BiologyCBSE Class 12 · Organisms and Populations

Population Growth: Exponential and Logistic Models

Updated 15 October 2026 · Fact-checked

A population's size changes through natality, mortality, immigration and emigration. With unlimited resources it grows exponentially, dN/dt = rN, giving a J-shaped curve. With limited resources it grows logistically, dN/dt = rN(K − N)/K, giving an S-shaped (sigmoid) curve that levels off at the carrying capacity K. The logistic model is more realistic.

Both curves start alike, but the J-curve keeps soaring while the S-curve flattens at the carrying capacity K.

Illustration drawn for QuestionBuddy.

Understand Population Growth: Exponential and Logistic

Population attributes. A population has features an individual does not:

Birth rate and death rate, expressed per capita (per individual). Example: 20 lotus plants last year + 8 new = 28; birth rate = 8/20 = 0.4 offspring per lotus per year. If 4 of 40 fruit flies die in a week, death rate = 4/40 = 0.1 per fruit fly per week.
Sex ratio (e.g. 60% females, 40% males).
Age distribution: plotted as an age pyramid (pre-reproductive, reproductive and post-reproductive groups). Shapes: expanding (triangular, broad base: growing), stable (bell-shaped), declining (urn-shaped, narrow base).
Population density (N): usually numbers, but sometimes percent cover or biomass is more meaningful (e.g. one huge banyan tree vs 200 carrot grass plants). Relative density can be enough, and density can be estimated indirectly, e.g. tigers from pug marks and faecal pellets.

Four processes change population size.

Increase: natality (B) (births) and immigration (I) (arrivals from elsewhere).
Decrease: mortality (D) (deaths) and emigration (E) (departures).

N(t+1) = N(t) + [(B + I) − (D + E)]. Under normal conditions births and deaths matter most; immigration can matter when a new habitat is being colonised.

1. Exponential growth (unlimited resources). When food and space are unlimited, each species can realise its full innate potential to grow (Darwin noted this). With per capita birth rate b and death rate d:
dN/dt = (b − d) × N = rN, where r = b − d is the intrinsic rate of natural increase, a key measure of how factors affect growth. Examples: Norway rat r = 0.015; flour beetle r = 0.12; human population of India in 1981, r = 0.0205. Integrated: ** $N_{t} = N_{0} e^{r t}$ (e = 2.71828). The plot of N against time is J-shaped**; any species growing like this would soon reach enormous numbers.

2. Logistic growth (limited resources). No natural population has unlimited resources, so individuals compete, and the 'fittest' survive and reproduce. A habitat can support only a maximum number: its carrying capacity (K). Growth shows a lag phase, then acceleration, then deceleration, and finally an asymptote at K. The plot is sigmoid (S-shaped). This is the Verhulst-Pearl logistic growth:
dN/dt = rN (K − N)/K.

When N is small, (K − N)/K ≈ 1 and growth is nearly exponential.
When N = K/2, the growth rate is greatest.
When N = K, growth stops (dN/dt = 0).

Since resources for most animals are finite and become limiting sooner or later, the logistic model is more realistic.

Life history variation. Populations evolve to maximise reproductive fitness (Darwinian fitness, high r) in their habitat: some breed once in a lifetime (Pacific salmon, bamboo), others many times (most birds and mammals); some produce many small offspring (oysters, pelagic fishes), others few large ones (birds, mammals).

Flow chart: N(t) → + births + immigrants − deaths − emigrants → N(t+1); resources unlimited → J-curve (rN); resources limited → S-curve (rN(K − N)/K) → plateau at K.

Draw and label: growth curves. Checklist: axes with population density (N) on the y-axis and time (t) on the x-axis; curve (a) rising ever more steeply (J-shape); curve (b) an S-shape levelling off at a dashed horizontal line. Labels: exponential (dN/dt = rN), logistic (dN/dt = rN(K − N)/K), carrying capacity (K), lag phase, acceleration, deceleration, asymptote.

