MGU Mathematics Series — Trigonometry and Precalculus
Part III: Advanced Algebraic Functions
Exponential · Logarithmic · Inverse · Models · Calculus Readiness
Chapter 9 of 14
MGU Mathematics Series  ·  Trigonometry and Precalculus  ·  Library Textbook Edition

Chapter 9:
Exponential and
Logarithmic Functions

Growth, decay, the number e, inverse relationships, logarithmic properties, equations, scales, and models — the function pair that governs finance, science, and calculus

■ Math Level: Trigonometry and Precalculus ■

■ 14 Sections  •  Part III of 5 ■

Follows Chapter 8: Polynomial and Rational Functions  ·  Prepares students for Chapter 10: Sequences, Series, and Mathematical Induction  ·  Bridges to Calculus I

Chapter 9 — Front Matter

Purpose and Overview

Chapter 9 reviews and deepens exponential and logarithmic functions. Students who have met these functions briefly in Algebra II will now study them with greater precision, connect them to inverse function theory from Chapter 2, and apply them to serious mathematical models in finance, biology, chemistry, physics, and data science.

This chapter treats exponential and logarithmic functions as inverse perspectives on the same underlying structure. Both grow out of the idea that some quantities change by multiplication rather than addition. An exponential function describes multiplicative growth or decay directly. A logarithm inverts that process, asking: what exponent produces a given result?

Together, these two function families model more of the real world than any other pair in precalculus. They describe population growth, radioactive half-life, compound interest, pH, earthquake intensity, sound volume, stellar brightness, and the spread of information. Before calculus, students must understand their graphs, their algebra, and their connection to each other.

This chapter prepares students for Calculus I, where derivatives and integrals of exponential and logarithmic functions play a central role in modeling change.

Opening Question

A hospital needs to determine how long a radioactive tracer injected into a patient will remain at detectable levels. The tracer has a half-life of six hours, meaning that every six hours, exactly half of the remaining material decays. If the initial dose is 80 milligrams, how much remains after 24 hours? After one week? At what point does the remaining amount fall below 0.1 milligrams? These questions cannot be answered with linear or polynomial functions. By the end of this chapter, students will be able to build the model, evaluate it at any time, and solve for the moment when a threshold is crossed.

14Sections
14Objectives
24Key Terms
15+Worked Examples

Learning Objectives

  1. 1Evaluate and interpret exponential functions with integer, rational, and real exponents.
  2. 2Identify exponential growth and decay from equations, graphs, and context.
  3. 3Explain why the number e arises naturally and evaluate expressions involving e.
  4. 4Interpret and apply continuous growth and decay models.
  5. 5Define the logarithm as the inverse of an exponential function.
  6. 6Evaluate common and natural logarithms and interpret their meaning.
  7. 7Apply the product, quotient, power, and change-of-base properties of logarithms.
  8. 8Solve exponential equations by taking logarithms of both sides.
  9. 9Solve logarithmic equations by converting to exponential form.
  10. 10Graph exponential and logarithmic functions, identifying key features.
  11. 11Read and interpret logarithmic scales such as the Richter scale, decibel scale, and pH.
  12. 12Build and interpret exponential models in finance, science, and population studies.
  13. 13Recognize and correct common errors in exponential and logarithmic reasoning.
  14. 14Connect exponential and logarithmic functions to the derivative and integral in Calculus I.
Chapter 9 — Contents

Sections at a Glance

Sections 9.1 – 9.4
Exponential Functions: Growth, Decay, and the Number e
Exponential Functions · Growth and Decay Models · Euler's Number · Continuous Models
Section 9.1

Exponential Functions Review

What Makes a Function Exponential

An exponential function places the variable in the exponent. This is a fundamental difference from polynomial functions, where the variable is in the base. The function f(x) = x² raises a variable base to a fixed exponent. The function f(x) = 2ˣ raises a fixed base to a variable exponent. These two structures grow in completely different ways.

