Unit 4 · Chapter 4.7

4.7Exponential and Logarithmic Models

Apply exponential growth and decay, continuous compound interest, logistic growth, and Newton's Law of Cooling to real-world problems. Identify the right model, find the growth constant, and answer questions about the quantity over time.

Every field that tracks change over time — biology, finance, medicine, physics, environmental science — relies on exponential and logarithmic models. Understanding these models is the bridge between pure math and real-world problem solving.

Essential Question

How do exponential and logarithmic models translate real-world phenomena — population growth, radioactive decay, compound interest, and disease spread — into equations we can analyze and solve?

Lesson Overview

Real-world quantities that grow or decay at a rate proportional to their current size are modeled by exponential functions. When growth is unlimited, we use A(t) = A₀eᵏᵗ. When growth is constrained by a maximum (carrying capacity), we use the logistic model. Finance uses the continuous compound interest formula A = Pe^(rt). Cooling and heating follow Newton's Law of Cooling. Quantities that grow slowly after an initial surge are often modeled with a logarithmic function. In every case, the key skill is identifying which model applies, setting up the equation, finding the growth/decay constant k, and then answering questions about the quantity at a given time.

Exponential & Logarithmic Models — ReferenceExponential GrowthA(t) = A₀·eᵏᵗ, k > 0Exponential DecayA(t) = A₀·eᵏᵗ, k < 0Continuous CompoundA = Pe^(rt)Logistic Growthf(t) = c / (1 + ae^(−bt))Logarithmic Modelf(t) = a + b·ln(t)

Five model types — identify which applies before setting up any equation

Finding k — The Key Step

  • You need two data points to find k
  • Plug both into A(t) = A₀eᵏᵗ
  • Divide the two equations to eliminate A₀
  • Solve for k using ln
  • k > 0 → growth; k < 0 → decay

Model Selection Guide

  • Unlimited growth: A(t) = A₀eᵏᵗ
  • Radioactive / drug decay: A(t) = A₀eᵏᵗ, k < 0
  • Money / continuous interest: A = Pe^(rt)
  • Population with limit: logistic model
  • Slow growth after surge: f(t) = a + b·ln(t)
  • Cooling / heating: Newton's Law
Growth (k> 0)Decay (k< 0)0tAA₀Key Derived FormulasHalf-Life (decay)When A(t) = A₀/2:t½ = −ln(2)/kor equivalently:t½ = ln(2)/|k|Doubling Time (growth)When A(t) = 2A₀:t₂ = ln(2)/kRule of 70 (approx):t₂ ≈ 70 / (100r%)c (carrying capacity)Slow startRapid growthLevels offtfT_s (ambient)T₀ (initial)T(t) = T_s + (T₀−T_s)eᵏᵗtT

Worked Examples

Example 1

A bacterial culture starts with 500 cells and doubles every 3 hours. Write an exponential growth model and find the population after 10 hours.

Model: A(t) = A₀·eᵏᵗ with A₀ = 500

Find k using the doubling condition: A(3) = 1000

500·e^(3k) = 1000 → e^(3k) = 2 → 3k = ln(2) → k = ln(2)/3 ≈ 0.2310

Model: A(t) = 500·e^(0.2310t)

At t = 10: A(10) = 500·e^(0.2310·10) = 500·e^2.310 ≈ 500·10.077 ≈ 5039 cells

Answer:A(t) = 500·e^(ln(2)/3 · t); A(10) ≈ 5039 cells
Example 2

Carbon-14 has a half-life of 5730 years. A fossil contains 30% of its original carbon-14. How old is the fossil?

Model: A(t) = A₀·eᵏᵗ. Use half-life to find k.

A(5730) = A₀/2 → e^(5730k) = 1/2 → 5730k = ln(1/2) = −ln(2)

k = −ln(2)/5730 ≈ −0.0001210

Set A(t) = 0.30·A₀: 0.30·A₀ = A₀·e^(kt) → 0.30 = e^(kt)

ln(0.30) = kt → t = ln(0.30)/k = −1.2040/(−0.0001210) ≈ 9950 years

Answer:The fossil is approximately 9950 years old.
Example 3

$8000 is invested at 4.5% annual interest compounded continuously. How long until the investment doubles?

Model: A = Pe^(rt) with P = 8000, r = 0.045

Double: A = 16000 → 16000 = 8000·e^(0.045t)

e^(0.045t) = 2 → 0.045t = ln(2)

t = ln(2)/0.045 = 0.6931/0.045 ≈ 15.4 years

Rule of 70 check: 70/4.5 ≈ 15.6 years ✓ (close approximation)

Answer:The investment doubles in approximately 15.4 years.
Example 4

A cup of coffee at 90°C is placed in a 22°C room. After 5 minutes it cools to 70°C. Find the temperature after 20 minutes.

Newton's Law of Cooling: T(t) = T_s + (T₀ − T_s)·eᵏᵗ

T_s = 22, T₀ = 90: T(t) = 22 + 68·eᵏᵗ

Use T(5) = 70: 70 = 22 + 68·e^(5k) → 48 = 68·e^(5k) → e^(5k) = 48/68

5k = ln(48/68) → k = ln(48/68)/5 ≈ −0.0693

T(20) = 22 + 68·e^(−0.0693·20) = 22 + 68·e^(−1.386) = 22 + 68·0.25 = 22 + 17 = 39°C

Answer:T(20) ≈ 39°C
Example 5

A logistic model for a town's population is P(t) = 50000/(1 + 24·e^(−0.15t)), where t is years since 2000. Find the population in 2020 and the carrying capacity.