Population Growth: Exponential and Logistic: key facts

Population change: $N_{t + 1} = N_{t} + [(B + I) - (D + E)]$; B births, I immigrants, D deaths, E emigrants.
Per capita rates: $b = \frac{births}{N}, d = \frac{deaths}{N}$; Lotus: 8/20 = 0.4 per year; fruit flies: 4/40 = 0.1 per week.
Exponential growth: $\frac{d N}{d t} = r N, r = b - d, N_{t} = N_{0} e^{r t}$; J-shaped curve.
Logistic growth: $\frac{d N}{d t} = r N (\frac{K - N}{K})$; S-shaped curve; growth stops at N = K; fastest at N = K/2.
Values of r (NCERT): $Norway rat 0.015; flour beetle 0.12; India (1981) 0.0205$; r = intrinsic rate of natural increase.

How to answer Population Growth: Exponential and Logistic questions

Population numericals use three formulas. Identify which one the question needs, then substitute carefully.

1For birth or death rate, divide the number of births or deaths by the starting population, and state the unit (per individual per time).
2For change in numbers, use N(t+1) = N(t) + (B + I) − (D + E).
3For exponential growth, compute r = b − d, then dN/dt = rN; for longer times use $N_{t} = N_{0} e^{r t}$ .
4For logistic growth, compute the factor (K − N)/K first, then multiply by rN.
5For graphs, J-shape = exponential (no limit); S-shape = logistic (levels at K); mark lag, acceleration, deceleration, asymptote.
6For age pyramids, read the base: broad = expanding, medium = stable, narrow = declining.

Population Growth: Exponential and Logistic in real life

The idea, met in an everyday situation. Answer the question in your own words, then compare with the model answer. The one-line hook is what to remember.

Logistic growth and carrying capacity

Suppose a lake stocked with 200 fish has an estimated carrying capacity of 2000 fish, and the fish population's intrinsic rate of natural increase is r = 0.5 per year.

Compare the growth rate (dN/dt) when the population is 200, 1000 and 1900. At which size would harvesting give the most sustainable yield, and why?

Try it first, then show the solutionSolution

N = 200: dN/dt = 0.5 × 200 × (2000 − 200)/2000 = 100 × 0.9 = 90 fish per year.
N = 1000: dN/dt = 0.5 × 1000 × (1000/2000) = 500 × 0.5 = 250 fish per year.
N = 1900: dN/dt = 0.5 × 1900 × (100/2000) = 950 × 0.05 = 47.5 fish per year.
Growth is greatest at N = K/2 = 1000; keeping the stock near 1000 and harvesting about the yearly growth gives the most sustainable catch.

Answer: 90, 250 and 47.5 fish per year; maximum at N = K/2 = 1000, the best level for sustainable harvest.

Memory hook: Half full grows fastest; full stops growing.

Tricks to remember Population Growth: Exponential and Logistic

Short memory aids. Say each one aloud once, then use it on a question the same day.

BIDE
Births and Immigrants add; Deaths and Emigrants subtract.
Use it when: N(t+1) equation.
J for Jumping, S for Settling
Exponential jumps up (J); logistic settles at K (S).
Use it when: Matching curves to models.
K is the ceiling
(K − N)/K is the 'room left'; no room, no growth.
Use it when: Logistic formula.

Population Growth: Exponential and Logistic: solved examples

Ordered from board level to HOTS. Try each one before you open the solution.

Example 1: Birth rate · 1 mark · Very short answerBoard level

A pond had 20 lotus plants last year. Through reproduction 8 new plants were added this year. Calculate the birth rate.

Try it first, then show the solutionSolution

Birth rate = new individuals ÷ original population = 8/20.

Answer: 0.4 offspring per lotus per year.

Example 2: Population after a year · 2 marks · Very short answerBoard level

Suppose a deer population of 500 has 120 births, 80 deaths, 30 immigrants and 20 emigrants in a year. Find the population at the end of the year and the per capita birth and death rates.

Try it first, then show the solutionSolution

N(t+1) = 500 + (120 + 30) − (80 + 20) = 500 + 150 − 100 = 550 (1 mark).
b = 120/500 = 0.24; d = 80/500 = 0.16 per deer per year (1 mark).

Answer: 550 deer; b = 0.24, d = 0.16 per deer per year.

Example 3: Exponential vs logistic · 5 marks · Long answerHarder board level

A population is growing in a habitat where food is limited. (a) Sketch its growth curve and name its four phases. (b) Write its growth equation and contrast it with the equation and curve expected if resources were unlimited. (c) Which of the two models is more realistic for animal populations, and why?

