Precalculus · F.BF.A.1

Logistic growth: when growth levels off

Two hundred elk are released into a valley that can feed about five thousand. The herd grows fast at first, but it cannot keep multiplying forever. What function grows like an exponential early on and then levels off?

You need

Building an exponential model from two pointsPrecalculus Horizontal asymptotesAlgebra 2

Where you are 6 in Exponential growth and decay

Functions13

Exponential functions6

Exponential growth and decay6

Doubling time and half life Growth and decay factors Building an exponential model from two points Linear versus exponential growth

Logistic growth: when growth levels off

Linear, polynomial and exponential growth compared

Why this exists

An exponential model, such as one built with Building an exponential model from two points, multiplies by the same factor every period, forever. That is a fine description of a small elk herd in an empty valley with plenty of grass. It cannot stay true. The valley holds only so much food, so as the herd grows, each year’s growth must slow, and the count must settle near some ceiling.

A model for this needs three features at once: nearly exponential growth while the count is small, slower growth as the count nears a ceiling, and a graph that flattens out at that ceiling instead of crossing it. The logistic model has all three. The same shape describes many other things that spread until they run out of room: a piece of gossip moving through a school, a new product reaching the people who might buy it.

Definition

A logistic model is a function of the form

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

with constants C > 0, A > 0 and k > 0. Here e is the base of the The number e and the natural exponential function.

C is the carrying capacity, the level the model levels off at.
A is fixed by the starting value: \displaystyle P(0) = \frac{C}{1 + A}.
k sets how fast the change happens. A larger k makes a steeper S.

The halfway time t^{*} is the time when \displaystyle P(t^{*}) = \frac{C}{2}.

The same family is often written with a base other than e, as \displaystyle \frac{C}{1 + Ab^{t}} with 0 < b < 1. Since b^{t} = e^{t \ln b}, this is the form above with k = -\ln b.

Our lead model, for the elk, is

P(t) = \frac{5000}{1 + 24e^{-0.35t}}

with t in years after the release. Here C = 5000, A = 24 and k = 0.35.

Intuition

Read the formula from the inside out. The term 24e^{-0.35t} is an exponential decay: it starts at 24 and shrinks by the factor e^{-0.35} \approx 0.705 every year. The herd is 5000 divided by one more than that shrinking term.

At the start the term is large, so the denominator is large and the herd is small: \displaystyle P(0) = \frac{5000}{25} = 200.
While the term is still much bigger than 1, the 1 hardly matters, and \displaystyle P(t) \approx \frac{5000}{24e^{-0.35t}} = \frac{5000}{24}e^{0.35t}. That is an exponential growth model.
When the term has shrunk to 1, the denominator is 2 and the herd is at half the carrying capacity.
Once the term is tiny, the denominator is barely above 1, and the herd sits just under 5000.

Figure 1The elk model. Slow, then fast, then slow again, with the steepest climb at half the carrying capacity.

Figure 2Early on, the logistic curve tracks the exponential \displaystyle \frac{5000}{24}e^{0.35t}. As the herd fills the valley, the two part ways.

Results

Start value. \displaystyle P(0) = \frac{C}{1 + A}, so a model with carrying capacity C and start P_0 has \displaystyle A = \frac{C - P_0}{P_0}.

Two horizontal asymptotes. As t grows, P(t) approaches C; as t runs far into the negatives, P(t) approaches 0. The lines y = C and y = 0 are Horizontal asymptotes. The model is increasing and stays strictly between 0 and C: it never reaches C.

The halfway time. When A > 1, \displaystyle t^{*} = \frac{\ln A}{k}. The graph is symmetric about the point \displaystyle \left(t^{*}, \frac{C}{2}\right): for every s,

P(t^{*} + s) + P(t^{*} - s) = C

The growth is fastest at t^{*} and slows on both sides of it. Here that is seen in a table of gains (Example 3); the proof that t^{*} is the exact moment of fastest growth needs calculus.

