Chalk−1

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.

N⁢(t)=50e0.4⁢tN⁢(5)≈369.45

Find the logistic constant

The logistic solution N(t) = K/(1 + Ae^(−rt)) needs A = (K − N₀)/N₀, which makes N(0) = 50.

500−5050=9

Write the logistic model

With K = 500, A = 9 and r = 0.4, evaluate at t = 5.

N⁢(t)=5001+9e−0.4⁢tN⁢(5)≈225.43

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.

0.4⁢(369.45)≈147.780.4⁢(225.43)(1−225.43500)≈49.52
0.4⁢(369.45)≈147.780.4⁢(225.43)(1−225.43500)≈49.52

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.

e0.4⁢t=90.4⁢t=ln9t=ln90.4t≈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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