MGU Mathematics Series — Algebra II
Part IV: Exponential, Logarithmic, and Sequence Models
Growth · Decay · Compound Interest · Half-Life · Doubling Time
Chapter 9 of 14
MGU Mathematics Series  ·  Algebra II  ·  Library Textbook Edition

Chapter 9:
Exponential
Functions

Exponential growth and decay, percent change models, compound interest, continuous growth, half-life, doubling time, and the foundation of logarithms — one of the most important function families in all of mathematics

■ Math Level: Algebra II ■

■ 14 Sections  •  Part IV of 5 ■

Follows Chapter 8: Radical Functions and Rational Exponents  ·  Prepares students for Chapter 10: Logarithmic Functions

Chapter 9 — Front Matter

Purpose and Overview

This chapter introduces exponential functions as one of the most important and widely applicable function families in mathematics. Unlike linear functions, which grow by adding a constant amount, exponential functions grow by multiplying by a constant factor. This distinction is not merely technical — it changes everything about how a quantity behaves over time. A population that doubles every ten years, a bank account earning compound interest, a radioactive substance losing half its mass at a steady rate, and a viral video accumulating views at an accelerating pace all follow exponential patterns.

Students who have completed Algebra I have seen exponential expressions in the context of exponent rules and basic growth patterns. This chapter builds on that foundation and develops the full structure of exponential functions: their equations, graphs, transformations, parameters, and real-world interpretations. By the end of this chapter, students should be able to identify an exponential function, graph it with and without transformations, interpret its parameters in context, and apply exponential models to problems in finance, science, and everyday life.

This chapter also lays the groundwork for Chapter 10, which introduces logarithmic functions as the inverse of exponential functions. The language and structure learned here will make logarithms far more accessible.

Opening Question

A town had a population of 5,000 people in the year 2000. Each year, the population grows by 8 percent. A neighboring town started with the same population but adds exactly 400 people each year.

Which town will be larger in the year 2030? By how much? Keep this question in mind as you work through the chapter. By Section 9.5, you will have the tools to answer it precisely — and the answer may surprise you.

14Sections
11Objectives
18Key Terms
15+Worked Examples

Learning Objectives

  1. 1Identify exponential functions and distinguish them from linear, quadratic, and power functions.
  2. 2Understand the roles of the initial value and the growth or decay factor.
  3. 3Write exponential equations from descriptions, tables, and two given points.
  4. 4Graph exponential functions, including transformations, and identify key features.
  5. 5Model percent growth and percent decay with exponential equations.
  6. 6Apply the compound interest formula to financial problems.
  7. 7Understand the idea of continuous exponential growth using Euler's number \(e\).
  8. 8Solve half-life and doubling time problems using appropriate exponential models.
  9. 9Apply exponential models to science, finance, and other real-world contexts.
  10. 10Recognize and correct common exponential function mistakes.
  11. 11Connect exponential functions to their inverse relationship with logarithms.
Chapter 9 — Contents

Sections at a Glance

Sections 9.1 – 9.3
Defining and Understanding Exponential Functions
What Makes a Function Exponential · Exponential Growth · Exponential Decay
Section 9.1

What Makes a Function Exponential

In Algebra I, you studied several important function families: linear functions, quadratic functions, and basic exponential patterns. In this chapter, you will develop exponential functions fully — understanding their structure, their behavior, and the wide variety of situations they model.

Recall that a linear function has the form \(f(x) = mx + b\). What characterizes a linear function is that each time the input increases by 1, the output increases by adding the same constant \(m\). This is called constant additive change.

An exponential function works differently. Each time the input increases by 1, the output is multiplied by the same constant. This is called constant multiplicative change, and it leads to a very different shape of growth.

Definition — Exponential Function

An exponential function is a function of the form \[f(x) = a \cdot b^x\] where \(a\) is a nonzero real number called the initial value, \(b\) is a positive real number with \(b \neq 1\) called the base, and \(x\) is the input variable appearing in the exponent.

Student Warning — Variable in the Exponent

The defining feature of an exponential function is that the variable \(x\) appears in the exponent, not the base. The function \(f(x) = x^3\) is a power function — the variable is the base. The function \(f(x) = 3^x\) is an exponential function — the variable is the exponent. These are completely different families with completely different behaviors. Do not confuse them.

Why the Base Matters

The behavior of an exponential function depends entirely on the value of the base \(b\).

Base ValueFunction TypeBehavior
\(b > 1\)Exponential growthOutput increases as \(x\) increases; graph rises steeply to the right
\(0 < b < 1\)Exponential decayOutput decreases as \(x\) increases; graph falls toward zero to the right
\(b = 1\)Not exponential\(f(x) = a \cdot 1^x = a\), a constant function — excluded from the definition
\(b \leq 0\)Not defined for all \(x\)Excluded; would produce undefined or imaginary values
Example 9.1.1
Identifying an Exponential Pattern from a Table
  • ProblemDetermine whether each table represents a linear or an exponential function.

    Table A: \(x\) = 0, 1, 2, 3, 4 and \(f(x)\) = 3, 6, 12, 24, 48
    Table B: \(x\) = 0, 1, 2, 3, 4 and \(g(x)\) = 5, 8, 11, 14, 17
  • ThinkFor Table A, compute the ratio of consecutive outputs. For Table B, compute the difference between consecutive outputs. A constant ratio means exponential; a constant difference means linear.
  • Solve Table A ratios: \(6/3 = 2\), \(12/6 = 2\), \(24/12 = 2\), \(48/24 = 2\). Constant ratio of 2 → exponential with base 2.

    Table B differences: \(8-5 = 3\), \(11-8 = 3\), \(14-11 = 3\), \(17-14 = 3\). Constant difference of 3 → linear.
  • CheckTable A equation: \(f(x) = 3 \cdot 2^x\). Verify: \(f(3) = 3 \cdot 2^3 = 3 \cdot 8 = 24\). ✓
Answer: Table A is exponential (\(f(x) = 3 \cdot 2^x\)). Table B is linear (\(g(x) = 3x + 5\)).
Practice 9.1Concept Check
  1. Classify each as a power function, an exponential function, a linear function, or none of these. Explain your reasoning.

    a. \(f(x) = 7^x\)    b. \(g(x) = x^7\)    c. \(h(x) = 7x\)    d. \(k(x) = (-3)^x\)
  2. A table shows outputs of 4, 8, 16, 32 as \(x\) goes from 0 to 3. Is this linear or exponential? Write the equation.
  3. A table shows outputs of 10, 7, 4, 1 as \(x\) goes from 0 to 3. Is this linear or exponential? Explain.
Section 9.2

Exponential Growth

When a quantity grows exponentially, it does not increase by the same amount each time period. It increases by the same factor each time period. The result is that growth accelerates: the larger the quantity becomes, the faster it grows.

