Logistic Functions: Growth with a Limit

A logistic function models growth that starts out nearly exponential, then slows down and levels off at a maximum called the carrying capacity. Its common form is $P(t) = \frac{C}{1 + ae^{-kt}}$ , where $C$ is the carrying capacity, $k \gt 0$ sets the growth speed, and $P(0) = \frac{C}{1 + a}$ . Growth is fastest when $P = \frac{C}{2}$ .

Before this lesson: You should know Exponential Growth and Decay, The Number e and Solving Exponential Equations.

Why Growth Levels Off

Exponential growth never stops, but real populations do. A pond only has so much food, and a school only has so many students to hear a rumor. As the population gets close to that limit, growth slows.

A logistic curve captures all three stages. At first, there is lots of room, so growth looks exponential. In the middle, growth is fastest. Near the limit $C$ , growth slows to almost nothing, and the curve flattens into a horizontal asymptote at $P = C$ .

Graph of P(t) = 500/(1 + 24e^(−0.3t)) for t from 0 to 30 in solid blue. It starts at 20, rises slowly, then quickly, then levels off just under the dashed orange line P = 500, the carrying capacity. A green dot marks the point of fastest growth at about t = 10.6, where P = 250. — The logistic curve P(t) = 500/(1 + 24e^(−0.3t)) starts at 20, grows fastest at P = 250 and levels off at the carrying capacity, 500.

Key Features of a Logistic Function

P(t) = \frac{C}{1 + ae^{-kt}}

Feature	Formula	Meaning
Carrying capacity	$C$	The level the function approaches; a horizontal asymptote
Initial value	$P(0) = \frac{C}{1 + a}$	The starting amount
Point of fastest growth	$t = \frac{\ln a}{k}$ , where $P = \frac{C}{2}$	The inflection point, where the curve switches from concave up to concave down
Growth constant	$k$	Larger $k$ makes a steeper S-curve

To find $a$ from a starting value, solve $P(0) = \frac{C}{1 + a}$ for $a$ : $a = \frac{C}{P(0)} - 1$ .

Worked Examples

Use a calculator for powers of $e$ and natural logs. Round to the nearest hundredth.

Example 1: Read the key features

For $P(t) = \frac{500}{1 + 24e^{-0.3t}}$ , find the carrying capacity, the initial value and the time of fastest growth.

C = 500

Read the carrying capacity: the number on top.

P(0) = \frac{500}{1 + 24e^{0}} = \frac{500}{1 + 24} = \frac{500}{25} = 20

Find the initial value: substitute t = 0.

t = \frac{\ln 24}{0.3} \approx \frac{3.178}{0.3} \approx 10.59

Find the time of fastest growth.

P = \frac{500}{2} = 250

At that time.

Answer: $C = 500$ , $P(0) = 20$ , fastest growth at $t \approx 10.59$ , when $P = 250$ . These match the graph above.

Example 2: Evaluate the function

For the same function, find $P(10)$ .

P(10) = \frac{500}{1 + 24e^{-0.3(10)}} = \frac{500}{1 + 24e^{-3}}

Substitute.

e^{-3} \approx 0.04979

Evaluate the power.

24(0.04979) \approx 1.1949

Multiply.

1 + 1.1949 = 2.1949

Add 1.

\frac{500}{2.1949} \approx 227.80

Divide.

Answer: $P(10) \approx 227.80$

Example 3: Solve for the time

For the same function, when does $P$ reach 400?

\frac{500}{1 + 24e^{-0.3t}} = 400

Set up.

500 = 400(1 + 24e^{-0.3t})

Multiply both sides by the denominator.

1.25 = 1 + 24e^{-0.3t}

Divide both sides by 400.

0.25 = 24e^{-0.3t}

Subtract 1.

e^{-0.3t} = \frac{0.25}{24} \approx 0.010417

Divide by 24.

-0.3t = \ln 0.010417 \approx -4.564

Take the natural log of both sides.

t \approx 15.21

Divide by −0.3.