Early and late growth. While Ae^{-kt} is large, \displaystyle P(t) \approx \frac{C}{A}e^{kt}, exponential growth with factor e^{k} per unit of time. The factor \displaystyle \frac{P(t+1)}{P(t)} starts near e^{k} and falls toward 1.

Time to reach a level. For 0 < y < C, the model equals y exactly once, at

t = \frac{1}{k}\ln\left(\frac{Ay}{C - y}\right)

Why it is true

Asymptotes and bounds. Since k > 0, e^{-kt} shrinks toward 0 as t grows, so the denominator 1 + Ae^{-kt} shrinks toward 1 and P(t) approaches \displaystyle \frac{C}{1} = C. As t runs into the negatives, e^{-kt} grows without bound, the denominator does too, and P(t) approaches 0. Because Ae^{-kt} > 0, the denominator is always above 1, so 0 < P(t) < C. As t increases the denominator decreases, and dividing C by a smaller positive number gives a larger result, so P increases.

The halfway time. \displaystyle P(t) = \frac{C}{2} means 1 + Ae^{-kt} = 2, so Ae^{-kt} = 1, so e^{kt} = A and \displaystyle t = \frac{\ln A}{k}. This is positive when A > 1, which is when the start is below \displaystyle \frac{C}{2}.

Symmetry. At t^{*} we have Ae^{-kt^{*}} = 1, so Ae^{-k(t^{*} + s)} = e^{-ks}. Write u = e^{-ks}. Then \displaystyle Ae^{-k(t^{*} - s)} = e^{ks} = \frac{1}{u}, and

\begin{aligned} P(t^{*} + s) + P(t^{*} - s) &= \frac{C}{1 + u} + \frac{C}{1 + \frac{1}{u}} \\ &= \frac{C}{1 + u} + \frac{Cu}{u + 1} = C \end{aligned}

So the rise in the s units of time after t^{*} matches the rise in the s units before it, mirrored. The curve gains the most over a stretch of time centered on t^{*}, as Example 3 shows in a table. That t^{*} is the exact moment of fastest growth is proved with calculus in Logistic growth.

Early growth and the factor. Dividing,

\frac{P(t+1)}{P(t)} = \frac{1 + Ae^{-kt}}{1 + Ae^{-kt}e^{-k}}

When Ae^{-kt} is large, the 1s hardly matter and the ratio is close to \displaystyle \frac{1}{e^{-k}} = e^{k}. When Ae^{-kt} is near 0, the ratio is close to \displaystyle \frac{1}{1} = 1.

Time to reach a level. From \displaystyle \frac{C}{1 + Ae^{-kt}} = y, multiply out: \displaystyle 1 + Ae^{-kt} = \frac{C}{y}, so \displaystyle Ae^{-kt} = \frac{C - y}{y} and \displaystyle e^{kt} = \frac{Ay}{C - y}. For 0 < y < C the right side is positive, so it has exactly one natural logarithm (Solving by taking a logarithm of both sides).

Worked examples

Example 1

Reading the elk model

Hinata, a wildlife biologist, uses \displaystyle P(t) = \frac{5000}{1 + 24e^{-0.35t}} for the elk. How many elk were released? Estimate the herd after 5, 10 and 20 years, and say what happens in the long run.

Solution

At t = 0, e^{0} = 1, so \displaystyle P(0) = \frac{5000}{1 + 24} = 200 elk were released.

For the others, evaluate the denominator first. At t = 5: e^{-1.75} \approx 0.1738, so the denominator is about 1 + 4.170 = 5.170 and P(5) \approx 967. At t = 10: e^{-3.5} \approx 0.0302, the denominator is about 1.725, and P(10) \approx 2899. At t = 20: e^{-7} \approx 0.000912, the denominator is about 1.022, and P(20) \approx 4893.

In the long run 24e^{-0.35t} shrinks toward 0, so the herd approaches the carrying capacity of 5000 elk without passing it. The herd grew by about 767 in the first five years, about 1932 in the next five, and about 1994 in the ten years after that: the growth is already slowing.

Example 2

When the herd reaches 4000

When does Hinata’s model reach 4000 elk?