The general form of an exponential function is f(x) = bˣ, where b > 0 and b ≠ 1. The base b must be positive because negative bases produce undefined results for fractional exponents. The base b = 1 produces the constant function f(x) = 1, which is not useful as a model of change.

Example 9.1.1
Evaluating an exponential function
  • ProblemLet f(x) = 3ˣ. Evaluate f(0), f(1), f(–1), f(2), and f(½).
  • ThinkSubstitute each input and simplify using exponent rules. The output must always be positive.
  • Solve f(0) = 3⁰ = 1 f(1) = 3¹ = 3 f(–1) = 3⁻¹ = 1/3 f(2) = 3² = 9 f(½) = 3^(1/2) = √3 ≈ 1.732
  • CheckAll outputs are positive, which is always true for exponential functions with positive bases.
Answer: f(0) = 1, f(1) = 3, f(–1) = 1/3, f(2) = 9, f(½) = √3
Key Properties of f(x) = bˣ (b > 0, b ≠ 1)

Domain: all real numbers (–∞, ∞)

Range: all positive real numbers (0, ∞)

y-intercept: (0, 1) always, since b⁰ = 1 for any valid base

x-intercept: none — exponential functions never cross the x-axis

Horizontal asymptote: y = 0 as x → –∞ when b > 1; as x → ∞ when 0 < b < 1

One-to-one: passes the horizontal line test; inverse functions exist

Student Warning — Exponential vs. Polynomial

Students sometimes confuse f(x) = 2ˣ (exponential) with f(x) = x² (polynomial). These are different functions with entirely different behaviors. For large positive values of x, exponential functions grow far faster than any polynomial.

Section 9.2

Growth and Decay Models

Exponential growth occurs when the base is greater than 1. Each unit increase in input multiplies the output by the base — multiplicative change, not additive change. The standard growth model is A(t) = A₀ · bᵗ, where A₀ is the initial value, b > 1 is the growth factor, and t is time. The growth factor can be written b = 1 + r, where r is the growth rate as a decimal.

Exponential decay occurs when 0 < b < 1. The decay factor can be written b = 1 – r. Two important measures describe the rate of change: doubling time T (for growth) and half-life H (for decay).

Model TypeStandard FormInterpretation
Exponential GrowthA(t) = A₀ · bᵗ, b > 1Multiplied by b each unit of time
Exponential DecayA(t) = A₀ · bᵗ, 0 < b < 1Multiplied by b (a fraction) each unit of time
Doubling TimeA(t) = A₀ · 2^(t/T)Quantity doubles every T units of time
Half-LifeA(t) = A₀ · (½)^(t/H)Quantity halves every H units of time
Example 9.2.1
Radioactive decay — using the half-life model
  • ProblemA radioactive tracer has a half-life of 6 hours. The initial dose is 80 mg. How much remains after 24 hours?
  • ThinkUse the half-life model. After every 6 hours, the amount is multiplied by ½. Twenty-four hours represents 24 ÷ 6 = 4 half-lives.
  • RepresentA(t) = 80 · (½)^(t/6)
  • SolveA(24) = 80 · (½)^(24/6) = 80 · (½)⁴ = 80 · (1/16) = 5
  • CheckAfter 4 halvings: 80 → 40 → 20 → 10 → 5. Matches.
Answer: After 24 hours, 5 milligrams remain.
Example 9.2.2
Population growth — using a growth rate
  • ProblemA city has population 200,000 in 2020 and grows at 3% per year. Predict the population in 2035.
  • ThinkGrowth rate r = 0.03, so growth factor b = 1.03. From 2020 to 2035 is 15 years.
  • RepresentP(t) = 200,000 · (1.03)ᵗ
  • SolveP(15) = 200,000 · (1.03)¹⁵ ≈ 200,000 · 1.5580 ≈ 311,600
  • CheckAt 3% annual growth, the population roughly doubles in about 23 years. A result above 300,000 after 15 years is reasonable.
Answer: The projected population in 2035 is approximately 311,600.
Interpretation Note

Exponential growth models assume a constant growth rate, which real populations often do not maintain. Models must always be interpreted with awareness of their assumptions and limitations.