Try it first, then show the solutionSolution

Exponential: resources unlimited; dN/dt = rN (r = b − d); J-shaped curve (1 mark).
Logistic: resources limited; dN/dt = rN(K − N)/K; S-shaped (sigmoid) curve (1 mark).
Logistic phases: lag, acceleration, deceleration, asymptote at carrying capacity K (1 mark).
Correctly labelled sketch of both curves with K marked (1 mark).
Logistic is more realistic because resources for most populations are finite and become limiting, leading to competition (1 mark).

Answer: J-curve (rN, unlimited) vs S-curve (rN(K − N)/K, limited); logistic is realistic.

Example 4: Computing r and growth · 2 marks · Very short answerHarder board level

In a population of 1000 flour beetles, the per capita birth rate is 0.20 and death rate 0.08 per week. Find r and the rate of increase this week. How does r compare with the NCERT value for flour beetles?

Try it first, then show the solutionSolution

r = b − d = 0.20 − 0.08 = 0.12; dN/dt = 0.12 × 1000 = 120 beetles per week (1 mark).
This matches the NCERT value of r = 0.12 for the flour beetle (1 mark).

Answer: r = 0.12; increase of 120 beetles per week; same as the NCERT value.

Example 5: Exponential projection · 3 marks · Short answerHOTS / competency

A bacterial culture of 1000 cells grows exponentially with r = 0.693 per hour. Using $N_{t} = N_{0} e^{r t}$ , find the population after 3 hours. (Take $e^{0.693} = 2$ .)

Try it first, then show the solutionSolution

$N_{3} = 1000 \times e^{0.693 \times 3} = 1000 \times (e^{0.693})^{3}$ (1 mark).
$= 1000 \times 2^{3} = 8000$ cells (1 mark).
So the population doubles every hour; this J-shaped growth cannot continue for long in a real flask because nutrients run out (1 mark).

Answer: 8000 cells (doubling every hour).

Example 6: Age pyramids · 3 marks · Short answerHarder board level

Describe the three kinds of age pyramids and what each tells about the future of the population.

Try it first, then show the solutionSolution

Expanding (triangular, broad base): many pre-reproductive individuals; population will grow (1 mark).
Stable (bell-shaped): pre-reproductive and reproductive groups similar; population size stays about the same (1 mark).
Declining (urn-shaped, narrow base): few young; population will decline (1 mark).

Answer: Triangular = growing; bell = stable; urn = declining.

Where students lose marks in Population Growth: Exponential and Logistic

The slips examiners see most often, and what to write instead.

Mistake: Rates without units
Students write only the number.
Wrong working
Birth rate = 0.4.
Right working
Birth rate = 0.4 offspring per lotus per year.
Do this: Always add 'per individual per time'.
Usually costs: ½ mark.
Mistake: Mixing up J and S curves
Curves are memorised without the reason.
Wrong working
Logistic growth gives a J-shaped curve.
Right working
Exponential → J-shaped; logistic → S-shaped (sigmoid).
Do this: J jumps; S settles.
Usually costs: 1 mark.
Mistake: Forgetting immigration and emigration
Students use only births and deaths.
Wrong working
N(t+1) = N(t) + B − D.
Right working
N(t+1) = N(t) + (B + I) − (D + E).
Do this: BIDE.
Usually costs: 1 mark.
Mistake: Wrong logistic factor
Students use N/K instead of (K − N)/K.
Wrong working
dN/dt = rN × N/K.
Right working
dN/dt = rN × (K − N)/K.
Do this: Factor = room left = (K − N)/K.
Usually costs: 1-2 marks.

Exam tips

Write the formula first, then substitute; examiners give a mark for the formula.
Label both axes and K on growth curve sketches.
Learn the NCERT values of r (0.015, 0.12, 0.0205) for MCQs.
Name the four phases of logistic growth: lag, acceleration, deceleration, asymptote.
Note: the organism-and-environment part (abiotic factors, responses, adaptations) is not in the 2026-27 syllabus.

Practice questions on Population Growth: Exponential and Logistic

Write each answer on paper first. Open the hint only if you are stuck, and the answer only when you have finished.

