Exponential Growth & Decay Explorer

model growth and decay with N(t) = N0 · e^(kt)

Worked example — Exponential Growth & Decay Explorer

N(t)	2225.5409
Doubling time	8.6643

These are the values the calculator shows when the page opens. Change any input above to recompute them.

Loading interactive simulation...

Lesson

The theory — Exponential Growth & Decay Explorer

Exponential change is what happens when the rate of change is proportional to the amount present — interest on a balance, decay of a nucleus, a population with room to spread. There is no separate formula for growth and for decay; there is one formula and a sign.

What each symbol means

N₀: the starting amount at t = 0. It scales the whole curve and changes nothing about its shape, which is why it does not appear in either of the two rate rows.
k: the continuous rate, and its sign decides everything. Positive and the amount grows without bound; negative and it falls towards zero without reaching it; exactly zero and nothing happens at all.
t: the elapsed time, in whatever unit k was quoted in. The two must match — a rate per year with a time in months is the commonest way to get a confidently wrong answer here.
N(t): the amount after that time. At N₀ = 1000, k = 0.08 and t = 10 it reads 2225.5409 — more than doubled, because the growth compounded continuously rather than once.

Where the formula comes from

"The rate is proportional to the amount" written down is dN/dt = kN. This is the only assumption; everything below follows from it.
Separating and integrating gives N(t) = N₀·e^(kt). The base e is not a choice of convenience — it is what the constant becomes when the growth is applied continuously rather than in steps.
For the doubling time, ask when N reaches 2N₀: e^(kt) = 2, so kt = ln 2 and t = ln 2 / k. The starting amount cancels, which is why doubling time is a property of the rate alone.
Half-life is the identical calculation with ½ in place of 2, giving t = ln 2 / |k| — the same expression, because halving at rate −k takes exactly as long as doubling at rate +k. At k = 0.08 the doubling row reads 8.6643; at k = 0.02, a quarter of the rate, it reads 34.6574, exactly four times as long.

How to read what you see

The two rate rows are mutually exclusive and the dash is doing real work. At k = 0.08 the doubling time reads 8.6643 and the half-life reads —; flip to k = −0.2 and the half-life reads 3.4657 while the doubling time is the one blanked out. A growing quantity has no half-life and a shrinking one has no doubling time, so the empty row is a statement rather than a gap.

Assumes: A rate that holds still for the whole span, and a quantity with nothing constraining it. Neither survives long in practice: populations meet a carrying capacity, and interest rates are renegotiated.
Breaks when: Set k to exactly 0 and both rate rows read — at once, which is the honest answer and worth pausing on. Nothing is broken: at zero rate the amount stays at 1000 for ever, so it never doubles and never halves, and ln 2 / 0 is undefined rather than infinite. The formula degenerates to a horizontal line, and the two blanks are the display refusing to divide by zero rather than failing to compute something.

Why dividing 70 by the percentage gives you the doubling time 🖖

The shortcut works because ln 2 = 0.693147, which is 70 in disguise once the rate is counted in percent rather than as a decimal. At 7% a year this tool reports a doubling time of 9.9021 while 70/7 says 10 — wrong by about one part in a hundred. At 8% it reads 8.6643 against 70/8 = 8.75, and the gap widens as the rate grows, because the approximation is only exact in the limit of small k. Bankers who quote 72 rather than 70 have not blundered: 72 divides cleanly by 2, 3, 4, 6, 8 and 9, and they are trading a little accuracy for arithmetic that can be done aloud.

Problems solved in full

A culture that starts at 1000 cells and grows at 8% per hour 5 steps
A culture starts at 1000 cells and grows at 8% per hour for 10 hours. Three people will give you three different answers and all three can defend theirs. Work out which one the exponential model makes, and how far apart they are.
1. The first answer treats 8% per hour as a fixed quantity: 80 cells an hour, 800 in ten hours. That is what "per hour" means for a salary, and it is what most people picture. It is also the only one of the three that is wrong on its own terms, because the cells produced in hour one are still dividing in hour two.
  
  $(i) N = N_{0} (1 + k t) = 1000 (1 + 0.8) = 1800$
2. The second fixes that by compounding once an hour — multiply by 1.08, ten times. Now the new cells contribute. This is a perfectly defensible model, and it is exactly right if the population happens to step at the top of each hour.
  
  $(ii) N = N_{0} (1 + k)^{t} = 1000 \times 1.0 8^{10} = 2158.92$
3. The third takes the limit. Cells do not wait for the clock, so the rate applies continuously, and that is the differential equation dN/dt = kN. Its solution is the exponential, and e^0.8 is the whole difference from the second answer.
  
  $(iii) \frac{d N}{d t} = k N ⟹ N = N_{0} e^{k t} = 1000 e^{0.8} = 2225.5409$
4. Doubling time comes out of the same solution, and N₀ cancels on the way. That is what makes exponential growth counter-intuitive: the time to double does not depend on how big the population already is, so a culture takes as long to go from a million to two million as it took to go from one thousand to two thousand.
  
  $N = 2 N_{0} ⟺ k t = ln 2 ⟹ t_{2} = \frac{ln 2}{0.08} = 8.6643$
5. The spread is the point. Continuous growth beats the naive reading by 24% and hourly compounding by 3%, over the same ten hours at the same quoted rate.
  