Solution

Set the model equal to 4000 and isolate the exponential.

\begin{aligned} \frac{5000}{1 + 24e^{-0.35t}} &= 4000 \\ 1 + 24e^{-0.35t} &= \tfrac{5000}{4000} = 1.25 \\ e^{-0.35t} &= \tfrac{0.25}{24} = \tfrac{1}{96} \end{aligned}

The second line multiplies both sides by the denominator and divides by 4000. Taking the natural logarithm of both sides gives -0.35t = -\ln 96, so \displaystyle t = \frac{\ln 96}{0.35} \approx 13.04. The herd reaches 4000 about 13 years after the release. The formula in Results gives the same: \displaystyle \frac{1}{0.35}\ln\left(\frac{24 \cdot 4000}{1000}\right) = \frac{\ln 96}{0.35}.

Example 3

The halfway time and the fastest growth

Find the halfway time of the elk model. Then use the herd’s yearly gains to see where the growth is fastest.

Solution

The halfway time is when 24e^{-0.35t} = 1, so \displaystyle t^{*} = \frac{\ln 24}{0.35} \approx 9.08 years. Then the herd is at 2500.

The gain in each year, P(t+1) - P(t), to the nearest elk:

years	gain
6 to 7	359
7 to 8	405
8 to 9	432
9 to 10	434
10 to 11	411
11 to 12	367

The gains climb, peak in the two years on either side of t^{*} \approx 9.08, and then fall. Before the halfway point there are few elk to have calves; after it there is little room left. Near half the carrying capacity the two limits are in balance, and the table shows the largest gains there. A table of whole years cannot pin down the exact moment of fastest growth; calculus shows it is exactly t^{*}, at P = 2500.

Example 4

The yearly factor shrinks

Asel wants to know how close the herd comes to an exponential model. Find \displaystyle \frac{P(t+1)}{P(t)} for t = 0, 10 and 19, and compare with e^{0.35}.

Solution

From P(0) = 200 and \displaystyle P(1) = \frac{5000}{1 + 24e^{-0.35}} \approx 279.1, the factor for the first year is about 1.396. The same computation gives about 1.142 from year 10 to year 11, and about 1.009 from year 19 to year 20.

An exponential with k = 0.35 would multiply by e^{0.35} \approx 1.419 every year. The first year’s factor is close to that, because 24e^{-0.35t} is still large and, as Why it is true shows, the factor is then near e^{k}. By year 19 the factor is almost 1: the herd is barely growing. A logistic model is an exponential whose factor slowly falls to 1.

Example 5

A model written with base 2

Describe \displaystyle g(x) = \frac{12}{1 + 3 \cdot 2^{-x}}: its value at 0, its horizontal asymptotes and its center. Rewrite it in the form with e.

Solution

\displaystyle g(0) = \frac{12}{1 + 3} = 3. As x grows, 2^{-x} shrinks toward 0, so g approaches 12; as x runs into the negatives, 2^{-x} grows without bound and g approaches 0. The horizontal asymptotes are y = 12 on the right and y = 0 on the left.

The center is where 3 \cdot 2^{-x} = 1, so 2^{x} = 3 and x = \log_2 3 \approx 1.585, with g = 6. Since 2^{-x} = e^{-x\ln 2}, the function is \displaystyle \frac{12}{1 + 3e^{-(\ln 2)x}}, with k = \ln 2 \approx 0.693.

The symmetry gives a check. The points x = 0 and x = 2\log_2 3 = \log_2 9 sit the same distance on either side of the center, and \displaystyle g(\log_2 9) = \frac{12}{1 + \frac{3}{9}} = 9. Indeed 3 + 9 = 12.

Figure 3Example 5. Two horizontal asymptotes, y = 0 on the left and y = 12 on the right, and a center halfway between them.

Example 6

Building a model for gossip

A piece of gossip starts with 12 of the 1200 students in a school. Declan finds that 3 days later, 300 students have heard it. Build a logistic model with carrying capacity 1200. When have half the students heard it? When have 1100?