Q1 · 1 mark · MCQBoard level
In the logistic growth equation, the population stops growing when:
(a) N = 0 (b) N = K/2 (c) N = K (d) r = K
Hint
(K − N)/K.
Answer
(c) N = K.
Q2 · 1 mark · Very short answerBoard level
In a lab population of 40 fruit flies, 4 died in a week. Calculate the death rate.
Hint
Per capita.
Answer
4/40 = 0.1 individuals per fruit fly per week.
Q3 · 1 mark · Assertion-reasonHarder board level
Assertion (A): The logistic growth model is considered more realistic than the exponential model.
Reason (R): Resources for most animal populations are finite and become limiting sooner or later.
(a) Both A and R are true and R is the correct explanation of A.
(b) Both A and R are true but R is not the correct explanation of A.
(c) A is true but R is false.
(d) A is false but R is true.
Hint
Carrying capacity.
Answer
(a).
Q4 · 2 marks · Very short answerHarder board level
A population with r = 0.4 and K = 500 has N = 100. Calculate dN/dt. What would dN/dt be if growth were exponential?
Hint
Compare rN with rN(K − N)/K.
Answer
Logistic: 0.4 × 100 × (500 − 100)/500 = 40 × 0.8 = 32 per unit time. Exponential: 0.4 × 100 = 40 per unit time.
Q5 · 4 marks · Case-basedHarder board level
Read and answer.
Suppose a graph shows the number of yeast cells in a flask over 20 hours: slow increase for 2 hours, rapid increase from 2 to 10 hours, slower increase from 10 to 14 hours, and a level line at about 650 cells (in thousands) after 14 hours.
(a) Name the type of growth. (1)
(b) Name the phases shown and the value at which it levels off. (1)
(c) Why does growth slow and stop? What would happen to the level line if more nutrients were added? (2)
Hint
Sigmoid; K.
Answer
(a) Logistic (sigmoid) growth. (b) Lag (0-2 h), acceleration (2-10 h), deceleration (10-14 h), asymptote at the carrying capacity K ≈ 650 thousand cells. (c) Nutrients and space become limiting and competition increases, so births fall and deaths rise until they balance. Adding nutrients raises the carrying capacity, so the population would grow again to a higher plateau.
Q6 · 3 marks · Short answerHOTS / competency
Suppose two islands each start with 50 rabbits. Island A has unlimited grass and no predators; Island B has limited grass. Sketch (describe) the expected growth curves for 10 years and explain why Island A's pattern cannot last forever.
Hint
J vs S.
Answer
Island A: J-shaped exponential curve, numbers rising ever faster (dN/dt = rN). Island B: S-shaped logistic curve, rising and then levelling at the island's carrying capacity. Island A's growth cannot last because grass and space will eventually become limiting; competition will then slow growth and numbers will level off (or crash), so it will turn into logistic-type growth.

Test your recall

Answer in your head or on paper first, then open the card. Pulling an answer out of memory makes it stick far better than re-reading.

1.What does r stand for?Show answer
Intrinsic rate of natural increase (r = b − d).
2.Shape of the exponential growth curve?Show answer
J-shaped.
3.What is K?Show answer
Carrying capacity: the maximum population a habitat can support.
4.At what N is logistic growth fastest?Show answer
N = K/2.
5.Name the four processes that change population density.Show answer
Natality, mortality, immigration, emigration.
6.Shape of the age pyramid of a declining population?Show answer
Urn-shaped.

Population Growth: Exponential and Logistic: frequently asked questions

What is exponential growth?

Growth when resources are unlimited: dN/dt = rN, giving a J-shaped curve.

What is logistic growth?

Growth when resources are limited: dN/dt = rN(K − N)/K, giving an S-shaped curve that levels off at the carrying capacity K.

What is carrying capacity?

The maximum number of individuals of a population that a habitat's resources can support.

How is population size calculated from births, deaths and migration?

N(t+1) = N(t) + [(B + I) − (D + E)], where B is births, I immigrants, D deaths and E emigrants.

Why is the logistic model more realistic?

Because in nature resources are limited, competition increases as numbers rise, and growth slows and stops at the carrying capacity.