / 2)^2
remaining = 200 × 0.25
remaining = 50 mg
Using the exponential form:
k = 0.693147 / 8 ≈ 0.08664 per day
remaining = 200 × exp(−0.08664 × 16)
remaining = 200 × exp(−1.3863)
remaining = 200 × 0.25
remaining = 50 mg
You want to know when 200 mg will decay to 25 mg, given the same 8-day half-life. Rearrange the formula to solve for time:
remaining = initial × (1 / 2)^(t / half_life)
remaining / initial = (1 / 2)^(t / half_life)
ln(remaining / initial) = (t / half_life) × ln(1 / 2)
t = half_life × ln(remaining / initial) / ln(1 / 2)
Because ln(1 / 2) = −ln(2), this simplifies to:
t = half_life × ln(initial / remaining) / ln(2)
Plugging in the numbers:
t = 8 × ln(200 / 25) / 0.693147
t = 8 × ln(8) / 0.693147
t = 8 × 2.07944 / 0.693147
t = 8 × 3
t = 24 days
Three half-lives (24 / 8 = 3) reduce the amount by a factor of eight, from 200 mg to 25 mg.
You have been on 200 mg every 7 days (half-life 8 days) and stop completely. How much remains after 30 days?
Because the protocol has run long enough to reach steady state, the body is holding roughly the average steady-state amount rather than a single 200 mg dose. For a long-running repeated schedule the average level is approximately:
average ≈ dose_per_period / (1 − exp(−k × period))
With k = 0.693147 / 8 ≈ 0.08664 and period = 7 days:
average ≈ 200 / (1 − exp(−0.08664 × 7))
average ≈ 200 / (1 − 0.543)
average ≈ 200 / 0.457
average ≈ 438 mg
That is the approximate reservoir at steady state. Now applying 30 days of pure decay:
remaining = 438 × exp(−0.08664 × 30)
remaining = 438 × exp(−2.599)
remaining = 438 × 0.0743
remaining ≈ 32.5 mg
This is a rough estimate — the true value depends on the exact timing of the last injection relative to the 30-day mark — but it illustrates how the same formula extends from single-dose questions to protocol-level questions.
If you have two concentration measurements at known times and the drug is known to follow first-order kinetics, you can compute the observed half-life directly:
k = ln(C₁ / C₂) / (t₂ − t₁)
half_life = ln(2) / k
Where C₁ and C₂ are the concentrations at times t₁ and t₂. In practice, clinical pharmacokineticists use many data points and non-linear regression to estimate k more accurately, because assay noise and two-compartment distribution can distort a single pair of readings.
Mixing units. If half-life is in hours, time must also be in hours. Converting one and not the other is the most frequent arithmetic error.
Forgetting that half-life is a rate, not a duration. It does not mean "the drug is gone after one half-life." After one half-life, 50% remains. After five half-lives, roughly 97% has cleared.
Applying the formula to zero-order drugs. Ethanol, phenytoin at high doses, and some chemotherapy agents saturate their elimination pathways. Their decay is not exponential, so a single half-life value will drift as the amount changes. See Why first-order kinetics applies to most injectables for the distinction.
Ignoring accumulation on repeated dosing. A single-dose half-life calculation tells you nothing about the steady-state level you reach when you keep injecting. The same formula is used, but you must superpose (add) the decay curves of every prior dose. That is what DoseCurve does automatically when you enter a schedule.
The half-life formula lets you translate a pharmacological property into a visual dose curve. Two practical consequences:
Frequency is the knob that smooths the curve; half-life is the property of the compound that determines how much smoothing each extra injection provides. The DoseCurve calculator builds this relationship into the plotted line so you can see the trade-off visually.
DoseCurve is an educational tool. Nothing on this site is medical advice. Always consult a qualified clinician before starting, changing or stopping any protocol.