Solution

The start fixes A: \displaystyle A = \frac{1200 - 12}{12} = 99. So \displaystyle N(t) = \frac{1200}{1 + 99e^{-kt}}, and the reading at day 3 fixes k:

\begin{aligned} \frac{1200}{1 + 99e^{-3k}} &= 300 \\ 1 + 99e^{-3k} &= 4 \\ e^{-3k} &= \tfrac{3}{99} = \tfrac{1}{33} \end{aligned}

So 3k = \ln 33 and \displaystyle k = \frac{\ln 33}{3} \approx 1.166.

Half the school, 600 students, is the halfway time: \displaystyle t^{*} = \frac{\ln 99}{k} = \frac{3\ln 99}{\ln 33} \approx 3.94 days.

For 1100: \displaystyle 1 + 99e^{-kt} = \frac{1200}{1100} = \frac{12}{11}, so \displaystyle e^{-kt} = \frac{1}{11 \cdot 99} = \frac{1}{1089}. Since 1089 = 33^2 and \displaystyle e^{-3k} = \frac{1}{33}, we get e^{-kt} = \left(e^{-3k}\right)^2 = e^{-6k}, so t = 6 days exactly. Getting from 12 to 300 took three days and from 300 to 1100 another three.

Example 7

A valley with too many elk

A second valley can feed 900 elk, and 1500 are released into it. Use \displaystyle P(t) = \frac{900}{1 + Ae^{-0.4t}}. Find A, describe the graph for t \geq 0, and find when the herd is down to 1000.

Solution

\displaystyle A = \frac{900 - 1500}{1500} = -0.4, so \displaystyle P(t) = \frac{900}{1 - 0.4e^{-0.4t}}. Check: \displaystyle P(0) = \frac{900}{0.6} = 1500.

Here A is negative, outside the definition, so the Results must be checked again. For t \geq 0, 0.4e^{-0.4t} is at most 0.4, so the denominator lies between 0.6 and 1 and P(t) > 900. As t grows the denominator rises toward 1, so the herd falls toward 900. The curve drops, steeply at first, and flattens onto y = 900 from above. There is no S and no halfway time.

Figure 4Example 7. A herd that starts above the carrying capacity shrinks toward it. There is no S shape.

For 1000: \displaystyle 1 - 0.4e^{-0.4t} = \frac{900}{1000} = 0.9, so \displaystyle e^{-0.4t} = \frac{0.1}{0.4} = 0.25 and \displaystyle t = \frac{\ln 4}{0.4} \approx 3.47 years.

Boundaries

The S shape needs 0 < P_0 < C, that is A > 0. A start above C gives -1 < A < 0 and a falling curve (Example 7); a start at exactly C gives A = 0 and a constant model. A start at 0 is not allowed: no herd, no growth.
There is a halfway time after the start only when \displaystyle P_0 < \frac{C}{2}. If \displaystyle \frac{C}{2} \leq P_0 < C, then 0 < A \leq 1, \ln A \leq 0, and the fastest growth is at or before t = 0.
The model never reaches C. Equations such as P(t) = C or P(t) = 0 have no solution. A real herd can sit at its ceiling; the model only approaches it.
The model is a model. Negative t gives values before the release that never happened, and a real herd swings with harsh winters. Whether a logistic curve fits real counts is a question for Judging and comparing models.
C is assumed fixed. If the valley’s food supply changes, so does C, and one formula no longer describes the whole history.

Where this goes next

Semi-log plots: straightening exponential data with logarithms straighten an exponential, which is one way to see when data stops behaving exponentially and starts to bend like a logistic curve. Judging and comparing models compares exponential, logistic and other models against real readings. In calculus, the logistic model returns as the solution of a growth law in which the rate is proportional both to the size and to the room left, and the fastest growth at \displaystyle \frac{C}{2} is proved there (Logistic growth). The way the yearly gains rise and then fall is the change in steepness studied in Concavity: when the rate of change itself changes.