Section 9.3

The Number e as an Introduction

The number e is not invented arbitrarily. It emerges naturally when studying what happens to compound interest as the compounding period grows shorter and shorter. If a bank offers 100% annual interest on $1 compounded n times per year, the account balance after one year is (1 + 1/n)ⁿ. As n grows without bound, this expression approaches a specific limit — the number e.

Compounding Periods (n)FormulaResult
1 (annually)(1 + 1/1)¹2.0000
2 (semi-annually)(1 + 1/2)²2.2500
4 (quarterly)(1 + 1/4)⁴2.4414
12 (monthly)(1 + 1/12)¹²2.6130
365 (daily)(1 + 1/365)³⁶⁵2.7146
∞ (continuously)lim(n→∞) (1 + 1/n)ⁿe ≈ 2.71828…
Euler's Number

e = lim(n→∞) (1 + 1/n)ⁿ ≈ 2.71828…

The number e is irrational — its decimal expansion never terminates or repeats. It is one of the most important constants in mathematics, alongside π. The function f(x) = eˣ is its own derivative, making it the natural function for modeling continuous change.

Example 9.3.1
Evaluating the natural exponential function
  • ProblemEvaluate e¹, e², and e⁻¹ to four decimal places.
  • Solve e¹ ≈ 2.7183 e² ≈ 7.3891 e⁻¹ = 1/e ≈ 0.3679
Answer: These are standard reference values used throughout the chapter.
MGU Library Connection
  • For a deeper explanation of limits and the definition of e, see MGU Calculus I, Chapter 1: Limits
  • MGU Library Dictionary entry: Euler's number
Section 9.4

Continuous Growth and Decay

In Section 9.2, growth was modeled by multiplying by a fixed factor per fixed time period — discrete compounding. The general compound interest formula for discrete compounding is A = P(1 + r/n)^(nt), where P is principal, r is annual rate, n is periods per year, and t is time in years.

As n increases — daily, hourly, by the minute — compounding approaches a limit. When compounding happens continuously, the formula becomes the continuous growth model:

The Continuous Growth and Decay Model

A(t) = A₀ · e^(rt)

A₀ is the initial value when t = 0. r is the continuous growth rate (positive for growth, negative for decay). When r > 0, the quantity grows. When r < 0, the quantity decays.

Example 9.4.1
Continuous compound interest
  • ProblemAn investment of $5,000 earns 4% annual interest compounded continuously. How much is the investment worth after 10 years?
  • RepresentA(t) = 5000 · e^(0.04t)
  • SolveA(10) = 5000 · e^(0.04 × 10) = 5000 · e^(0.4) ≈ 5000 · 1.4918 ≈ 7459
  • CheckAt 4% compounded annually: 5000(1.04)¹⁰ ≈ 7401. Continuous compounding gives slightly more. Reasonable.
Answer: The investment is worth approximately $7,459 after 10 years.
Example 9.4.2
Continuous decay
  • ProblemA sample of 200 grams decays continuously at 3% per year. How much remains after 20 years?
  • RepresentA(t) = 200 · e^(–0.03t)
  • SolveA(20) = 200 · e^(–0.6) ≈ 200 · 0.5488 ≈ 109.8
Answer: Approximately 109.8 grams remain after 20 years.
Student Warning — Discrete vs. Continuous Rates

The rate r in the continuous model is not the same as the growth rate in the discrete model. A discrete annual growth rate of 4% corresponds to a continuous rate of r = ln(1.04) ≈ 0.0392. For most problems in this course, use the stated rate directly, but be aware of this distinction.

Sections 9.5 – 9.7
Logarithmic Functions: Inverses, Properties, and Computation
Logarithmic Definition · Common and Natural Logs · Four Core Properties
Section 9.5

Logarithmic Functions Review

Exponential functions raise a base to a power. Logarithms answer the inverse question: given a result, what power was used? The logarithm base b of a number x is the exponent to which b must be raised to produce x:

The Fundamental Equivalence

log_b(x) = y <=> b^y = x

Read as: "the logarithm base b of x is y" means the same as "b raised to y equals x." Students should be able to move fluently in both directions.