A classic example is a population of bacteria that doubles every hour. If you start with 100 bacteria at hour 0, you have 200 at hour 1, 400 at hour 2, 800 at hour 3, and so on. Each hour, the population is multiplied by 2. In ten hours, the population exceeds 100,000. This acceleration is the hallmark of exponential growth.

Exponential Growth Form

\[f(x) = a \cdot b^x \quad \text{where } a > 0 \text{ and } b > 1\]

Here, \(a\) is the starting value (the output when \(x = 0\)), and \(b\) is the growth factor — the number by which you multiply at each step. Each increase of 1 in the input multiplies the output by \(b\).

Example 9.2.1
Writing an Exponential Growth Equation
  • ProblemA colony of bacteria starts with 200 cells and doubles every hour. Write a function \(f(t)\) that gives the number of bacteria after \(t\) hours.
  • ThinkThe starting value is 200. The growth factor (doubling) is 2. Each hour, the count is multiplied by 2, so the exponent is \(t\).
  • Solvef(t) = 200 · 2^t
  • CheckAt \(t = 0\): \(f(0) = 200 \cdot 1 = 200\). ✓ At \(t = 3\): \(f(3) = 200 \cdot 8 = 1600\). Doubling three times: \(200 \to 400 \to 800 \to 1600\). ✓
Answer: \(f(t) = 200 \cdot 2^t\), where \(t\) is measured in hours.
Example 9.2.2
Evaluating for a Specific Time
  • ProblemUsing the function from Example 9.2.1, how many bacteria are in the colony after 8 hours?
  • ThinkSubstitute \(t = 8\) into \(f(t) = 200 \cdot 2^t\).
  • Solvef(8) = 200 · 2^8 = 200 · 256 = 51,200
Answer: After 8 hours, there are 51,200 bacteria.
Practice 9.2Skill Practice
  1. Write: initial value 5, growth factor 3
  2. Write: initial value 100, triples each step
  3. Evaluate \(f(x) = 4 \cdot 3^x\) at \(x = 4\)
  4. Evaluate \(g(t) = 50 \cdot 2^t\) at \(t = 6\)
  5. A culture starts at 60 and quadruples daily. Write \(f(d)\). Find \(f(4)\).
  6. A savings pattern: \$500 doubles each year. Write \(A(t)\). Find \(A(5)\).
Section 9.3

Exponential Decay

When a quantity decays exponentially, it decreases by the same multiplicative factor at each step. Instead of multiplying by something greater than 1, you multiply by something between 0 and 1. The result is that the quantity gets smaller and smaller, approaching zero but never quite reaching it.

A common example is radioactive decay. A radioactive isotope may lose half its atoms in a fixed amount of time, called its half-life. If you start with 80 grams, after one half-life you have 40, then 20, then 10, then 5, and so on. The substance never completely disappears, but it becomes negligibly small. This approach toward zero without ever touching it is one of the defining features of exponential decay.

Exponential Decay Form

\[f(x) = a \cdot b^x \quad \text{where } a > 0 \text{ and } 0 < b < 1\]

The only structural difference from exponential growth is that the base \(b\) is a fraction between 0 and 1. Each increase of 1 in the input multiplies the output by \(b\), making the quantity smaller.

Example 9.3.1
Writing an Exponential Decay Equation
  • ProblemA cup of coffee starts at 180°F. Each minute, the temperature is 90% of the previous minute's temperature. Write an exponential function for the temperature after \(t\) minutes.
  • ThinkEach minute you keep 90% of the previous temperature, so you multiply by 0.9. The starting temperature is 180. The decay factor is 0.9.
  • SolveT(t) = 180 · (0.9)^t
  • CheckAt \(t = 0\): \(T(0) = 180\). ✓ At \(t = 1\): \(T(1) = 180 \cdot 0.9 = 162\). That is a 10% reduction. ✓ At \(t = 2\): \(T(2) = 180 \cdot 0.81 = 145.8\). ✓
Answer: \(T(t) = 180 \cdot (0.9)^t\), where \(t\) is measured in minutes.
Example 9.3.2
Evaluating a Decay Function
  • ProblemUsing the cooling function from Example 9.3.1, estimate the temperature of the coffee after 10 minutes.
  • ThinkSubstitute \(t = 10\) into \(T(t) = 180 \cdot (0.9)^t\).
  • Solve T(10) = 180 · (0.9)^10 (0.9)^10 ≈ 0.3487 T(10) ≈ 180 · 0.3487 ≈ 62.8°F
Answer: After 10 minutes, the coffee is approximately 62.8°F.
Interpretation Note — The Asymptote

Notice that the coffee never cools to exactly 0°F in this model. The function \(T(t) = 180 \cdot (0.9)^t\) approaches zero but never reaches it. This is a characteristic feature of exponential decay: the function has a horizontal asymptote at \(y = 0\). The graph gets closer and closer to the x-axis, but never touches it.

Practice 9.3Skill Practice
  1. Write: initial value 400, decay factor 0.5
  2. Write: initial value 1000, halves each step
  3. Evaluate \(f(x) = 300 \cdot (0.75)^x\) at \(x = 4\)
  4. Evaluate \(g(t) = 80 \cdot (0.6)^t\) at \(t = 3\)
  5. A car loses 15% of its value annually. Initial value \$20,000. Write \(V(t)\). Find \(V(5)\).
  6. A substance starts at 500 mg and loses 20% per hour. Write \(D(t)\). Find \(D(8)\).
Sections 9.4 – 9.6
Parameters, Percent Change, and Graphing
Initial Value and Growth Factor · Percent Models · Graphing Exponential Functions
Section 9.4

Initial Value and Growth Factor

Every basic exponential function \(f(x) = a \cdot b^x\) is completely determined by two parameters. The initial value \(a\) is the output when \(x = 0\), because \(b^0 = 1\) for any base. The base \(b\) is the growth or decay factor — the number by which the output is multiplied each time the input increases by 1. Finding both parameters is the key step in writing an exponential equation from data.