Answer: at about $t = 15.21$

Example 4: Build a model

A population has carrying capacity 1,000. It starts at 50 and reaches 200 at $t = 4$ . Find a logistic model.

a = \frac{C}{P(0)} - 1 = \frac{1000}{50} - 1 = 20 - 1 = 19

Find $a$ .

\frac{1000}{1 + 19e^{-4k}} = 200

Use the point $(4, 200)$ .

1 + 19e^{-4k} = \frac{1000}{200} = 5

Rearrange.

e^{-4k} = \frac{4}{19}

Subtract 1 and divide by 19.

-4k = \ln \frac{4}{19} \approx -1.558

Take the natural log.

k \approx 0.3895

Divide by −4.

Answer: $P(t) = \frac{1000}{1 + 19e^{-0.3895t}}$

Example 5: Word problem with a rumor

In a school of 1,200 students, the number who have heard a rumor after $t$ days is $N(t) = \frac{1200}{1 + 59e^{-0.8t}}$ . How many start out knowing it? When have half the students heard it? How many know after 7 days?

N(0) = \frac{1200}{1 + 59} = \frac{1200}{60} = 20

Start.

Half the school is $\frac{1200}{2} = 600$ , reached at $t = \frac{\ln 59}{0.8} \approx \frac{4.078}{0.8} \approx 5.10$

N(7) = \frac{1200}{1 + 59e^{-5.6}}

After 7 days.

59e^{-5.6} \approx 59(0.003698) \approx 0.2182

Evaluate.

\frac{1200}{1.2182} \approx 985.1

Divide.

Answer: 20 students at first; half the school by about 5.1 days; about 985 students after 7 days. The rumor spreads fastest around day 5.

Common Mistakes

Mistake	Example of the error	Fix
Calling C the starting value	Saying the population starts at 500	$C$ is the limit. The start is $\frac{C}{1 + a} = 20$ .
Dropping the negative exponent	Using $e^{0.3t}$ in the denominator	It is $e^{-kt}$ ; that is what makes the curve level off.
Wrong order of operations	$\frac{500}{1} + 24e^{-3}$	Find the whole denominator $1 + 24e^{-3}$ first, then divide.
Expecting exponential forever	Predicting 5,000 students will hear a rumor in a school of 1,200	A logistic model never passes $C$ .
Wrong fastest-growth point	Fastest growth at $P = C$	Growth is fastest at $P = \frac{C}{2}$ ; it slows to 0 near $C$ .

Related Methods

Compare with exponential models

For small $t$ , $\frac{C}{1 + ae^{-kt}}$ is close to $\frac{C}{a}e^{kt}$ , an exponential function. That is why exponential regression works well early on but overshoots later.

Logistic regression

Graphing calculators have a Logistic command that fits $y = \frac{c}{1 + ae^{-bx}}$ to data, the same way ExpReg fits exponential models.

In calculus

In Calculus 2, logistic functions come from the differential equation $\frac{dP}{dt} = kP(1 - \frac{P}{C})$ , which says growth slows as $P$ nears $C$ .

Practice Problems

Round answers to the nearest hundredth.

For $f(t) = \frac{80}{1 + 7e^{-0.5t}}$ , find the carrying capacity and $f(0)$ .
For the same $f$ , find $f(4)$ .
For the same $f$ , when does $f$ reach half the carrying capacity?
For the same $f$ , when does $f$ reach 60?
Write a logistic model with carrying capacity 300, starting value 15 and $k = 0.4$ .
A population has carrying capacity 600, starts at 100 and reaches 200 at $t = 2$ . Find $k$ .
What happens to $\frac{C}{1 + ae^{-kt}}$ as $t$ gets very large, and why?
Find the point of fastest growth for $g(t) = \frac{50}{1 + 9e^{-t}}$ .
What is the long-run value of $h(t) = \frac{12}{1 + 3e^{-2t}}$ ?
A lake is stocked with fish. The count after $t$ years is $F(t) = \frac{2000}{1 + 39e^{-0.5t}}$ . How many fish were stocked, and when will there be 1,500 fish?