Example 9.5.1
Converting between logarithmic and exponential forms
  • Problem AConvert log₂(32) = 5 to exponential form.
  • Answer A2⁵ = 32
  • Problem BConvert 10³ = 1000 to logarithmic form.
  • Answer Blog₁₀(1000) = 3
  • Problem CEvaluate log₃(81).
  • ThinkWhat power of 3 gives 81? Since 3⁴ = 81, the answer is 4.
Answers: 2⁵ = 32; log₁₀(1000) = 3; log₃(81) = 4

Logarithms as Inverse Functions

From Chapter 2, inverse functions reverse input-output relationships. If f(x) = bˣ, its inverse is f⁻¹(x) = log_b(x). This produces two cancellation laws used constantly when solving equations:

Cancellation Laws

b^(log_b(x)) = x    for all x > 0

log_b(bˣ) = x    for all real x

Student Warning — Domain of Logarithms

The domain of f(x) = log_b(x) is (0, ∞). The logarithm is undefined for zero and for negative inputs. If the argument inside a logarithm is an expression like –x, check whether that expression is positive — do not assume log_b(–x) is always undefined without examining the sign of x.

Section 9.6

Common and Natural Logarithms

LogarithmNotationBaseKey Values
Common logarithmlog(x)10log(10) = 1; log(100) = 2; log(0.1) = –1
Natural logarithmln(x)eln(1) = 0; ln(e) = 1; ln(e²) = 2; ln(1/e) = –1

The natural logarithm is the inverse of the natural exponential function. The cancellation laws become ln(eˣ) = x and e^(ln x) = x. The natural logarithm appears throughout calculus, physics, chemistry, biology, and economics because of its deep connection to rates of change.

Example 9.6.1
Evaluating common and natural logarithms without a calculator
  • ProblemEvaluate log(10000), log(0.001), ln(e⁵), and ln(1).
  • Solve log(10000) = log(10⁴) = 4 log(0.001) = log(10⁻³) = –3 ln(e⁵) = 5 ln(1) = ln(e⁰) = 0
Answers: 4, –3, 5, 0
Calculator Tip

When a logarithm has a base other than 10 or e, use the change of base formula (Section 9.7) to evaluate it on a calculator. The log button computes base 10; the ln button computes base e.

Section 9.7

Logarithmic Properties

Logarithmic properties allow complex expressions to be simplified, expanded, or condensed. They come directly from the laws of exponents and are essential for solving exponential and logarithmic equations.

The Four Core Properties (b > 0, b ≠ 1, M and N positive)

Product: log_b(MN) = log_b(M) + log_b(N)

Quotient: log_b(M/N) = log_b(M) – log_b(N)

Power: log_b(Mᵖ) = p · log_b(M)

Change of Base: log_b(x) = log(x)/log(b) = ln(x)/ln(b)

Student Warning — No Property for Sums

The product property says log_b(MN) = log_b(M) + log_b(N). It does not say log_b(M + N) = log_b(M) + log_b(N). There is no logarithm property for the log of a sum. This is one of the most frequent errors in logarithmic algebra.