Example 9.4.1
Finding the Equation from a Table Starting at x = 0
  • ProblemFind the exponential equation for the table: \(x\) = 0, 1, 2, 3 and \(f(x)\) = 12, 36, 108, 324.
  • ThinkBecause the table starts at \(x = 0\), the initial value is simply \(f(0) = 12\). Find the base by computing the ratio of consecutive outputs.
  • Solve b = f(1)/f(0) = 36/12 = 3 Verify: 108/36 = 3 and 324/108 = 3 ✓ f(x) = 12 · 3^x
  • Check\(f(2) = 12 \cdot 9 = 108\). ✓
Answer: \(f(x) = 12 \cdot 3^x\)
Example 9.4.2
Finding the Equation from Two Points (no x = 0)
  • ProblemAn exponential function passes through \((2,\, 20)\) and \((4,\, 80)\). Find the equation.
  • ThinkUse \(f(x) = a \cdot b^x\) and substitute both points to create a system. Divide the two equations to eliminate \(a\) and solve for \(b\) first.
  • Solve f(2) = a · b^2 = 20 ...(1) f(4) = a · b^4 = 80 ...(2) Divide (2) by (1): b^2 = 80/20 = 4, so b = 2 Substitute into (1): a · 4 = 20, so a = 5 f(x) = 5 · 2^x
  • Check\(f(2) = 5 \cdot 4 = 20\). ✓ \(f(4) = 5 \cdot 16 = 80\). ✓
Answer: \(f(x) = 5 \cdot 2^x\)
Practice 9.4Skill Practice
  1. Find the exponential equation: \(x\) = 0, 1, 2, 3 and \(f(x)\) = 7, 21, 63, 189.
  2. Find the exponential equation: \(x\) = 0, 1, 2, 3 and \(g(x)\) = 400, 200, 100, 50.
  3. An exponential function passes through \((1, 6)\) and \((3, 54)\). Find the equation.
  4. An exponential function passes through \((0, 10)\) and \((2, 90)\). Find the equation.
  5. An exponential function passes through \((2, 48)\) and \((5, 384)\). Find the equation.
Section 9.5

Percent Change and Exponential Models

In many real-world applications, growth or decay is described as a percent change per period. A savings account earns 5% interest per year. A town's population grows by 2% per year. A car loses 15% of its value each year. These descriptions connect naturally to the exponential form \(f(t) = a \cdot b^t\).

Percent Change and the Base

If a quantity grows by \(r\%\) per period (where \(r\) is written as a decimal), the growth factor is \(b = 1 + r\).

If a quantity decays by \(r\%\) per period, the decay factor is \(b = 1 - r\).

In either case: \[f(t) = a \cdot b^t\] where \(a\) is the initial value and \(t\) is the number of time periods.

Example 9.5.1
Population Growth at a Percent Rate
  • ProblemA city has a population of 48,000. Each year, the population grows by 2.5%. Write a function for the population after \(t\) years, and estimate the population after 10 years.
  • ThinkInitial value: 48,000. Growth rate: 2.5% = 0.025. Growth factor: \(1 + 0.025 = 1.025\).
  • Solve P(t) = 48,000 · (1.025)^t P(10) = 48,000 · (1.025)^10 ≈ 48,000 · 1.2801 ≈ 61,444
  • CheckThe population grew by about 13,444 people over 10 years. A 2.5% annual rate compounding is reasonable for modest steady growth. ✓
Answer: \(P(t) = 48{,}000 \cdot (1.025)^t\). After 10 years: approximately 61,444 people.
Example 9.5.2
Depreciation of a Car
  • ProblemA car is purchased for \$28,000. It loses 12% of its value each year. Write a function for the car's value after \(t\) years, and find the value after 5 years.
  • ThinkDecay rate: 12% = 0.12. Decay factor: \(1 - 0.12 = 0.88\). Initial value: 28,000.
  • Solve V(t) = 28,000 · (0.88)^t V(5) = 28,000 · (0.88)^5 ≈ 28,000 · 0.5277 ≈ $14,776
Answer: \(V(t) = 28{,}000 \cdot (0.88)^t\). After 5 years: approximately \$14,776.
Returning to the Opening Question

Town A: starts at 5,000, grows 8% per year → \(A(t) = 5{,}000 \cdot (1.08)^t\)

Town B: starts at 5,000, adds 400 per year → \(B(t) = 5{,}000 + 400t\)

After 30 years: \(A(30) \approx 5{,}000 \cdot 10.063 \approx 50{,}313\)   and   \(B(30) = 5{,}000 + 12{,}000 = 17{,}000\).

Town A ends up with more than three times the population of Town B. Even a modest 8% annual growth, applied repeatedly, vastly outpaces any constant additive rate. This is the power of multiplicative growth.

Section 9.6

Graphing Exponential Functions

To understand the graphs of exponential functions, it helps to begin with two parent cases and identify their key features. All exponential graphs share a common structure, and these parent cases establish the pattern.

\(x\)\(f(x) = 2^x\) (Growth)\(g(x) = (1/2)^x\) (Decay)
\(-3\)\(1/8\)\(8\)
\(-2\)\(1/4\)\(4\)
\(-1\)\(1/2\)\(2\)
\(0\)\(1\)\(1\)
\(1\)\(2\)\(1/2\)
\(2\)\(4\)\(1/4\)
\(3\)\(8\)\(1/8\)
Key Features of \(f(x) = a \cdot b^x\) with \(a > 0\)

Domain: All real numbers \((-\infty,\, \infty)\)

Range: All positive real numbers \((0,\, \infty)\) when \(a > 0\)

y-intercept: \((0,\, a)\), since \(f(0) = a \cdot b^0 = a \cdot 1 = a\)

Horizontal asymptote: \(y = 0\) — the graph approaches but never touches the x-axis

No x-intercepts: The graph never crosses the x-axis

Growth (\(b > 1\)): Rises steeply right, flattens left. Always increasing.

Decay (\(0 < b < 1\)): Falls steeply right, rises left. Always decreasing.

Diagram Instruction

Draw two side-by-side coordinate planes. On the left, graph \(f(x) = 2^x\), showing the curve rising steeply to the right and flattening toward zero on the left. Mark the y-intercept at \((0, 1)\) and the horizontal asymptote at \(y = 0\). On the right, graph \(g(x) = (1/2)^x\), falling steeply right and rising on the left. Mark the y-intercept at \((0, 1)\) and the asymptote at \(y = 0\). Note that both graphs pass through \((0, 1)\) and that the two are reflections of each other across the y-axis.