Exercises

Reading a logistic model

For \displaystyle P(t) = \frac{800}{1 + 15e^{-0.6t}}, find the carrying capacity, P(0) and the halfway time.
For \displaystyle f(x) = \frac{30}{1 + 4e^{-x}}, find f(0), both horizontal asymptotes, and the x where f(x) = 15.
For \displaystyle P(t) = \frac{2400}{1 + 11e^{-0.25t}}, find P(4) and P(12) to the nearest whole number.
For \displaystyle h(x) = \frac{9}{1 + 2(0.5)^{x}}, find h(0), the horizontal asymptotes and the center. What is k in the form with e?
Write the logistic model with carrying capacity 1500, start 100 and k = 0.5.
Write the logistic model with carrying capacity 320, start 40 and k = 0.9, and find P(2) to one decimal place.

Solving for a time

When does \displaystyle P(t) = \frac{800}{1 + 15e^{-0.6t}} reach 600?
When does Hinata’s elk model reach 1000 elk?
When does Hinata’s elk model reach 4900 elk?
For \displaystyle f(x) = \frac{30}{1 + 4e^{-x}}, solve f(x) = 24.
For \displaystyle h(x) = \frac{9}{1 + 2(0.5)^{x}}, solve h(x) = 8 exactly.

Building a model

A model has carrying capacity 2000, starts at 50 and is at 400 when t = 4. Find A and k.
Gossip starts with 9 of the 900 students in another school, and 90 have heard it after 1 day. Build a logistic model and find when 810 have heard it.
The model \displaystyle P(t) = \frac{C}{1 + 9e^{-0.5t}} starts at 40. Find C and the halfway time.
A model has carrying capacity 600, starts at 40 and has halfway time 5. Find A and k.

Shape and rates

Find the average rate of change of Hinata’s elk model from year 2 to year 4 and from year 16 to year 18.
For the elk model, find P(t^{*} + 2) and P(t^{*} - 2) to the nearest elk and add them.
A valley that can feed 400 deer receives 500. With k = 0.3, write the model and find when the herd is down to 420.
Show that \displaystyle f(x) = \frac{10}{1 + e^{-x}} satisfies f(x) + f(-x) = 10 for every x.

Reasoning

Explain why, when 0 < P(0) < C, the model is never above C and never below 0.
Show that \displaystyle \frac{C}{1 + Ae^{-kt}} = \frac{C}{1 + e^{-k(t - t^{*})}} with \displaystyle t^{*} = \frac{\ln A}{k}. What does this say about two herds with the same C and k but different starts?
Two readings fix an exponential model. Explain why they do not fix a logistic model when the carrying capacity is unknown, and what extra fact would.
What happens to the halfway time and to the graph if k is doubled and C and A stay the same?
Declan says the elk herd grows 35\% a year because k = 0.35. Asel says that is wrong in two ways. Explain both.

