Knowing two measurements of a decaying quantity is enough to recover its decay constant. The continuous decay model x(t) = x₀ × e^(-kt) rearranges to k = -ln(x(t) / x₀) / t. Once you have k, the per-period decay rate is r = 1 - e^(-k) and the half-life is t½ = ln(2) / k. If you already know the half-life and want to project a remaining amount at a specific future time instead, use the half-life calculator.
Worked example
A radioactive sample reads 1,000 counts per minute at noon and 720 cpm 4 hours later. k = -ln(0.72) / 4 = 0.0822 per hour. Half-life = 0.693 / 0.0822 = 8.43 hours. Once you have that k value, feed it into the exponential decay calculator to project the remaining count at any future time, not just at the moment you took the second reading.
When to use this calculator
Use it when you have a known starting amount and one later observation: medication blood levels, sensor calibration drift, environmental contaminant clearance, cooling temperature differentials, or any process where you suspect first-order kinetics but don't yet know the rate.
From k to per-period rate
If you prefer to report the loss as "X% per hour" rather than a continuous constant, convert with r = 1 - e^(-k). A k of 0.05 per hour reports as 4.88% per hour, slightly less than 5% because compounding within the period eats a fraction.
A sensor drift example
Sensor calibration drift often shows up as a slowly changing signal rather than a sudden failure. If a sensor reads 5.00 V when new and 4.82 V after 1,000 hours of continuous operation, the implied decay constant is k = -ln(4.82/5.00) / 1000 ≈ 3.67 × 10⁻⁵ per hour. Annualized over 8,760 hours in a year, that corresponds to roughly 27.5% drift per year, information a maintenance team can use to schedule recalibration before the reading drifts outside its accuracy specification.
Checking your answer
A quick sanity check on any computed k is to plug it back into x(t) = x₀ × e^(-kt) and confirm it reproduces x₁. Using k ≈ 0.1022 from the 100-to-60-over-5-hours example above, x(5) = 100 × e^(-0.1022 × 5) = 100 × e^(-0.511) ≈ 100 × 0.5999 ≈ 60.0, matching the original later reading and confirming the arithmetic is correct.
If the later reading x₁ turns out larger than x₀, the computed k comes out negative, which simply means the quantity grew instead of decayed over the interval. In that case, take the absolute value of k and switch to the exponential growth calculator instead, since a negative decay constant and a positive growth rate describe the same upward curve.
Measurement error and small samples
Small measurement errors matter more than they first appear because the formula uses a logarithm. Suppose the later reading of 60 actually carries a plus or minus 2 measurement uncertainty. Using 58 instead of 60 changes k from 0.1022 to -ln(0.58)/5 ≈ 0.1089 per hour, a swing of about 6.6%. Repeated measurements or a longer elapsed time both tighten the estimate as a result.
The units of k always match the units of 1/t, so a decay constant computed from hours must be converted before comparing it to one computed from days. A k of 0.12 per hour equals 0.12 × 24 = 2.88 per day, not 0.12 per day, and mixing units like this is one of the most common mistakes when comparing decay constants pulled from different sources.
FAQ
How do I find the decay constant from two data points?
Take the natural log of the ratio between the later reading and the starting reading, then divide by negative the elapsed time: k = -ln(x₁/x₀) / t. For example, with x₀ = 100, x₁ = 60 and t = 5 hours, k = -ln(0.6) / 5, which works out to about 0.1022 per hour. No calculus is needed, just one logarithm and one division.
What is the relationship between k and r?
The continuous decay constant k and the per-period decay rate r describe the same decline in two different units. They convert with r = 1 - e^(-k) and, going the other way, k = -ln(1 - r). A decay constant of k = 0.1 per hour corresponds to a per-period rate of about 9.52%, meaning the quantity falls by roughly that fraction each hour.
Can the decay constant be derived from half-life?
Yes, directly. The formula is k = ln(2) / t½, where t½ is the half-life in whatever time unit you are working with. A substance with a 6-hour half-life has k = 0.693 / 6, which equals 0.1155 per hour. This shortcut skips the two-measurement calculation entirely whenever the half-life is already published or known.
Why use the natural logarithm?
Continuous decay follows the differential equation dx/dt = -kx, and its solution is the exponential x = x₀ × e^(-kt). Since exponentials and natural logarithms are inverse operations, undoing that e^(-kt) term to solve for k or t requires taking the natural log of both sides. There is no way to isolate the exponent using ordinary algebra alone.
What if I have more than two data points?
With three or more measurements, exponential regression gives a more reliable estimate than picking just two points. Take the natural log of every y-value, fit a straight line to ln(y) versus x by least squares, and the resulting slope is the decay (or growth) constant while e raised to the intercept recovers the initial value. This averages out measurement noise better than a single pair.
Where is decay constant used?
The decay constant appears wherever a first-order process loses a fixed fraction of itself per unit time: radioactive isotope characterization, drug elimination in pharmacokinetics, capacitor discharge through a resistor, contaminant clearance from soil or water, and Newton's law cooling curves. Engineers often report the reciprocal quantity instead, the time constant τ = 1/k, which has units of time.