Example 9.7.1
Expanding a logarithmic expression
  • ProblemExpand log₂(8x³/y) using logarithm properties.
  • ThinkUse the quotient property first, then product property, then power property.
  • Solve log₂(8x³/y) = log₂(8x³) – log₂(y) = log₂(8) + log₂(x³) – log₂(y) = 3 + 3·log₂(x) – log₂(y)
Answer: 3 + 3·log₂(x) – log₂(y)
Example 9.7.2
Condensing a logarithmic expression
  • ProblemWrite 2·ln(x) + ln(x–1) – ln(5) as a single logarithm.
  • ThinkApply the power property to the coefficient, then use product and quotient properties.
  • Solve 2·ln(x) + ln(x–1) – ln(5) = ln(x²) + ln(x–1) – ln(5) = ln(x²(x–1)) – ln(5) = ln(x²(x–1)/5)
Answer: ln(x²(x–1)/5)
Example 9.7.3
Using the change of base formula
  • ProblemEvaluate log₇(200) to four decimal places.
  • Solvelog₇(200) = ln(200)/ln(7) = 5.2983/1.9459 ≈ 2.7228
  • CheckVerify that 7^2.7228 ≈ 200 on a calculator.
Answer: log₇(200) ≈ 2.7228
Sections 9.8 – 9.10
Solving Equations and Graphing
Exponential Equations · Logarithmic Equations · Graphs and Transformations
Section 9.8

Solving Exponential Equations

Strategy for Exponential Equations

1. Isolate the exponential expression on one side.

2. Take the natural logarithm (or common logarithm) of both sides.

3. Apply the power property to bring the exponent down to a coefficient.

4. Solve for the variable algebraically.

5. Check the result — exponential equations do not produce extraneous solutions, but the computation should be verified.

Example 9.8.1
Basic exponential equation
  • ProblemSolve 3ˣ = 50.
  • Think50 is not a power of 3, so take the natural logarithm of both sides.
  • Solve ln(3ˣ) = ln(50) x · ln(3) = ln(50) x = ln(50)/ln(3) ≈ 3.9120/1.0986 ≈ 3.561
  • Check3^3.561 ≈ 50. Verify on a calculator. ✓
Answer: x ≈ 3.561
Example 9.8.2
Exponential equation with same base
  • ProblemSolve 5 · 2^(3x–1) = 80.
  • ThinkIsolate the exponential first. Then check whether both sides can be written with the same base.
  • Solve 2^(3x–1) = 16 Since 16 = 2⁴, set exponents equal: 3x – 1 = 4 3x = 5 x = 5/3
  • Check5 · 2^(3(5/3)–1) = 5 · 2⁴ = 5 · 16 = 80. ✓
Answer: x = 5/3
Example 9.8.3
Solving for a continuous growth rate
  • ProblemA population doubles in 12 years. Find the continuous growth rate r using the model P(t) = P₀·e^(rt).
  • ThinkWhen the population doubles, P(12) = 2P₀. Substitute and solve for r.
  • Solve 2P₀ = P₀ · e^(12r) 2 = e^(12r) ln(2) = 12r r = ln(2)/12 ≈ 0.6931/12 ≈ 0.0578
Answer: The continuous growth rate is approximately 5.78% per year. The formula r = ln(2)/T gives the rate when T is the doubling time.
Section 9.9

Solving Logarithmic Equations

Strategy for Logarithmic Equations

1. Isolate the logarithmic expression (or condense multiple logs into one).

2. Convert from logarithmic form to exponential form: log_b(x) = y means b^y = x.

3. Solve for the variable algebraically.

4. Check that all arguments of logarithms in the original equation are positive. Reject any extraneous solutions.

Example 9.9.1
Basic logarithmic equation
  • ProblemSolve log₄(x) = 3.
  • Solvex = 4³ = 64
  • Checklog₄(64) = 3 because 4³ = 64. The argument 64 is positive. ✓
Answer: x = 64
Example 9.9.2
Equation requiring condensing — checking for extraneous solutions
  • ProblemSolve log(x) + log(x – 3) = 1.
  • ThinkCondense the left side using the product property, then convert to exponential form.
  • Solve log(x(x–3)) = 1 x(x–3) = 10 x² – 3x – 10 = 0 (x–5)(x+2) = 0 x = 5 or x = –2
  • Checkx = 5: log(5) and log(2) are defined. ✓   x = –2: log(–2) is undefined. ✗ Reject x = –2.
Answer: x = 5. The solution x = –2 is extraneous and is rejected.
Example 9.9.3
Natural logarithm equation
  • ProblemSolve ln(2x + 1) = 4.
  • Solve 2x + 1 = e⁴ 2x = e⁴ – 1 x = (e⁴ – 1)/2 ≈ (54.598 – 1)/2 ≈ 26.799
  • CheckThe argument 2(26.799) + 1 = 54.598 > 0. ✓
Answer: x = (e⁴ – 1)/2 ≈ 26.799
Student Warning — Extraneous Solutions

After solving a logarithmic equation, always substitute back to check that every logarithm in the original equation has a positive argument. Extraneous solutions are common when the equation involves expressions like x, x – k, or x² inside logarithms.