Carrying capacity c = 50000 (numerator of logistic model)

t = 2020 − 2000 = 20

P(20) = 50000/(1 + 24·e^(−0.15·20)) = 50000/(1 + 24·e^(−3))

e^(−3) ≈ 0.0498 → 24·0.0498 ≈ 1.195

P(20) = 50000/(1 + 1.195) = 50000/2.195 ≈ 22779

Answer:P(2020) ≈ 22,779; carrying capacity = 50,000

Guided Practice

Guided Problem 1

A population of 2000 rabbits grows to 3500 in 4 years. Write an exponential growth model and predict the population after 10 years.

Hint: Use A(4) = 3500 with A₀ = 2000 to find k. Then evaluate A(10).

Guided Problem 2

A radioactive substance decays from 80 grams to 50 grams in 12 years. Find the half-life.

Hint: Find k using A(12) = 50, A₀ = 80. Then set A(t) = 40 (half of 80) and solve for t.

Guided Problem 3

$5000 is invested at 3.2% compounded continuously. What is the balance after 8 years?

Hint: Use A = Pe^(rt) with P = 5000, r = 0.032, t = 8.

Guided Problem 4

A turkey at 165°F is placed in a 70°F room. After 30 minutes it cools to 140°F. Find the temperature after 90 minutes.

Hint: Set up Newton's Law: T(t) = 70 + 95·eᵏᵗ. Use T(30) = 140 to find k, then evaluate T(90).

Guided Problem 5

The logistic model P(t) = 8000/(1 + 15·e^(−0.2t)) models fish in a lake. Find the initial population and the population after 10 years.

Hint: Initial population: P(0). For P(10), substitute t = 10 and simplify.

Key Vocabulary

Exponential Growth Model

A(t) = A₀eᵏᵗ with k > 0. Models quantities that increase at a rate proportional to their current size: bacteria, investments, populations.

Exponential Decay Model

A(t) = A₀eᵏᵗ with k < 0. Models quantities that decrease proportionally: radioactive decay, drug concentration, depreciation.

Continuous Compound Interest

A = Pe^(rt). P = principal, r = annual rate (decimal), t = time in years. The limit of compounding as the frequency approaches infinity.

Half-Life

The time for a decaying quantity to reach half its current value. t½ = ln(2)/|k|. Used in radioactive dating and pharmacology.

Doubling Time

The time for a growing quantity to double. t₂ = ln(2)/k. Rule of 70: t₂ ≈ 70/(annual % rate).

Logistic Growth Model

f(t) = c/(1 + ae^(−bt)). Models growth limited by a carrying capacity c. Produces an S-shaped curve.

Carrying Capacity

The maximum sustainable population or value in a logistic model. The horizontal asymptote as t → ∞.

Newton's Law of Cooling

T(t) = T_s + (T₀ − T_s)eᵏᵗ. Models how an object's temperature approaches the surrounding temperature T_s over time.

Check Your Understanding

Interactive Practice — 5 Questions

1

A population starts at 1000 and grows to 1500 in 5 years. What is the growth constant k (to four decimal places)?

2

Carbon-14 has a half-life of 5730 years. What fraction remains after 11460 years?

3

For the logistic model P(t) = 12000/(1 + 5e^(−0.3t)), what is the carrying capacity?

4

$10000 is invested at 5% compounded continuously for 6 years. Which expression gives the balance?

5

A substance decays with k = −0.0347. What is its half-life (to the nearest year)?

Independent Practice

Independent Practice

1

A city's population was 45,000 in 2010 and 58,000 in 2020. Write an exponential model and predict the population in 2030.

2

A drug has a half-life of 4 hours. A patient takes a 200 mg dose. How much remains after 10 hours?

3

$12,000 is invested at 6% compounded continuously. How long until the balance reaches $20,000?

4

A thermometer reads 5°C and is placed in a 25°C room. After 10 minutes it reads 15°C. Find the temperature after 25 minutes.

5

The logistic model P(t) = 30000/(1 + 29·e^(−0.4t)) models a social media platform's users (in thousands). Find the initial users, users after 5 years, and the carrying capacity.

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Common Mistakes

Using A(t) = A₀·(1 + r)^t instead of A(t) = A₀·eᵏᵗ when the problem says "continuous" growth or decay.

Continuous models use base e. Discrete (annual/monthly) models use (1 + r)^t. Read the problem carefully to identify which applies.

Forgetting to convert the percentage rate to a decimal: using r = 5 instead of r = 0.05 in A = Pe^(rt).

Always convert percent to decimal before substituting. 5% → r = 0.05.

Confusing the carrying capacity with the initial value in a logistic model.

In f(t) = c/(1 + ae^(−bt)), the carrying capacity is c (numerator). The initial value is f(0) = c/(1 + a).

Setting up Newton's Law as T(t) = T₀·eᵏᵗ — forgetting to subtract the ambient temperature.

Newton's Law is T(t) = T_s + (T₀ − T_s)·eᵏᵗ. The object approaches T_s, not zero.

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Math Tips

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Always identify A₀ (initial value) and the given condition (a second point) before solving for k.

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Half-life and doubling time are inverses of each other in structure: both use t = ln(2)/|k|. Memorize this one formula for both.

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The Rule of 70: doubling time ≈ 70 ÷ (annual % rate). Quick mental check for finance problems.

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In Newton's Law of Cooling, k is always negative (the object cools toward T_s). If you get a positive k, recheck your setup.

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For logistic models: P(0) = c/(1 + a). As t → ∞, P → c. These two facts let you quickly sanity-check any logistic answer.