Logarithmic and Exponential Models

Real-world applications including compound interest, population growth, and radioactive decay

Written and reviewed by the Study Mondo Education TeamLast updated April 18, 2026

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Logarithmic and Exponential Models

Exponential Growth Model

The general exponential growth model is:

$A(t) = A_0 e^{kt}$

where:

$A(t)$ = amount at time $t$
$A_0$ = initial amount (at $t = 0$ )
$k$ = growth rate constant ( $k > 0$ for growth)
$t$ = time
$e \approx 2.71828$ (Euler's number)

Finding the Growth Rate $k$

If you know two data points, you can solve for $k$ :

Substitute known values
Divide by $A_0$
Take natural log of both sides
Solve for $k$

Exponential Decay Model

Same formula, but $k < 0$ (negative):

$A(t) = A_0 e^{-kt}$

where $k > 0$ represents the decay rate.

Half-Life Formula

The half-life is the time it takes for half the substance to decay.

$A(t) = A_0\left(\frac{1}{2}\right)^{t/h}$

where $h$ is the half-life.

Relationship to $k$ : $h = \frac{\ln(2)}{k}$

Compound Interest Models

Compound Interest (n times per year)

$A = P\left(1 + \frac{r}{n}\right)^{nt}$

where:

$P$ = principal (initial investment)
$r$ = annual interest rate (as decimal)
$n$ = number of times compounded per year
$t$ = time in years

Continuous Compounding

$A = Pe^{rt}$

This is the limit as $n \to \infty$ (compounding infinitely often).

Doubling Time

The time it takes for a quantity to double:

$2A_0 = A_0e^{kt}$ $2 = e^{kt}$ $\ln(2) = kt$ $t = \frac{\ln(2)}{k}$

Doubling time: $t_d = \frac{\ln(2)}{k} \approx \frac{0.693}{k}$

Population Growth Models

Unlimited Growth (Malthusian)

$P(t) = P_0e^{kt}$

Assumes unlimited resources (not realistic long-term).

Logistic Growth

$P(t) = \frac{L}{1 + Ce^{-kt}}$

where:

$L$ = carrying capacity (maximum sustainable population)
$C$ = constant based on initial conditions
$k$ = growth rate

The population approaches $L$ as $t \to \infty$ .

Newton's Law of Cooling

Temperature of an object over time:

$T(t) = T_s + (T_0 - T_s)e^{-kt}$

where:

$T(t)$ = temperature at time $t$
$T_s$ = surrounding (ambient) temperature
$T_0$ = initial temperature
$k$ = cooling rate constant

Logarithmic Models

Some phenomena grow logarithmically:

$y = a + b\ln(x)$

Examples:

Earthquake intensity (Richter scale)
Sound intensity (decibels)
pH scale (acidity)

Richter Scale

Earthquake magnitude: $M = \log\left(\frac{I}{I_0}\right)$

where $I$ is intensity and $I_0$ is reference intensity.

An increase of 1 on the Richter scale means 10 times more intense!

Solving Applied Problems

General Strategy:

Identify the type of model needed
Write the equation with known values
Solve for the unknown (often $k$ first, then answer the question)
Check if the answer makes sense in context

📚 Practice Problems

1Problem 1medium

❓ Question:

A population of bacteria grows from 100 to 500 in 3 hours. Assuming exponential growth $P(t) = P_0e^{kt}$ , find the growth rate $k$ and predict the population after 5 hours.

💡 Show Solution

Solution:

Part 1: Find $k$

Given: $P_0 = 100$ , $P(3) = 500$ , $t = 3$

Step 1: Write the equation. $500 = 100e^{3k}$

Step 2: Divide by 100. $5 = e^{3k}$

Step 3: Take natural log. $\ln(5) = 3k$

Step 4: Solve for $k$ . $k = \frac{\ln(5)}{3} \approx \frac{1.609}{3} \approx 0.536 \text{ per hour}$

Part 2: Find $P(5)$

$P(5) = 100e^{0.536(5)}$ $= 100e^{2.68}$ $\approx 100(14.58)$ $\approx 1458$

Answers:

Growth rate: $k \approx 0.536$ per hour
Population after 5 hours: approximately 1,458 bacteria

2Problem 2easy

❓ Question:

Invest $5,000 at 6% annual interest compounded monthly. How much will you have after 10 years?

💡 Show Solution

Solution:

Given:

$P = 5000$ (principal)
$r = 0.06$ (6% as decimal)
$n = 12$ (monthly compounding)
$t = 10$ years

Step 1: Write the compound interest formula. $A = P\left(1 + \frac{r}{n}\right)^{nt}$

Step 2: Substitute values. $A = 5000\left(1 + \frac{0.06}{12}\right)^{12(10)}$

$= 5000\left(1 + 0.005\right)^{120}$

$= 5000(1.005)^{120}$

Step 3: Calculate. $(1.005)^{120} \approx 1.8194$

$A \approx 5000(1.8194) \approx 9097$

Answer: You will have approximately $9,097 after 10 years.