Section 9.10

Exponential and Logarithmic Graphs

The Graph of f(x) = bˣ

Graph instruction: Draw the graph of f(x) = 2ˣ for x from –3 to 3. Plot the points (–3, 1/8), (–2, 1/4), (–1, 1/2), (0, 1), (1, 2), (2, 4), (3, 8). Connect with a smooth increasing curve. Mark the y-intercept at (0, 1). Draw a dashed horizontal line along the x-axis to show the asymptote y = 0.

Featuref(x) = bˣ, b > 1f(x) = bˣ, 0 < b < 1
Domain(–∞, ∞)(–∞, ∞)
Range(0, ∞)(0, ∞)
y-intercept(0, 1)(0, 1)
BehaviorIncreasingDecreasing
Horizontal asymptotey = 0 as x → –∞y = 0 as x → ∞

The Graph of f(x) = log_b(x)

Because the logarithm is the inverse of the exponential function, its graph is the reflection of the exponential graph across the line y = x. Graph instruction: Draw the graph of f(x) = log₂(x) for x from 1/8 to 8. Plot the points (1/8, –3), (1/4, –2), (1/2, –1), (1, 0), (2, 1), (4, 2), (8, 3). Connect with a smooth increasing curve. Mark the x-intercept at (1, 0). Draw a dashed vertical line along the y-axis to show the asymptote x = 0.

Transformations

The transformation rules from Chapter 2 apply directly. For f(x) = a · b^(x–h) + k: a is a vertical stretch or reflection; h is a horizontal shift; k is a vertical shift that relocates the horizontal asymptote to y = k. For f(x) = a · log_b(x – h) + k: h is a horizontal shift that relocates the vertical asymptote to x = h; k is a vertical shift.

Example 9.10.1
Graphing a transformed exponential function
  • ProblemDescribe the graph of f(x) = 3^(x–2) + 1.
  • ThinkThis is the graph of y = 3ˣ shifted right 2 units and up 1 unit.
  • Key Features Horizontal asymptote: y = 1 (shifted from y = 0) y-intercept: f(0) = 3⁻² + 1 = 1/9 + 1 ≈ 1.11 Passes through (2, 2): 3⁰ + 1 = 2 Domain: (–∞, ∞); Range: (1, ∞)
  • Graph DescriptionA smooth increasing curve that approaches y = 1 from above as x → –∞, passes through (2, 2), and rises steeply for large x.
Answer: Asymptote y = 1; rightward and upward shift of y = 3ˣ.
Sections 9.11 – 9.12
Logarithmic Scales and Real-World Models
Richter Scale · Decibels · pH · Compound Interest · Newton's Law of Cooling · Carbon Dating
Section 9.11

Logarithmic Scales

Many phenomena span enormous ranges of values. Earthquake energies vary by more than a trillion times. Sound intensities from a whisper to a jet engine vary by around a quadrillion times. pH values in chemistry reflect hydrogen ion concentrations differing by factors of a billion. On a standard linear scale, the small values would be invisible. Logarithmic scales solve this by compressing large ranges — on a logarithmic scale, each equal step represents multiplication by the same factor.

