Logistic Model (AI HL)
A logistic model describes growth that starts slowly, accelerates, then levels off as it approaches a fixed ceiling - the carrying capacity. It's the standard shape for populations limited by space or resources, and for products adopted across a saturating market. This page covers the formula, what the constants mean, and the mistakes that most often cost marks. It's part of the broader Logistic & Other Models topic.
24 questions on this sub-topic.
The logistic formula
Covered under IB syllabus reference AHL2.9: logistic models \(f(x) = \dfrac{L}{1+Ce^{-kx}}\), with \(L, C, k > 0\). The horizontal asymptote at \(f(x)=L\) is often called the carrying capacity - used where growth faces a natural restriction, such as a population on an island, bacteria in a petri dish, or the height of a growing seedling.
Logistic model
\(f(x) = \dfrac{L}{1+Ce^{-kx}}\)
Given in the formula booklet. \(L\) is the carrying capacity, \(C\) and \(k\) shape how fast the curve rises to it.
Exponential "other" model
\(f(x)=ke^{rx}+c\)
Also in the formula booklet. Used for bounded exponential growth or decay that isn't S-shaped, such as Newton's law of cooling.
Need the differential-equation form \(\frac{dP}{dt}=kP\left(1-\frac{P}{L}\right)\) (HL enrichment, not in the booklet) or the full syllabus table? See Logistic & Other Models.
Worked examples
A population follows \(P(t) = \dfrac{500}{1 + 9e^{-0.4t}}.\) Find the initial population.
Worked solution
At \(t = 0\): \(P(0) = \dfrac{500}{1 + 9}\) M1
\(= \dfrac{500}{10} = 50.\) A1
For \(P(t) = \dfrac{2000}{1 + 24 e^{-0.5t}}\), find the time at which the population reaches \(90\%\) of capacity.
Worked solution
\(90\%\) of 2000 is 1800: \(1 + 24 e^{-0.5t} = \dfrac{2000}{1800}\) M1
\(= 1.111.\) A1
\(24 e^{-0.5t} = 0.111 \Rightarrow e^{-0.5t}\) M1
\(= 0.004630.\) A1
\(t \approx 10.7\) yr. A1
A logistic model has carrying capacity 1200 and its inflection point (half capacity, 600) occurs at \(t = 5.\) Write the model in the form \(\dfrac{1200}{1 + C e^{-kt}}\) given also that \(P(0) = 120.\)
(a) State \(C.\)
(b) State \(k.\)
Worked solution
(a) From \(P(0) = 120\): \(\dfrac{1200}{1+C} = 120 \Rightarrow C\) M1
\(= 9.\) A1
(b) Inflection (\(P=600\)) at \(t=5\): \(1 + 9 e^{-5k} = 2 \Rightarrow e^{-5k}\) M1
\(= \tfrac19.\) A1
\(k = \dfrac{\ln 9}{5} \approx 0.439.\) A1
Model: \(P(t) = \dfrac{1200}{1 + 9 e^{-0.439t}}.\) A1
Common mistakes
- Confusing a saturation model with a logistic model. \(S(t)=L(1-e^{-kt})\) is steepest at \(t=0\) and has no S-shape - don't apply the "fastest growth at half capacity" rule to it.
- Assuming growth is fastest right at the start. A full logistic curve is slowest near \(t=0\) and fastest at the inflection point, where \(P=L/2\) - not immediately.
- Treating the carrying capacity as a value that gets reached. \(P\) gets arbitrarily close to \(L\) as \(t\to\infty\), but the model never actually equals \(L\) for any finite \(t\) - useful when a question asks you to comment on the model's limitations.
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24 logistic-model questions, marked instantly like the real exam.
Quick answers
What is the logistic model formula?
\(f(x) = \dfrac{L}{1+Ce^{-kx}}\), with \(L, C, k > 0\). It's given in the formula booklet, so you don't need to memorise it.
What is the carrying capacity in a logistic model?
The carrying capacity is \(L\), the horizontal asymptote the model approaches as \(x\) (or \(t\)) increases - the value the quantity gets arbitrarily close to but never actually reaches.