Biology · Introductory biology · Worked example
Compare exponential and logistic growth
A herd of 50 deer has a per-capita growth rate r = 0.4 per year. Predict its size after 5 years with the exponential model, then with the logistic model and a carrying capacity K = 500.
Write the exponential model
N₀ = 50 and r = 0.4 per year, so N(t) = 50e^(0.4t) with t in years. After 5 years the exponent is 0.4 × 5 = 2.
Find the logistic constant
The logistic solution N(t) = K/(1 + Ae^(−rt)) needs A = (K − N₀)/N₀, which makes N(0) = 50.
Write the logistic model
With K = 500, A = 9 and r = 0.4, evaluate at t = 5.
Compare the growth rates at 5 years
The exponential herd is adding rN = 0.4 × 369.45 ≈ 148 deer a year and speeding up. The logistic herd adds rN(1 − N/K) ≈ 49.5 deer a year, close to its largest possible rate, rK/4 = 50, because it is near K/2.
Look further ahead
After 10 years the exponential model predicts about 2,730 deer, while the logistic herd has about 429 and is leveling off toward 500. The two models agree while the herd is small and part ways as crowding starts to matter.
Result
After 5 years: about 369 deer with exponential growth and about 225 with logistic growth. The logistic herd then levels off toward K = 500.
Your turn
When does the logistic herd reach half its carrying capacity, 250 deer?
Show the answer and explanation
After about 5.49 years.
Set 500/(1 + 9e^(−0.4t)) = 250, so 1 + 9e^(−0.4t) = 2 and 9e^(−0.4t) = 1, which means e^(0.4t) = 9. Then 0.4t = ln 9 and t = ln 9/0.4 ≈ 5.49.
Keep exploring
The population growth tool opens with the logistic model for these deer. Switch the model to exponential to compare the curves, or change K and r.
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