3Problem 3hard

❓ Question:

Carbon-14 has a half-life of 5,730 years. If a fossil contains 25% of its original carbon-14, how old is the fossil?

💡 Show Solution

Solution:

Given:

Half-life: $h = 5730$ years
Current amount: $A(t) = 0.25A_0$ (25% remaining)

Step 1: Write the half-life formula. $A(t) = A_0\left(\frac{1}{2}\right)^{t/h}$

Step 2: Substitute known values. $0.25A_0 = A_0\left(\frac{1}{2}\right)^{t/5730}$

Step 3: Divide by $A_0$ . $0.25 = \left(\frac{1}{2}\right)^{t/5730}$

Step 4: Recognize that $0.25 = \frac{1}{4} = \left(\frac{1}{2}\right)^2$ . $\left(\frac{1}{2}\right)^2 = \left(\frac{1}{2}\right)^{t/5730}$

Step 5: Set exponents equal. $2 = \frac{t}{5730}$

Step 6: Solve for $t$ . $t = 2(5730) = 11,460 \text{ years}$

Alternative method using logarithms:

$\ln(0.25) = \frac{t}{5730}\ln(0.5)$

$t = \frac{5730 \ln(0.25)}{\ln(0.5)} = 11,460 \text{ years}$

Answer: The fossil is approximately 11,460 years old.

4Problem 4medium

❓ Question:

A bacteria population starts at 500 and doubles every 3 hours. Write an exponential model and find the population after 12 hours.

💡 Show Solution

Step 1: Use exponential growth model: P(t) = P₀ · 2^(t/d) where P₀ = initial amount, d = doubling time

Step 2: Identify values: P₀ = 500 d = 3 hours

Step 3: Write the model: P(t) = 500 · 2^(t/3)

Step 4: Find population at t = 12: P(12) = 500 · 2^(12/3) = 500 · 2⁴ = 500 · 16 = 8000

Answer: Model: P(t) = 500 · 2^(t/3); Population after 12 hours: 8000

5Problem 5hard

❓ Question:

The magnitude M of an earthquake is given by M = log(I/I₀), where I is the intensity and I₀ is a reference intensity. How many times more intense is an earthquake of magnitude 7 compared to one of magnitude 5?

💡 Show Solution

Step 1: Set up equations for both magnitudes: M₁ = 7 = log(I₁/I₀) M₂ = 5 = log(I₂/I₀)

Step 2: Convert to exponential form: 7 = log(I₁/I₀) → I₁/I₀ = 10⁷ 5 = log(I₂/I₀) → I₂/I₀ = 10⁵

Step 3: Find the ratio I₁/I₂: I₁ = I₀ · 10⁷ I₂ = I₀ · 10⁵

I₁/I₂ = (I₀ · 10⁷)/(I₀ · 10⁵) = 10⁷/10⁵ = 10² = 100

Step 4: Interpretation: An earthquake of magnitude 7 is 100 times more intense than an earthquake of magnitude 5.

Answer: 100 times more intense

Explain using:

⚠️ Common Mistakes: Logarithmic and Exponential Models

Avoid these 4 frequent errors

🌍 Real-World Applications: Logarithmic and Exponential Models

See how this math is used in the real world

📝 Worked Example: Related Rates — Expanding Circle

Problem:

A stone is dropped into a still pond, creating a circular ripple. The radius of the ripple is increasing at a rate of $2$ cm/s. How fast is the area of the circle increasing when the radius is $10$ cm?

2Write the relationship between variables

3Differentiate both sides with respect to time

4Substitute known values

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❓ Frequently Asked Questions

What is Logarithmic and Exponential Models?▾

Real-world applications including compound interest, population growth, and radioactive decay

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Logarithmic and Exponential Models

Try the Interactive Version!

Logarithmic and Exponential Models

Exponential Growth Model

Finding the Growth Rate kkk

Exponential Decay Model

Half-Life Formula

Compound Interest Models

Compound Interest (n times per year)

Continuous Compounding

Doubling Time

Population Growth Models

Unlimited Growth (Malthusian)

Logistic Growth

Newton's Law of Cooling

Logarithmic Models

Richter Scale

Solving Applied Problems

📚 Practice Problems

1Problem 1medium

❓ Question:

2Problem 2easy

❓ Question:

3Problem 3hard

❓ Question:

4Problem 4medium

❓ Question:

5Problem 5hard

❓ Question:

⚠️ Common Mistakes: Logarithmic and Exponential Models

🌍 Real-World Applications: Logarithmic and Exponential Models

📝 Worked Example: Related Rates — Expanding Circle

Practice Lab

Practice with Flashcards

Browse All Topics

📌 Related Topics in Exponential and Logarithmic Functions

❓ Frequently Asked Questions

Finding the Growth Rate $k$