  $\frac{2225.54}{1800} = 1.236, \frac{2225.54}{2158.92} = 1.031$
Answer

The tool prints 2225.5409 and a doubling time of 8.6643 hours. The three candidate answers were 1800, 2158.92 and 2225.54 — the first and last differ by 425 cells, more than 40% of the starting population, and none of that gap comes from disagreeing about the rate. It comes entirely from disagreeing about when growth is credited. It is also where the "rule of 70" comes from: 70/8 = 8.75 against a true 8.6643, about 1% high, because ln 2 = 0.693 and somebody preferred a numerator with a lot of divisors.

$N (10) = 1000 e^{0.8} = 2225.5409, t_{2} = \frac{ln 2}{k} = 8.6643$
How long one cell lives, and when the curve stops counting 6 steps
The decay preset leaves 135.3353 of the original 1000 cells after ten hours, halving every 3.4657. Work out how long a single cell lives on average, and decide at what point the curve stops describing a culture at all.
1. N(t) = 1000e^(−0.2t), so at t = 10 the exponent is −2 and the panel reads 135.3353.
  
  $N (t) = N_{0} e^{k t} = 1000 e^{- 0.2 \times 10} = 1000 e^{- 2} = 135.3353$
2. The half-life comes from e^(−0.2T) = ½, giving T = ln 2 / 0.2 = 3.4657 in the row the doubling time has vacated. Ten hours is 2.8854 of those, and 1000 × 2^(−2.8854) is 135.3353 again — the same value reached the other way.
  
  $e^{- 0.2 T} = \frac{1}{2} \Rightarrow T = \frac{ln 2}{0.2} = 3.4657; \frac{10}{3.4657} = 2.8854, 1000 \times 2^{- 2.8854} = 135.3353$
3. The half-life is not the constant in the exponent. Write the same curve as N = N₀e^(−t/τ) and τ = 1/|k| = 5 hours. At five hours the count is 1000/e = 367.8794.
  
  $N = N_{0} e^{- t / τ}, τ = \frac{1}{∣ k ∣} = 5; N (τ) = \frac{1000}{e} = 367.8794$
4. τ is the average lifetime of one cell. The chance a given cell is still there at time t is e^(−t/τ), and integrating that from zero to infinity gives τ exactly. So τ/T = 1/ln 2 = 1.4427 for every exponential decay there is: the mean life always runs about 44% past the half-life, because the half-life is the median and a few long survivors pull the mean beyond it.
  
  $Pr (alive at t) = e^{- t / τ} ⟹ \overset{ˉ}{t} = \int_{0}^{\infty} e^{- t / τ} d t = τ = 5, \frac{τ}{T} = \frac{1}{ln 2} = 1.4427$
5. Now the emptying. The curve never reaches zero, but the flask does. The expected number of survivors is exactly N(t), so the last cell is expected to go where the curve crosses 1: t = τ ln N₀ = 5 ln 1000 = 34.5388 hours.
  
  $N (t) = 1 ⟺ t = τ ln N_{0} = 5 ln 1000 = 34.5388$
6. Somewhere between the two the readout stops being a count. At t = 30 the curve gives 2.4788, and no culture has ever held 2.4788 cells. That figure is an average over many repeats of the experiment; a single flask holds 0, 1, 2 or 3.
  
  $N (30) = 2.4788$
Answer

A cell lives 5 hours on average against a half-life of 3.4657, and the ratio 1/ln 2 is the same for carbon-14 and for caesium-137. The flask is expected to be empty at 34.5388 hours, which is not a new calculation but the curve's own crossing of N = 1. Starting with twice as many cells pushes that out by 5 ln 2 = 3.4657 hours, one extra half-life, whatever the starting count was — which is the doubling-time argument from the other preset with the sign reversed. Read N(t) as a count while it is large and as an expectation once it is small. The formula does not change; what it is telling you does.

$t_{last} = τ ln N_{0} ⟹ N_{0} \to 2 N_{0} buys τ ln 2 = T = 3.4657 hours$

References (2)

Where the exponential decay law was first established as a physical law rather than a curve fit: E. Rutherford & F. Soddy, “XLI. The cause and nature of radioactivity.—Part I.” The London, Edinburgh, and Dublin Philosophical Magazine and Journal of Science 4(21), 370–396, 1902.
The second half of the same argument, which introduces the transformation theory the half-life belongs to: E. Rutherford & F. Soddy, “LXIV. The cause and nature of radioactivity.—Part II.” The London, Edinburgh, and Dublin Philosophical Magazine and Journal of Science 4(23), 569–585, 1902.

By the 42Math team · Updated 2026-05-21

Reviewed against published sources. Corrections welcome. Editorial policy · Corrections

Example problems

steady growth - k = 0.08 turns 1000 into 2225.5409 by t = 10, doubling every 8.6643; the half-life row reads —
steady decay - k = −0.2 leaves 135.3353 of the original 1000, halving every 3.4657, and the doubling-time row goes blank instead
slow growth - k = 0.02 needs 34.6574 to double — exactly four times the 8.6643 of k = 0.08, because doubling time is ln(2)/k