Show answers

C = 80

Read the carrying capacity: the number on top.

f(0) = \frac{80}{1 + 7} = \frac{80}{8} = 10

Find the initial value: substitute 0.

Answer: $C = 80$ , $f(0) = 10$

f(4) = \frac{80}{1 + 7e^{-2}}

Substitute 4.

7e^{-2} \approx 7(0.1353) \approx 0.9473

Evaluate the power with a calculator.

\frac{80}{1.9473} \approx 41.08

Divide.

Answer: about 41.08

Half of $C$ is 40, reached at $t = \frac{\ln 7}{0.5}$
$\ln 7 \approx 1.946$ , so $t \approx \frac{1.946}{0.5} \approx 3.89$
Answer: about $t = 3.89$

$\frac{80}{1 + 7e^{-0.5t}} = 60$ gives $1 + 7e^{-0.5t} = \frac{80}{60} = \frac{4}{3}$

e^{-0.5t} = \frac{1}{21}

Subtract 1 and divide by 7.

-0.5t = \ln \frac{1}{21} \approx -3.045

Take the natural log of both sides.

t \approx 6.09

Divide both sides by −0.5.

Answer: about $t = 6.09$

a = \frac{300}{15} - 1 = 20 - 1 = 19

Find a: capacity over initial value, minus 1.

Answer: $P(t) = \frac{300}{1 + 19e^{-0.4t}}$

a = \frac{600}{100} - 1 = 5

Find a: capacity over initial value, minus 1.

$\frac{600}{1 + 5e^{-2k}} = 200$ gives $1 + 5e^{-2k} = 3$ , so $e^{-2k} = 0.4$

-2k = \ln 0.4 \approx -0.9163

Take the natural log of both sides.

k \approx 0.4581

Divide both sides by −2.

Answer: $k \approx 0.46$

As $t$ grows, $e^{-kt}$ approaches 0
So the denominator approaches $1 + 0 = 1$
Answer: the function approaches $C$ , the carrying capacity

Fastest growth at $t = \frac{\ln 9}{1} \approx 2.20$
$g = \frac{50}{2} = 25$ there
Answer: about $(2.20, 25)$

As $t$ grows, $3e^{-2t}$ approaches 0
Answer: $h$ approaches $\frac{12}{1} = 12$

10.

F(0) = \frac{2000}{1 + 39} = \frac{2000}{40} = 50

Find how many fish were stocked: substitute 0.

$\frac{2000}{1 + 39e^{-0.5t}} = 1500$ gives $1 + 39e^{-0.5t} = \frac{4}{3}$ , so $e^{-0.5t} = \frac{1}{117}$

-0.5t = \ln \frac{1}{117} \approx -4.762

Take the natural log of both sides.

t \approx 9.52

Divide both sides by −0.5.

Answer: 50 fish stocked; 1,500 fish after about 9.52 years

FAQ

What is the difference between exponential and logistic growth?

Exponential growth keeps speeding up forever. Logistic growth starts out like exponential growth but slows down and levels off at the carrying capacity.

What is the carrying capacity?

It is the largest value the model approaches, the number $C$ in the numerator. On the graph, it is a horizontal asymptote.

Where does a logistic function grow fastest?

At half the carrying capacity, $P = \frac{C}{2}$ , which happens at $t = \frac{\ln a}{k}$ . This is the inflection point of the S-shaped curve.

What are real examples of logistic growth?

Animal populations with limited food, the spread of a rumor or disease, and sales of a new product in a market of fixed size all follow logistic patterns.

Summary

A logistic function $\frac{C}{1 + ae^{-kt}}$ makes an S-shaped curve.
$C$ is the carrying capacity, a horizontal asymptote.
The initial value is $\frac{C}{1 + a}$ , so $a = \frac{C}{P(0)} - 1$ .
Growth is fastest at $P = \frac{C}{2}$ , when $t = \frac{\ln a}{k}$ .
Solve for $t$ by clearing the fraction, isolating $e^{-kt}$ and taking $\ln$ .

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