Example 9.6.1
Graphing an Exponential Function
  • ProblemGraph \(f(x) = 3 \cdot 2^x\) by completing a table and identifying key features.
  • Solve x = -2: f(-2) = 3 · (1/4) = 0.75 x = -1: f(-1) = 3 · (1/2) = 1.5 x = 0: f(0) = 3 · 1 = 3 x = 1: f(1) = 3 · 2 = 6 x = 2: f(2) = 3 · 4 = 12 x = 3: f(3) = 3 · 8 = 24
  • Checky-intercept: \((0, 3)\). Asymptote: \(y = 0\). Growth: yes (\(b = 2 > 1\)). The graph rises steeply right and approaches zero left.
Answer: Exponential growth curve through \((0, 3)\); horizontal asymptote at \(y = 0\); domain all reals; range \((0, \infty)\).
Practice 9.6Graphing and Features
  1. Graph each function. For each, identify the y-intercept, state growth or decay, and describe the asymptote.

    a. \(f(x) = 4^x\)   b. \(g(x) = (0.5)^x\)   c. \(h(x) = 2 \cdot 3^x\)   d. \(k(x) = 5 \cdot (0.8)^x\)
  2. Two exponential functions share the same base \(b = 2\) but one has \(a = 1\) and the other has \(a = 10\). Describe how their graphs differ and how they are the same.
  3. Explain why the graphs of \(f(x) = 2^x\) and \(g(x) = (1/2)^x\) are reflections of each other across the y-axis.
Sections 9.7 – 9.9
Transformations, Compound Interest, and Continuous Growth
Transformations · Compound Interest Formula · The Natural Base e
Section 9.7

Transformations of Exponential Functions

Just as the transformation rules from Chapter 1 apply to all function families, they apply to exponential functions as well. The general transformed exponential function is:

General Transformed Exponential Function

\[f(x) = a \cdot b^{x - h} + k\]

a — vertical stretch or compression; if negative, reflection across x-axis

h — horizontal shift (right if \(h > 0\), left if \(h < 0\))

k — vertical shift; moves the horizontal asymptote from \(y = 0\) to \(y = k\)

b — growth or decay factor (unchanged by transformations)

Example 9.7.1
Vertical Shift — Moving the Asymptote
  • ProblemCompare \(f(x) = 2^x\) and \(g(x) = 2^x + 3\). Describe the transformation and identify the new asymptote and y-intercept of \(g\).
  • ThinkAdding 3 to the entire function shifts the graph 3 units upward. Every output increases by 3, including the asymptote.
  • Solve Transformation: shift up 3 units New asymptote: y = 3 (was y = 0) New y-intercept: g(0) = 2^0 + 3 = 1 + 3 = 4
Answer: Shift up 3. Asymptote: \(y = 3\). y-intercept: \((0, 4)\).
Example 9.7.2
Full Transformation Analysis
  • ProblemDescribe the transformation from \(y = 3^x\) to \(g(x) = -2 \cdot 3^{x+1} + 4\). Identify the y-intercept and horizontal asymptote of \(g\).
  • ThinkRead the parameters: \(a = -2\), \(h = -1\) (shift left 1), \(k = 4\) (shift up 4).
  • Solve a = -2: vertical stretch by 2, reflected across x-axis h = -1: shift left 1 unit k = 4: shift up 4 units Asymptote: y = 4 y-intercept: g(0) = -2·3^(0+1) + 4 = -2·3 + 4 = -6 + 4 = -2
  • CheckThe graph is reflected, stretched, shifted left 1, shifted up 4. The y-intercept is \((0, -2)\) and the asymptote is \(y = 4\). ✓
Answer: Reflect across x-axis, stretch vertically by 2, shift left 1, shift up 4. Asymptote: \(y = 4\). y-intercept: \((0, -2)\).
Diagram Instruction

Draw a coordinate plane showing \(y = 3^x\) as a dashed curve and \(g(x) = -2 \cdot 3^{x+1} + 4\) as a solid curve. Mark the horizontal asymptote of \(g\) at \(y = 4\) with a labeled dashed line. Mark the y-intercept of \(g\) at \((0, -2)\). Show that the solid curve falls steeply to the right (because of the negative coefficient) and approaches \(y = 4\) from below on the left side of the plane.

Practice 9.7Transformations
  1. Describe each transformation from \(y = 2^x\) and state the new y-intercept and horizontal asymptote.

    a. \(f(x) = 2^x + 6\)   b. \(f(x) = 2^{x-3}\)   c. \(f(x) = -2^x\)   d. \(f(x) = 3 \cdot 2^{x+1} - 4\)
  2. Write the equation of an exponential function with base 2, shifted right 4 units and up 5 units.
  3. A function has a horizontal asymptote at \(y = -2\) and passes through \((0, 1)\). It is based on \(y = 4^x\). Write the equation.
Section 9.8

Compound Interest

When money is deposited in a savings account, it earns interest. With simple interest, the interest is calculated only on the original principal — you earn the same fixed dollar amount each period. With compound interest, interest is calculated on the principal plus any interest already earned, which means each period's interest is larger than the last. Compound interest produces exponential growth.

Compound Interest Formula

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

A = final amount (balance after \(t\) years)

P = principal (initial deposit)

r = annual interest rate as a decimal (e.g., 6% → 0.06)

n = number of compounding periods per year

t = number of years

Compounding Frequency\(n\)
Annually1
Semiannually2
Quarterly4
Monthly12
Daily365
Example 9.8.1
Annual Compounding
  • ProblemYou deposit \$2,000 in an account that earns 6% annual interest, compounded annually. How much do you have after 5 years?
  • Think\(P = 2000\), \(r = 0.06\), \(n = 1\), \(t = 5\).
  • Solve A = 2000(1 + 0.06/1)^(1·5) = 2000 · (1.06)^5 (1.06)^5 ≈ 1.3382 A ≈ 2000 · 1.3382 ≈ $2,676.45
Answer: After 5 years, the balance is approximately \$2,676.45.
Example 9.8.2
Monthly Compounding — Same Rate, More Periods
  • ProblemYou deposit \$2,000 at 6% annual interest, compounded monthly. How much do you have after 5 years?
  • Think\(P = 2000\), \(r = 0.06\), \(n = 12\), \(t = 5\). Note that \(r/n = 0.06/12 = 0.005\) and \(nt = 60\).
  • Solve A = 2000(1 + 0.005)^60 = 2000 · (1.005)^60 (1.005)^60 ≈ 1.3489 A ≈ 2000 · 1.3489 ≈ $2,697.70
  • CheckMonthly compounding gives about \$21 more than annual compounding. More frequent compounding always yields a higher balance for the same rate. ✓
Answer: After 5 years with monthly compounding, the balance is approximately \$2,697.70.
Example 9.8.3
Finding the Principal (Working Backward)
  • ProblemHow much should you deposit now at 4% annual interest compounded quarterly to have \$10,000 in 8 years?
  • Think\(A = 10{,}000\), \(r = 0.04\), \(n = 4\), \(t = 8\). Solve for \(P\).
  • Solve 10,000 = P · (1.01)^32 (1.01)^32 ≈ 1.3749 P = 10,000 / 1.3749 ≈ $7,273
Answer: You should deposit approximately \$7,273 now.
Student Warning — Converting Percent to Decimal

Always convert the interest rate to a decimal before using the formula. A rate of 6% becomes \(r = 0.06\), not \(r = 6\). Using \(r = 6\) in the formula would give an absurd result: a growth factor of \((1 + 6/1) = 7\), meaning the balance would multiply by 7 each year — a 600% annual return, not 6%.

