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Modeling with
Exponential & Logarithmic Equations

Learn to translate real-world growth and decay into powerful mathematical models using \( y = ab^x \) and \( y = a\ln(x) + c \).

Algebra 2

What Is Exponential Growth?

A quantity grows exponentially when its rate of change is proportional to its current value. The general model is:

If \( b > 1 \), the function models growth. If \( 0 < b < 1 \), it models decay.

Visualizing Growth vs Decay

Growth: \( b > 1 \)
Curve rises steeply to the right: population growth, compound interest
Decay: \( 0 < b < 1 \)
Curve falls toward zero: radioactive decay, depreciation

What Is a Logarithm?

A logarithm answers the question: "To what exponent must I raise the base to get this value?"

Logarithms are the inverse of exponentials. If exponential functions model rapid growth, logarithmic functions model the time needed to reach a given level.

The Natural Base \( e \)

Many real-world models use the natural base \( e \approx 2.71828 \), which arises from continuous compounding.

Continuous Growth
\( y = ae^{kt} \) where \( k > 0 \)
Continuous Decay
\( y = ae^{kt} \) where \( k < 0 \)
Natural Log
\( \ln(x) = \log_e(x) \), the inverse of \( e^x \)

The Core Modeling Framework

Every exponential/logarithmic model follows the same pattern: identify the initial value, determine the rate, then write the equation.

1
Identify the type: Is the quantity growing, decaying, or have you been given a logarithmic relationship?
2
Find initial value \( a \): This is the value when \( x = 0 \) (or \( t = 0 \)).
3
Find the rate/base: Use a second data point to solve for \( b \) or \( k \).
4
Write the model: Plug \( a \), \( b \) (or \( k \)) into \( y = ab^x \) or \( y = ae^{kt} \).

Exponential vs Logarithmic Graphs

When to Use Exponential vs Logarithmic

Exponential Model Logarithmic Model
Equation form \( y = ab^x \) or \( y = ae^{kt} \) \( y = a\ln(x) + c \)
Use when Output grows/decays rapidly Output grows quickly then levels off
Solving for unknown Take \( \log \) of both sides Exponentiate both sides
Real-world example Bacteria population doubling Earthquake magnitude scales

Step-by-Step: Building a Model

1
Read the problem: Identify the initial value and any given data points.
2
Choose the form: Percent change per period → \( y = a(1 + r)^t \). Continuous → \( y = ae^{kt} \).
3
Substitute known values: Plug in \( (x_1, y_1) \) to find \( b \) or \( k \).
4
Solve using logarithms: Isolate the exponential, then apply \( \ln \) or \( \log \).
5
Verify: Check that your model reproduces the original data points.

Example 1: Bacterial Growth

A colony starts with 500 bacteria and doubles every 3 hours. Write a model for the population \( P(t) \) after \( t \) hours.

1
Initial value: \( a = 500 \) (population at \( t = 0 \))
2
Growth factor: Doubles every 3 hours → \( b = 2 \), period = 3
3
Model: \( P(t) = 500 \cdot 2^{t/3} \)

Bacterial Growth Graph

Example 2: Radioactive Decay

A substance has a half-life of 5 years. If you start with 200 grams, find the mass \( M(t) \) after \( t \) years and determine when only 25 grams remain.

1
Model: \( M(t) = 200 \cdot \left(\frac{1}{2}\right)^{t/5} \)
2
Solve for \( t \): Set \( 25 = 200 \cdot (0.5)^{t/5} \)
3
\( \frac{25}{200} = (0.5)^{t/5} \) → \( 0.125 = (0.5)^{t/5} \)
4
\( \ln(0.125) = \frac{t}{5}\ln(0.5) \) → \( t = \frac{5\ln(0.125)}{\ln(0.5)} = \) 15 years

Example 3: Continuous Compounding

You invest $1,000 at 6% annual interest, compounded continuously. How long until the investment reaches $2,500?

1
\( 2500 = 1000 \cdot e^{0.06t} \)
2
\( 2.5 = e^{0.06t} \) → take \( \ln \) of both sides
3
\( \ln(2.5) = 0.06t \) → \( t = \frac{\ln(2.5)}{0.06} \approx \) 15.27 years

Example 4: Logarithmic Model

The magnitude \( M \) of an earthquake with energy \( E \) (in joules) is modeled by: \( M = \frac{2}{3}\log_{10}\!\left(\frac{E}{10^{4.4}}\right) \). An earthquake releases \( E = 10^{12} \) joules. Find its magnitude.

1
Substitute: \( M = \frac{2}{3}\log_{10}\!\left(\frac{10^{12}}{10^{4.4}}\right) \)
2
Simplify: \( M = \frac{2}{3}\log_{10}(10^{7.6}) = \frac{2}{3}(7.6) \)
3
Result: \( M \approx \) 5.07

Common Mistakes to Avoid

✕ Confusing Rate and Factor
A 5% growth rate means \( b = 1.05 \), not \( b = 0.05 \). The growth factor is \( 1 + r \).
✕ Forgetting the Ln Step
To isolate \( t \) in \( ab^t = c \), you must take \( \ln \) of both sides. You cannot just divide by \( b \).
✕ Wrong Log Base
Using \( \ln \) when the model uses base 10 (or vice versa). Match the logarithm to the base of your exponential.

Model Summary Reference

Scenario Model Solve for Time
% growth per period \( y = a(1+r)^t \) \( t = \frac{\ln(y/a)}{\ln(1+r)} \)
% decay per period \( y = a(1-r)^t \) \( t = \frac{\ln(y/a)}{\ln(1-r)} \)
Continuous growth/decay \( y = ae^{kt} \) \( t = \frac{\ln(y/a)}{k} \)
Half-life \( y = a\!\left(\frac{1}{2}\right)^{t/h} \) \( t = \frac{h\ln(y/a)}{\ln(0.5)} \)
Doubling time \( y = a \cdot 2^{t/d} \) \( t = \frac{d\ln(y/a)}{\ln 2} \)

Question 1 of 5

A population of 1,200 bacteria triples every 4 hours. Which equation models the population \( P(t) \) after \( t \) hours?

Question 2 of 5

A car depreciates continuously at a rate of 12% per year. If it costs $25,000 new, what is its value after 5 years using the continuous decay model?

Question 3 of 5

Solve for \( t \):   \( 800 = 200 \cdot e^{0.05t} \)

Question 4 of 5

A substance has a half-life of 8 years. If you begin with 100 grams, which expression gives the amount remaining after \( t \) years?

Question 5 of 5

The loudness in decibels is modeled by \( L = 10\log_{10}\!\left(\frac{I}{I_0}\right) \). If one sound is 1,000 times more intense than another, how many decibels louder is it?