Answers

C = 800; \displaystyle P(0) = \frac{800}{16} = 50; \displaystyle t^{*} = \frac{\ln 15}{0.6} \approx 4.51.
\displaystyle f(0) = \frac{30}{5} = 6; y = 30 on the right and y = 0 on the left; f(x) = 15 when 4e^{-x} = 1, so x = \ln 4 \approx 1.39.
\displaystyle P(4) = \frac{2400}{1 + 11e^{-1}} \approx 476 and \displaystyle P(12) = \frac{2400}{1 + 11e^{-3}} \approx 1551.
\displaystyle h(0) = \frac{9}{3} = 3; asymptotes y = 9 on the right and y = 0 on the left; the center is where 2(0.5)^{x} = 1, at x = 1, h = 4.5; k = -\ln 0.5 = \ln 2 \approx 0.693.
\displaystyle A = \frac{1500 - 100}{100} = 14, so \displaystyle P(t) = \frac{1500}{1 + 14e^{-0.5t}}.
\displaystyle A = \frac{320 - 40}{40} = 7, so \displaystyle P(t) = \frac{320}{1 + 7e^{-0.9t}}, and \displaystyle P(2) = \frac{320}{1 + 7e^{-1.8}} \approx 148.3.
\displaystyle 1 + 15e^{-0.6t} = \frac{4}{3}, so \displaystyle e^{-0.6t} = \frac{1}{45} and \displaystyle t = \frac{\ln 45}{0.6} \approx 6.34.
1 + 24e^{-0.35t} = 5, so \displaystyle e^{-0.35t} = \frac{1}{6} and \displaystyle t = \frac{\ln 6}{0.35} \approx 5.12 years.
\displaystyle 1 + 24e^{-0.35t} = \frac{50}{49}, so \displaystyle e^{-0.35t} = \frac{1}{1176} and \displaystyle t = \frac{\ln 1176}{0.35} \approx 20.2 years.
1 + 4e^{-x} = 1.25, so \displaystyle e^{-x} = \frac{1}{16} and x = \ln 16 \approx 2.77.
\displaystyle 1 + 2(0.5)^{x} = \frac{9}{8}, so \displaystyle (0.5)^{x} = \frac{1}{16} = (0.5)^4 and x = 4.
\displaystyle A = \frac{2000 - 50}{50} = 39; 1 + 39e^{-4k} = 5 gives \displaystyle e^{-4k} = \frac{4}{39}, so \displaystyle k = \frac{1}{4}\ln\frac{39}{4} \approx 0.569.
\displaystyle A = \frac{900 - 9}{9} = 99; 1 + 99e^{-k} = 10 gives \displaystyle e^{-k} = \frac{1}{11}, so k = \ln 11 \approx 2.40 and \displaystyle N(t) = \frac{900}{1 + 99e^{-2.40t}}. For 810: \displaystyle e^{-kt} = \frac{1}{891}, so \displaystyle t = \frac{\ln 891}{\ln 11} \approx 2.83 days.
\displaystyle \frac{C}{10} = 40, so C = 400; \displaystyle t^{*} = \frac{\ln 9}{0.5} \approx 4.39.
\displaystyle A = \frac{600 - 40}{40} = 14; \displaystyle \frac{\ln 14}{k} = 5, so \displaystyle k = \frac{\ln 14}{5} \approx 0.528.
\displaystyle \frac{P(4) - P(2)}{2} \approx \frac{722.7 - 387.1}{2} \approx 167.8 elk per year, and \displaystyle \frac{P(18) - P(16)}{2} \approx \frac{4788.9 - 4592.4}{2} \approx 98.3 elk per year.
With t^{*} \approx 9.08: P(t^{*} + 2) \approx 3341 and P(t^{*} - 2) \approx 1659, and 3341 + 1659 = 5000, as the symmetry says.
\displaystyle A = \frac{400 - 500}{500} = -0.2, so \displaystyle P(t) = \frac{400}{1 - 0.2e^{-0.3t}}. For 420: \displaystyle 0.2e^{-0.3t} = \frac{1}{21}, so \displaystyle e^{-0.3t} = \frac{5}{21} and \displaystyle t = \frac{\ln 4.2}{0.3} \approx 4.78 years.
\displaystyle f(-x) = \frac{10}{1 + e^{x}} = \frac{10e^{-x}}{e^{-x} + 1}, multiplying top and bottom by e^{-x}, so \displaystyle f(x) + f(-x) = \frac{10 + 10e^{-x}}{1 + e^{-x}} = 10.
P(0) < C forces A > 0, so Ae^{-kt} > 0 and the denominator is above 1 for every t; C divided by a number above 1 is positive and less than C.
e^{kt^{*}} = A, so Ae^{-kt} = e^{kt^{*}}e^{-kt} = e^{-k(t - t^{*})}. Every such model is one curve, \displaystyle \frac{C}{1 + e^{-kt}}, shifted right by t^{*}: two herds with the same C and k trace the same S, one later than the other.
A logistic model has three unknowns, C, A and k, and two readings give only two equations. Knowing C (from the food supply, say) or a third reading would fix it.
\displaystyle t^{*} = \frac{\ln A}{2k} is halved. The new model is P(2t), so the graph is squeezed toward the vertical axis by a factor of 2: the same S, run twice as fast, with the same asymptotes.
First, the early factor is e^{0.35} \approx 1.419, about 42\% a year, not 35\%. Second, that holds only while the herd is small; the yearly factor falls toward 1 as the herd nears 5000 (Example 4).