MGU Library Connection
  • For more detailed applications of compound interest in personal finance, annuities, and loan amortization, see the MGU Finance module on Time Value of Money
  • Exponential models studied here are foundational to all financial mathematics
  • The relationship between compounding frequency and the natural base \(e\) is developed fully in Calculus I
Section 9.9

Continuous Growth as an Introduction

In Section 9.8, you saw that more frequent compounding produces slightly more interest. What happens if you compound interest continuously — at every possible instant, not just once a month or once a day? Mathematically, this is expressed by taking the compound interest formula to the limit as \(n \to \infty\). The result introduces one of the most important constants in all of mathematics: Euler's number, denoted \(e\).

\[e \approx 2.71828182845\ldots\]

The number \(e\) is irrational: its decimal expansion never repeats and never ends. Like \(\pi\), it appears throughout mathematics and science because of its natural connection to growth, change, and calculus.

Continuous Growth and Decay Formula

\[A = Pe^{rt}\]

A = final amount   P = initial amount   e ≈ 2.718   r = rate as a decimal (positive for growth, negative for decay)   t = time in years

Example 9.9.1
Continuous Compounding
  • ProblemYou invest \$3,000 at 5% annual interest, compounded continuously. How much do you have after 4 years?
  • ThinkUse the continuous formula. \(P = 3000\), \(r = 0.05\), \(t = 4\).
  • Solve A = 3000 · e^(0.05 · 4) = 3000 · e^0.2 e^0.2 ≈ 1.2214 A ≈ 3000 · 1.2214 ≈ $3,664.21
Answer: After 4 years of continuous compounding at 5%, the balance is approximately \$3,664.21.
Scope Note

A full treatment of the number \(e\), its origin from the compounding limit, and its central role in calculus belongs in Precalculus and Calculus I. At this level, students should recognize the continuous growth formula, know that \(e \approx 2.718\), and be able to apply it in straightforward problems. The derivative of \(e^x\) is \(e^x\) itself — a remarkable property you will encounter in Calculus I.

MGU Library Connection
  • The number \(e\) and the exponential function \(e^x\) are treated in depth in the MGU Precalculus text
  • In Calculus I, students discover that \(\frac{d}{dx}(e^x) = e^x\) — making it uniquely important in differential equations, probability, and mathematical physics
Sections 9.10 – 9.12
Half-Life, Doubling Time, and Applied Models
Half-Life · Doubling Time · Exponential Models in Science and Finance
Section 9.10

Half-Life

The half-life of a substance is the amount of time it takes for the quantity to reduce to half its original amount. Half-life is used most commonly in nuclear physics to describe radioactive decay, but the concept applies anywhere a quantity decreases exponentially at a constant rate. The exponent in the half-life model counts the number of half-life periods that have passed — not the raw time.

Half-Life Model

\[Q(t) = Q_0 \cdot \left(\frac{1}{2}\right)^{t/h}\]

Q(t) = amount remaining at time \(t\)   Q₀ = initial amount   h = half-life (same units as \(t\))   t/h = number of half-lives elapsed

Example 9.10.1
Radioactive Decay
  • ProblemA sample contains 400 milligrams of a radioactive isotope with a half-life of 5 years. Write a function for the amount remaining after \(t\) years, and find the amount remaining after 20 years.
  • Think\(Q_0 = 400\), \(h = 5\). After 20 years, exactly \(20/5 = 4\) half-lives have passed.
  • Solve Q(t) = 400 · (1/2)^(t/5) Q(20) = 400 · (1/2)^4 = 400 · (1/16) = 25 mg
  • CheckFour halvings from 400: \(400 \to 200 \to 100 \to 50 \to 25\). ✓
Answer: \(Q(t) = 400 \cdot \left(\dfrac{1}{2}\right)^{t/5}\). After 20 years, 25 mg remain.
Example 9.10.2
Carbon Dating — Setting Up the Equation
  • ProblemCarbon-14 has a half-life of approximately 5,730 years. A fossil contains 30% of its original Carbon-14. Set up the equation to find the age of the fossil. (Do not solve for \(t\) — that requires logarithms.)
  • ThinkWe want \(Q(t)/Q_0 = 0.30\). Substitute into the half-life model and cancel \(Q_0\).
  • Solve 0.30 · Q₀ = Q₀ · (1/2)^(t/5730) 0.30 = (1/2)^(t/5730)
Answer: The equation \(0.30 = \left(\dfrac{1}{2}\right)^{t/5730}\) cannot be solved with algebra alone. We will solve it in Chapter 10 using logarithms.
Student Warning — The Exponent Is t/h, Not t or h

The exponent in the half-life model is \(\dfrac{t}{h}\) — the number of half-lives that have elapsed. A common error is writing just \(h\) or just \(t\) as the exponent. If you do that, you are computing the amount after exactly one half-life, not after \(t\) time units. Always divide the elapsed time by the half-life to count the number of periods.

Section 9.11

Doubling Time

The doubling time of a growing quantity is the amount of time it takes for the quantity to double. Just as half-life measures the speed of decay, doubling time measures the speed of growth. The model is structured the same way: the exponent counts the number of doubling periods that have passed.

Doubling Time Model

\[Q(t) = Q_0 \cdot 2^{t/d}\]

Q₀ = initial amount   d = doubling time   t = elapsed time (same units as \(d\))   t/d = number of doublings elapsed

Example 9.11.1
Population Doubling
  • ProblemA city of 25,000 people has a population doubling time of 20 years. Write a function for the population after \(t\) years, and find the population after 60 years.
  • Think\(Q_0 = 25{,}000\), \(d = 20\). Sixty years is \(60/20 = 3\) doubling periods.
  • Solve P(t) = 25,000 · 2^(t/20) P(60) = 25,000 · 2^3 = 25,000 · 8 = 200,000
  • CheckThree doublings from 25,000: \(25{,}000 \to 50{,}000 \to 100{,}000 \to 200{,}000\). ✓
Answer: \(P(t) = 25{,}000 \cdot 2^{t/20}\). After 60 years: 200,000 people.
The Rule of 72

A famous approximation in finance and science: to estimate the doubling time of a quantity growing at \(r\%\) per period, divide 72 by the rate: \(\text{Doubling time} \approx \dfrac{72}{r}\). For example, at a 6% annual growth rate, the doubling time is approximately \(72/6 = 12\) years. This is a useful mental shortcut, widely used in personal finance. The exact answer requires logarithms (Chapter 10).

