biologylabEXPLORE THE LIVING WORLDWorksheets Find the population after any number of years, how long it takes to reach a target at a given percentage growth rate, the doubling time, or the growth rate from two counts. Then compare an exponential J-curve with a logistic S-curve that levels off at the carrying capacity. Every answer shows the formula with your numbers in it.
r is the per-capita growth rate (births minus deaths per individual per time unit) written as a decimal: 5% per hour is 0.05. Exponential growth assumes unlimited resources, so it describes only the early, fast phase of a real population.
Pick the question your problem asks: how long until the population reaches a number, what it will be after t years, how long it takes to double, or what growth rate turns one count into another. Percentages compound once a year; the continuous-rate answer is shown alongside.
Enter the rate as a percentage per time period: 12 for 12% per year, −3 for a 3% decline. “Grows by 12% per year” means compound growth, N₀(1 + 0.12)^t; if your problem gives a continuous rate r and uses e, read the continuous answer shown alongside. Any time unit works as long as the rate and the time use the same one.
The questions students search for most, worked the way a teacher expects to see them. Load any of the first six into the solver above with its preset button.
Answer: about 21 years.
Answer: about 12 years.
Answer: about 6 years.
Answer: 8,800 people.
Answer: 1,086,400 people.
Answer: about 53 years.
Answer: 0.1 per year, a 10% annual increase (next year about 550 deer).
Answer: about 35 years.
Both models start from the same idea: the change in population size, dN/dt, depends on how many individuals are already there. Exponential growth keeps the per-capita rate r fixed no matter how crowded things get, which is only realistic while food, space and mates are plentiful. Logistic growth multiplies r by the “room left” factor (1 − N/K), so growth fades as the population nears the carrying capacity K. Switch between the two tabs above with the same numbers to see how much difference that single factor makes.
| Feature | Exponential growth | Logistic growth |
|---|---|---|
| Equation | dN/dt = rN, so N(t) = N₀e^(rt) | dN/dt = rN(1 − N/K), so N(t) = K / (1 + ((K − N₀)/N₀)e^(−rt)) |
| Curve shape | J-shaped: steeper and steeper | S-shaped (sigmoid): levels off at K |
| Key assumption | Unlimited resources, no crowding | Resources are finite, so per-capita growth falls as N rises |
| Growth rate dN/dt | Keeps rising with N | Largest at N = K/2 (equal to rK/4), zero at N = K |
| Doubling time | Constant: ln 2 / r | Only while N is far below K; doublings take longer near K |
| Typical examples | Bacteria in fresh broth, a species newly introduced to an island, the human population in the 20th century | Yeast in a flask, sheep in Tasmania, most lab cultures once the log phase ends |
A population that “grows by 12% per year” is multiplied by 1.12 once a year: N = N₀(1 + r)^t, the same compound-interest formula banks use. Solve it for time with logs, t = ln(N/N₀) ÷ ln(1 + r). Ecology textbooks also write growth as a continuous per-capita rate, N = N₀e^(rt), which compounds every instant; then t = ln(N/N₀) ÷ r. The same r gives a slightly faster rise in the continuous model, so 20,000 people reach 210,000 in 20.75 years at 12% compounded yearly but in 19.59 years at a continuous r = 0.12. If a question uses e or calls r the intrinsic or per-capita rate, use the continuous formula; if it says “percent per year,” use the compound one.
r is the birth rate minus the death rate per individual per unit of time, written as a decimal. A value of 0.1 per hour means each individual adds, on average, 0.1 new individuals every hour. Because growth compounds continuously, the population multiplies by e^0.1 ≈ 1.105 each hour, a 10.5% rise rather than exactly 10%. A negative r describes a shrinking population, and r = 0 means births and deaths balance. Keep r and t in the same time unit: a rate per day paired with a time in hours gives nonsense.
For exponential growth the doubling time is T₂ = ln 2 / r ≈ 0.693 / r. If you know the growth rate as a percentage, divide 69.3 by it (usually rounded to the “rule of 70”): a population growing at 1.1% per year doubles in about 63 years, and one growing at 2% doubles in about 35. The world population grew fastest, at roughly 2.1% per year, in the late 1960s, a pace that doubles a population every 33 years; the rate has since fallen below 1%.
| Growth rate per year | Doubling time (ln 2 / r) |
|---|---|
| 0.5% | about 139 years |
| 1% | about 69 years |
| 1.1% | about 63 years |
| 2% | about 35 years |
| 3% | about 23 years |
| 7% | about 10 years |
K is set by whatever runs out first: food, nesting sites, water, or the reach of predators and disease. Sheep introduced to Tasmania in the early 1800s followed a textbook S-curve, leveling off near 1.7 million by the 1850s. Populations can also overshoot: 25 reindeer released on St. Paul Island, Alaska, in 1911 grew to about 2,000 by 1938, stripped the lichen they depended on, and crashed to 8 animals by 1950. K is not a fixed number either; it moves with the seasons, rainfall and the arrival of competitors, which is why real curves wobble around the smooth line this calculator draws.
Wondering about the why? Start here.
Exponential growth, N(t) = N₀·e^(rt), assumes unlimited resources and gives a J-shaped curve that multiplies by the same factor every time unit. Logistic growth adds a carrying capacity K, dN/dt = rN(1 − N/K), so the curve is S-shaped and levels off as N approaches K. With N₀ = 100, r = 0.1 per hour and t = 10 hours, the exponential model gives about 272 individuals while the logistic model with K = 1,000 gives about 232.
Read: population growth modelsFor exponential growth the doubling time is T₂ = ln 2 / r ≈ 0.693 / r, where r is the per-capita growth rate as a decimal. A rate of 0.1 per hour doubles the population every 6.93 hours; the human population growing at 1.1% per year (r = 0.011) doubles in about 63 years. In the logistic model the same formula only applies while N is far below K.
Count bacterial doublingsCarrying capacity, K, is the largest population an environment can sustain over time given its food, space, and other limits. In the logistic model growth is fastest at exactly N = K/2 and slows to zero as N reaches K; a population that starts above K shrinks back toward it. The calculator reports how long it takes to reach K/2 and marks K on the graph as a dashed line.
See populations that cycle insteadDivide the target by the starting population, take the natural log, and divide by ln(1 + r) for a percentage rate compounded yearly: t = ln(N/N₀) ÷ ln(1 + r). A town of 20,000 growing 12% per year needs ln(10.5) ÷ ln(1.12) = 2.3514 ÷ 0.11333 = 20.75 years, about 21 years, to reach 210,000. If the rate is a continuous r in N = N₀e^(rt), use t = ln(N/N₀) ÷ r instead, which gives 19.59 years.
Solve your own word problemSubtract deaths from births and divide by the population size: r = (B − D) ÷ N. A population of 500 deer with 120 births and 70 deaths in a year has r = 50 ÷ 500 = 0.1 per individual per year, a 10% increase. From two counts t years apart, the continuous per-capita rate is r = ln(N/N₀) ÷ t and the yearly percentage is (N/N₀)^(1/t) − 1.
Read: population growth modelsThe rule of 70 estimates doubling time by dividing 70 by the percentage growth rate: a population growing 2% per year doubles in about 70 ÷ 2 = 35 years. It works because the exact doubling time is ln 2 ÷ r and ln 2 ≈ 0.693. The estimate is within about 2% for rates up to 10% per year; at 1.3% per year it gives 53.85 years against an exact 53.66 years compounded yearly or 53.32 years for a continuous rate.
Find a doubling time