Half-Life & Doubling Time

You’ll be able to

Derive a decay model from a stated half-life and use it to predict remaining amount
Compute doubling time from a growth rate and vice versa
Explain why half-life and doubling time do not depend on the starting amount

A fixed time for a fixed fraction

The defining feature of exponential change is that a fixed multiplier applies over a fixed interval, whatever the current amount. So the time to fall to half — the half-life — is the same whether you start with 10 grams or 10 tons. The same holds upward: doubling time is independent of the starting population. This is exactly what fails for linear change, where halving takes longer the more you have.

Half-life and doubling-time models

A(t) = A₀·(1/2)^(t/h) with half-life h · P(t) = P₀·2^(t/T) with doubling time T

The exponent t/h counts *how many half-lives have elapsed*. That reading makes the formula reconstructable from scratch.

Worked example

Carbon-14 has a half-life of about 5,730 years. A sample retains 34% of its original C-14. Estimate its age.

1.Model: A(t) = A₀(1/2)^(t/5730).
2.Retaining 34% means A(t)/A₀ = 0.34, so (1/2)^(t/5730) = 0.34.
3.Take logs: (t/5730)·ln(0.5) = ln(0.34).
4.t/5730 = ln(0.34)/ln(0.5) ≈ (−1.0788)/(−0.6931) ≈ 1.5565.
5.t ≈ 1.5565 × 5730 ≈ 8,919 years.

Answer: The sample is roughly 8,900 years old. The intermediate value 1.5565 is meaningful in itself: about one and a half half-lives have passed, which fits — one half-life would leave 50% and two would leave 25%, bracketing the observed 34%.

Tip

Always compute the number of half-lives (t/h) as a checkpoint. If the answer says 3.4 half-lives but the sample retains 60% of its material, something is wrong — 3.4 half-lives would leave under 10%.

Doubling time from a percentage rate

Given growth at rate r per period, the doubling time T solves (1 + r)^T = 2, so T = ln 2/ln(1 + r). For continuous growth at rate k the arithmetic is cleaner still: e^(kT) = 2 gives T = ln 2/k ≈ 0.693/k. Both say the same thing — doubling time and growth rate are inversely related, so doubling the rate halves the time.

Doubling time

T = ln 2 / ln(1 + r) for periodic growth · T = ln 2 / k ≈ 0.693/k for continuous growth

The familiar "rule of 70" is this formula in disguise: 100·ln 2 ≈ 69.3, so T ≈ 70/(percent rate) is a first-order approximation, good to a few percent for small rates.

Checkpoint

A population grows 4% per year. Approximately how long until it doubles?

Why the starting amount cancels

Set A₀·b^(t+h) = ½·A₀·b^t and the A₀ divides out of both sides immediately, leaving b^h = ½ — an equation with no A₀ in it. So the half-life depends only on the base, never on how much you started with. This is the algebraic content of "exponential," and it is why half-life is quoted as a property of an isotope rather than of a particular sample.

Worked example

A drug leaves the bloodstream with a half-life of 6 hours. A 400 mg dose is given. When does the amount fall below 50 mg, and how much remains after 24 hours?

1.Model: A(t) = 400(1/2)^(t/6) mg.
2.After 24 hours: 24/6 = 4 half-lives, so A = 400(1/2)⁴ = 400/16 = 25 mg.
3.For the threshold, note 50 mg is 400/8 = 400(1/2)³, which is exactly 3 half-lives.
4.3 half-lives is 18 hours, where A = 50 exactly.
5.So the amount is below 50 mg for t > 18 hours.

Answer: 25 mg remains after 24 hours, and the amount drops below 50 mg after 18 hours. Recognizing 50 as 400 divided by a power of 2 turns the threshold question into mental arithmetic — no logarithms needed.

Checkpoint

After 3 half-lives, what fraction of the original amount remains?

Answer the 2 checkpoints as you read.