Example 9.11.2
Using the Rule of 72
  • ProblemAn investment grows at 9% per year. Estimate how long it takes to double.
  • SolveDoubling time ≈ 72/9 = 8 years
Answer: The investment approximately doubles every 8 years.
Section 9.12

Exponential Models in Science and Finance

At this point you know several forms of exponential models. Choosing the right form depends on what information is given and what the context describes. The table below summarizes all forms studied in this chapter.

Model TypeFormulaUse When…
Basic form\(f(t) = a \cdot b^t\)Base is given directly
Percent growth\(f(t) = a(1 + r)^t\)Percent growth rate \(r\) is given
Percent decay\(f(t) = a(1 - r)^t\)Percent decay rate \(r\) is given
Compound interest\(A = P\left(1 + \frac{r}{n}\right)^{nt}\)Finance; compounding \(n\) times/year
Continuous growth/decay\(A = Pe^{rt}\)Continuous compounding; science models
Half-life\(Q(t) = Q_0 \cdot (1/2)^{t/h}\)Half-life \(h\) is given
Doubling time\(Q(t) = Q_0 \cdot 2^{t/d}\)Doubling time \(d\) is given
Example 9.12.1
Choosing and Applying a Model — Revenue Growth
  • ProblemA company's revenue grows at a constant rate of 15% per year. In its first year, the company earned \$500,000. Estimate the revenue in year 7 (interpreting year 1 as \(t = 0\)).
  • ThinkPercent growth model. Initial value \$500,000, growth rate 15% = 0.15, so growth factor is 1.15. Year 7 is \(t = 6\).
  • Solve R(t) = 500,000 · (1.15)^t R(6) = 500,000 · (1.15)^6 ≈ 500,000 · 2.313 ≈ $1,156,535
Answer: By year 7, the company's revenue is approximately \$1,156,535.
Example 9.12.2
Bacteria — Tripling Every 90 Minutes
  • ProblemA bacterial culture starts with 800 cells and triples every 90 minutes. How many bacteria are present after 6 hours?
  • ThinkFirst convert: 6 hours = 360 minutes. The tripling period is 90 minutes, so the number of tripling periods is \(360/90 = 4\). Use a doubling-time style model adapted for tripling.
  • Solve Q(360) = 800 · 3^(360/90) = 800 · 3^4 = 800 · 81 = 64,800
Answer: After 6 hours, there are 64,800 bacteria.
Example 9.12.3
Drug Concentration
  • ProblemA patient receives 500 mg of a drug. The drug is eliminated at 20% per hour. Write a function for the amount remaining, and find the amount after 8 hours.
  • ThinkDecay rate 20% → decay factor \(1 - 0.20 = 0.80\).
  • Solve D(t) = 500 · (0.80)^t D(8) = 500 · (0.80)^8 ≈ 500 · 0.1678 ≈ 83.9 mg
Answer: After 8 hours, approximately 83.9 mg of the drug remain in the body.
Practice 9.10 – 9.12Half-Life, Doubling, and Applications
  1. A substance has a half-life of 8 years and starts at 960 grams. Find the amount remaining after 8, 24, and 40 years.
  2. A bacterial population doubles every 4 hours and starts at 300. Write the function and find the population after 16 hours and after 24 hours.
  3. Use the Rule of 72 to estimate the doubling time for growth rates of 3%, 8%, and 12% per year.
  4. A smartphone depreciates at 25% per year. A new model costs \$900. Find its value after 3 years.
  5. A town had 12,000 people in 2010 and grows at 1.8% per year. Predict the population in 2035.
  6. An investor deposits \$6,000 at 4.5% compounded monthly. Find the balance after 7 years.
  7. A researcher has 1,000 mg of a substance with a half-life of 30 days. Find the amount remaining after 120 days.
  8. Compare the final balance of \$2,500 invested for 6 years at 4% interest compounded: (a) annually, (b) monthly, (c) continuously.
Sections 9.13 – 9.14
Common Mistakes and Preparing for Logarithms
Eight Recurring Errors · The Unsolved Equation Problem · What Logarithms Solve
Section 9.13

Common Exponential Function Mistakes

Mistake 1 — Confusing Exponential and Linear Functions
Outputs 3, 6, 12, 24 → student writes f(x) = 3x (linear)

Correction: A constant ratio (each output multiplied by 2) means exponential, not linear. Check differences for linear; check ratios for exponential. The equation is \(f(x) = 3 \cdot 2^x\).

Mistake 2 — Multiplying the Coefficient into the Base
4 · 2³ = 8³ = 512

Correction: \(4 \cdot 2^3 = 4 \cdot 8 = 32\). The coefficient \(a\) does not merge into the base. Evaluate the power first, then multiply by the coefficient.

Mistake 3 — Forgetting to Convert Percent to Decimal
5% growth rate → f(t) = 100 · (1 + 5)^t = 100 · 6^t

Correction: 5% = 0.05 as a decimal. The growth factor is \(1 + 0.05 = 1.05\), and the function is \(f(t) = 100 \cdot (1.05)^t\).

Mistake 4 — Confusing Growth Factor and Growth Rate
"The growth factor is 1.08, so the growth rate is 108%."

Correction: If the growth factor is \(b = 1.08\), the growth rate is \(r = b - 1 = 0.08 = 8\%\). The growth factor is always 1 more than the rate (when \(r\) is expressed as a decimal).

Mistake 5 — Misidentifying the Initial Value
Table starts at x = 2 with f(2) = 20 → student writes a = 20

Correction: The initial value is \(f(0)\), not the first listed value. If the table starts at \(x = 2\), work backward or use two points to find \(a\) properly.

Mistake 6 — Believing the Function Eventually Equals Zero
"The decay function eventually reaches zero after enough time."

Correction: Exponential functions with positive initial values never actually equal zero. They approach zero as a horizontal asymptote but never reach it. The domain includes all real numbers; the range for positive \(a\) is always \((0, \infty)\).

Mistake 7 — Using Simple Interest Instead of Compound
Compound interest problem: A = P + P · r · t

Correction: Compound interest uses \(A = P\left(1 + \dfrac{r}{n}\right)^{nt}\). Simple interest grows linearly; compound interest grows exponentially. The problem will specify which type.

Mistake 8 — Wrong Exponent in the Half-Life Formula
Q(t) = 400 · (1/2)^h instead of Q(t) = 400 · (1/2)^(t/h)

Correction: The exponent must be \(t/h\) — the number of half-lives elapsed. Plugging in just \(h\) gives the amount after exactly one half-life, regardless of what \(t\) is. The denominator \(h\) converts the time into the number of periods.

Section 9.14

Preparing for Logarithms

You have now developed a thorough understanding of exponential functions. You can write them, graph them, transform them, identify their parameters, and apply them to finance, science, and real-world models.

But one class of problems has appeared repeatedly in this chapter that you could not finish. In Example 9.10.2 and related problems, equations like

\[0.30 = \left(\frac{1}{2}\right)^{t/5730} \qquad \text{or} \qquad 1000 \cdot (1.06)^t = 5000\]

arose naturally, and you could not solve for \(t\) because the variable is locked in the exponent. No algebraic tool studied so far can isolate \(t\) when it appears as an exponent.

This is precisely the problem that logarithms were invented to solve. A logarithm is the inverse operation of exponentiation — it answers the question: "What power must I raise this base to in order to get this number?" Just as subtraction undoes addition and division undoes multiplication, a logarithm undoes an exponent.

Connecting to Chapter 10

If \(2^t = 8\), then \(\log_2(8) = t = 3\). The logarithm base 2 of 8 is the exponent that makes \(2^t = 8\) true.

In Chapter 10, you will define logarithms, evaluate them, learn their properties, graph logarithmic functions, and use them to solve every exponential equation you could not finish here — including the carbon dating problem, the doubling time problem, and the compound interest problem in which you want to find the time, not the balance.

Practice 9.14Readiness Check
  1. Set up (but do not solve) an exponential equation for each situation. Explain in words what would be needed to solve for \(t\).

    a. A bacteria population starts at 500 and triples each day. When will it reach 40,500?
    b. A \$1,000 investment grows at 7% annually. When will it reach \$2,000?
    c. A substance with a half-life of 25 years starts at 800 mg. When will it reach 100 mg?
  2. A student says: "I know that \(3^t = 81\), and I can see that \(t = 4\) because \(3^4 = 81\)." Explain how this reasoning uses the idea of a logarithm informally.
  3. Why can't ordinary algebraic operations (adding, subtracting, multiplying, dividing, taking roots) isolate a variable that appears in an exponent?
Chapter 9 — Back Matter

Chapter Summary

Key Terms Review

exponential function
A function of the form \(f(x) = a \cdot b^x\) where \(a \neq 0\), \(b > 0\), and \(b \neq 1\). The variable appears in the exponent.
initial value
The value of the function when \(x = 0\); the coefficient \(a\) in \(f(x) = a \cdot b^x\). Found by evaluating \(f(0) = a \cdot 1 = a\).
growth factor
The base \(b\) when \(b > 1\). Each unit increase in \(x\) multiplies the output by \(b\). Larger than 1 for growth.
decay factor
The base \(b\) when \(0 < b < 1\). Each unit increase in \(x\) multiplies the output by \(b\), reducing it. Between 0 and 1 for decay.
exponential growth
A pattern in which a quantity increases by a constant multiplicative factor over equal time intervals. Characterized by \(b > 1\).
exponential decay
A pattern in which a quantity decreases by a constant multiplicative factor over equal time intervals. Characterized by \(0 < b < 1\).
percent change
The rate of increase or decrease expressed as a percentage of the previous value. Growth rate \(r\) gives factor \(1+r\); decay rate gives factor \(1-r\).
compound interest
Interest calculated on both the principal and the accumulated interest from prior periods. Uses the formula \(A = P(1 + r/n)^{nt}\).
principal
The initial amount of money on which interest is calculated, denoted \(P\) in the compound interest formula.
natural base
Euler's number \(e \approx 2.71828\). Used as the base in continuous growth models and central to calculus.
continuous growth
An exponential model using \(A = Pe^{rt}\), representing growth that compounds at every instant rather than at discrete intervals.
half-life
The time required for a quantity undergoing exponential decay to reduce to half its original amount. Model: \(Q(t) = Q_0 \cdot (1/2)^{t/h}\).
doubling time
The time required for a quantity undergoing exponential growth to double. Model: \(Q(t) = Q_0 \cdot 2^{t/d}\). Approximated by the Rule of 72.
horizontal asymptote
A horizontal line that a graph approaches but never reaches. For basic exponential functions, the asymptote is \(y = 0\). Vertical shifts move it to \(y = k\).
Rule of 72
An approximation: doubling time \(\approx 72/r\), where \(r\) is the annual percent growth rate. Useful for quick mental calculations in finance.
transformation
A shift, reflection, stretch, or compression applied to a parent function. The general form is \(a \cdot b^{x-h} + k\).
compounding period
The frequency with which interest is calculated and added: annually (\(n=1\)), quarterly (\(n=4\)), monthly (\(n=12\)), daily (\(n=365\)).
logarithm
The inverse operation of exponentiation. Answers the question: "What exponent gives this value?" Introduced fully in Chapter 10.

Concept Review Questions

  1. What is the structural difference between a power function like \(f(x) = x^4\) and an exponential function like \(f(x) = 4^x\)? Why does this difference matter for the behavior of each function?
  2. How do you determine from a table of values whether a relationship is linear or exponential? What do you look for in each case?
  3. What does the initial value tell you about an exponential function? What does the base tell you?
  4. Explain in words why an exponential growth function eventually exceeds any linear function, even if the linear function appears larger at first.
  5. What does the horizontal asymptote of an exponential function represent, and why does the graph never cross it for positive initial values?
  6. When you transform \(y = 2^x\) to \(y = 2^x + 5\), what happens to the horizontal asymptote? What happens to the y-intercept?
  7. In the compound interest formula \(A = P\left(1 + r/n\right)^{nt}\), what does each variable represent? What happens to \(A\) as \(n\) increases, holding all else constant?
  8. What is the difference between the growth rate \(r\) and the growth factor \(b\) in a percent change model? How are they related?
  9. Explain in words what the half-life of a substance is. If a substance has a half-life of 10 years and you start with 640 grams, how many grams remain after 50 years?
  10. Why do the half-life and compound interest problems where you solve for \(t\) require logarithms? What makes those equations different?

Skill Practice

Part AWriting and Evaluating Exponential Functions
  1. Write: initial value 6, growth factor 4
  2. Write: initial value 100, decay factor 0.7
  3. Write: initial value 200, triples each step
  4. Write: initial value 50, halves each step
  5. Evaluate \(f(x) = 3 \cdot 5^x\) for \(x = 3\)
  6. Evaluate \(g(x) = 100 \cdot (0.6)^x\) for \(x = 4\)
  7. Evaluate \(h(t) = 8 \cdot 2^t\) for \(t = 0, 1, 2, 3\)
  8. Find \(a\) and \(b\): passes through \((0, 5)\) and \((1, 20)\)
  9. Find equation: passes through \((0, 3)\) and \((2, 75)\)
  10. Find equation: passes through \((1, 6)\) and \((3, 54)\)
Part BPercent Change Models
  1. Population 20,000, grows 3.5%/yr. Write \(P(t)\).
  2. Car value \$35,000, depreciates 10%/yr. Write \(V(t)\).
  3. Population model from #1: find \(P(10)\).
  4. Car model from #2: find \(V(6)\).
  5. Growth factor is 1.12. What is the growth rate?
  6. Decay factor is 0.85. What is the decay rate?
  7. Growth rate is 4.5%. What is the growth factor?
  8. Decay rate is 8%. What is the decay factor?
Part CCompound Interest
  1. \$1,000 at 4% annually for 10 years
  2. \$5,000 at 6% quarterly for 3 years
  3. \$800 at 3.5% monthly for 5 years
  4. \$3,000 at 5% continuously for 4 years
  5. Find \(P\) needed to reach \$20,000 in 10 years at 5% annual
  6. Compare annually vs. monthly: \$2,500 at 4% for 6 years
Part DHalf-Life and Doubling Time
  1. Half-life 8 yr, start 960 g: amount after 24 yr
  2. Half-life 12 yr, start 600 g: amount after 36 yr
  3. Doubles every 4 hr, start 300: after 16 hr
  4. Doubles every 4 hr, start 300: after 24 hr
  5. Rule of 72: doubling time at 3%/yr
  6. Rule of 72: doubling time at 12%/yr

Error Analysis

Each problem shows a student's work containing an error. Identify the mistake and write the correct solution.

Problem 1
A car worth $25,000 depreciates at 10%/yr → V(t) = 25,000 · (10)^t

What is wrong? What should the equation be?

Problem 2
Population starts at 400, grows by factor 1.05 per year. After 10 years: P = 400 + 1.05^10

What is wrong? What should the calculation be?

Problem 3
f(x) = 5 · 2^x has initial value 10 because at x = 1, f(1) = 10.

Explain the error. What is the correct initial value?

Problem 4
Half-life 12 years, start 600 g. After 24 years: Q = 600 · (1/2)^12

Identify the error. Write the correct calculation and answer.

Problem 5
$3,000 at 8% compounded monthly for 5 years: A = 3000(1 + 0.08)^(12·5)

What is wrong with the formula? Write the correct calculation.

Chapter 9 Checkpoint

Chapter 9 Checkpoint
Complete without assistance — assesses all major skills and concepts from Chapter 9
Part A — Concepts
  1. Is \(f(x) = x^5\) an exponential function? Is \(g(x) = 5^x\)? Explain the difference in a complete sentence.
  2. What value of \(b\) would make \(f(x) = a \cdot b^x\) a constant function? Why is this excluded from the definition?
  3. Explain why exponential functions (with positive initial values) have a horizontal asymptote but no x-intercept.
  4. How does the compound interest formula change when interest is compounded continuously rather than \(n\) times per year?
Part B — Skills
  1. Find the exponential equation passing through \((0, 8)\) and \((3, 216)\).
  2. A quantity starts at 500 and decays by 5% per year. Write the function and find the value after 12 years.
  3. Graph \(f(x) = 4 \cdot (1/3)^x\). Label the y-intercept and horizontal asymptote. State the domain and range.
  4. Describe all transformations from \(y = 2^x\) to \(y = -3 \cdot 2^{x-4} + 1\). Identify the new horizontal asymptote.
  5. You deposit \$6,000 at 4.5% annual interest compounded monthly. How much do you have after 7 years?
  6. A substance has a half-life of 15 hours. You start with 240 mg. How much remains after 45 hours?
Part C — Applications
  1. A city has 75,000 residents and is growing at 3.2% per year. Predict the population after 20 years.
  2. A new laptop depreciates at 18% annually. Its original price was \$1,200. Find its value after 4 years.
  3. A virus spreads so the number of infections doubles every 6 days. On day 0, there are 50 confirmed cases. How many cases are predicted on day 30?
  4. An investment of \$10,000 earns 6% interest compounded quarterly. How much is the investment worth after 25 years?
  5. A scientist starts with 1,000 mg of Carbon-14 (\(t_{1/2} = 5{,}730\) yr). Write the function and find the amount remaining after 11,460 years.
Part D — Error Analysis
  1. A student evaluates \(f(x) = 6 \cdot 2^x\) and says the initial value is 12 because "that's what \(f(1)\) gives." Identify and correct the error.
  2. A student writes \(V(t) = 15{,}000 \cdot (1 + 0.8)^t\) for a car that depreciates at 20% per year. What is wrong? Write the correct equation.

Answers may be placed in the answer key.

Looking Ahead
Chapter 10: Logarithmic Functions

Every major topic in this chapter points toward the same gap: when the variable appears in the exponent of an exponential equation, standard algebraic methods cannot isolate it. Equations like \(2^t = 256\), \(\left(\frac{1}{2}\right)^{t/5730} = 0.30\), and \(1000 \cdot (1.06)^t = 5000\) are all solved by the same tool: the logarithm.

A logarithm answers the question: "What power must I raise this base to in order to get this number?" In Chapter 10, you will learn to define \(\log_b(x)\) as the inverse of \(b^x\), evaluate logarithms with common base 10 and natural base \(e\), use the product, quotient, and power properties of logarithms, solve exponential equations by taking logarithms of both sides, graph logarithmic functions as reflections of exponential functions, and interpret logarithmic scales including the pH scale, the Richter scale, and the decibel scale.

Exponential and logarithmic functions are a matched pair. Everything you learned in this chapter is the foundation for understanding the next. You are ready to continue.

MGU Library Connection
  • Exponential and Logarithmic Rules — Appendix I of this textbook
  • Algebra I, Chapter 9: Exponents and Exponential Patterns — prerequisite foundation
  • Chapter 10 of this textbook: Logarithmic Functions — the inverse function pair
  • MGU Precalculus: Exponential and Logarithmic Functions in Depth (including \(e^x\), \(\ln x\), and exponential equations)
  • MGU Finance Module: Time Value of Money, Compound Interest, and Annuities
  • MGU Physics: Radioactive Decay, Nuclear Half-Life, and Carbon Dating
  • Calculus I: The derivative of \(e^x\) and the natural logarithm as an integral
  • MGU Dictionary entries: exponential function, base, growth factor, half-life, Euler's number, asymptote