ScaleFormulaReference
Richter ScaleM = log(A/A₀)A₀ = reference amplitude
Decibel Scaleβ = 10 · log(I/I₀)I₀ = 10⁻¹² W/m² (threshold of hearing)
pHpH = –log([H⁺])[H⁺] in moles per liter; neutral pH = 7
Key Insight — Logarithmic Scales

A magnitude 7 earthquake is not twice as powerful as a magnitude 6. Because the Richter scale is logarithmic, each whole-number increase represents a tenfold increase in amplitude. A difference of 2 units means 10² = 100 times the amplitude.

Example 9.11.1
Comparing earthquake magnitudes
  • ProblemHow many times larger in amplitude is a magnitude 8.2 earthquake than a magnitude 6.0 earthquake?
  • ThinkThe difference in magnitudes is 8.2 – 6.0 = 2.2. The amplitude ratio is 10 raised to that difference.
  • SolveAmplitude ratio = 10^2.2 ≈ 158
Answer: The magnitude 8.2 earthquake has a seismic wave amplitude approximately 158 times that of the magnitude 6.0 earthquake.
Example 9.11.2
Computing pH
  • ProblemA solution has [H⁺] = 4.3 × 10⁻⁵ mol/L. Find the pH.
  • Solve pH = –log(4.3 × 10⁻⁵) = –(log(4.3) + log(10⁻⁵)) = –(0.6335 – 5) = 4.37
Answer: pH ≈ 4.37, indicating an acidic solution (below 7).
Section 9.12

Exponential Models in Finance and Science

Compound Interest

Compound Interest Formulas

Discrete: A = P(1 + r/n)^(nt)

Continuous: A = Pe^(rt)

P = principal; r = annual rate as a decimal; n = periods per year; t = time in years; A = final amount

Example 9.12.1
Comparing compounding methods
  • Problem$10,000 is invested at 5% annual interest for 20 years. Compare annual, monthly, and continuous compounding.
  • Solve Annual (n=1): 10000(1.05)²⁰ ≈ $26,533 Monthly (n=12): 10000(1 + 0.05/12)²⁴⁰ ≈ $27,126 Continuous: 10000 · e^(0.05×20) ≈ $27,183
  • InterpretContinuous compounding yields the highest return, but the difference between frequent discrete compounding and continuous compounding is small.
Answer: Annual → $26,533; Monthly → $27,126; Continuous → $27,183.

Newton's Law of Cooling

Newton's Law of Cooling describes how a warm object cools toward the temperature of its environment:

T(t) = T_env + (T₀ – T_env) · e^(–kt)

T_env is the surrounding temperature, T₀ is the initial temperature of the object, k > 0 is a cooling constant, and t is time.

Example 9.12.2
Newton's law of cooling — finding k and predicting temperature
  • ProblemCoffee at 90°C cools in a room at 22°C. After 10 minutes it is 70°C. Find k and the temperature after 30 minutes.
  • RepresentT(t) = 22 + 68 · e^(–kt)
  • Find k 70 = 22 + 68 · e^(–10k) 48 = 68 · e^(–10k) e^(–10k) = 48/68 = 12/17 –10k = ln(12/17) ≈ –0.3483 k ≈ 0.03483
  • Predict T(30) = 22 + 68 · e^(–0.03483 × 30) ≈ 22 + 68(0.3518) ≈ 45.9°C
Answer: k ≈ 0.0348; after 30 minutes the coffee is approximately 45.9°C.

Radioactive Decay and Half-Life

Half-Life and Continuous Decay Rate

H = ln(2)/r <=> r = ln(2)/H

If the half-life H is known, the continuous decay rate r is ln(2)/H ≈ 0.6931/H.

Example 9.12.3
Carbon-14 dating
  • ProblemCarbon-14 has a half-life of approximately 5,730 years. A fossil contains 30% of the original carbon-14. Estimate its age.
  • Step 1r = ln(2)/5730 ≈ 0.6931/5730 ≈ 0.0001210
  • Step 2 0.30·A₀ = A₀ · e^(–0.0001210·t) 0.30 = e^(–0.0001210·t) ln(0.30) = –0.0001210·t –1.2040 = –0.0001210·t t = 1.2040/0.0001210 ≈ 9,950
Answer: The fossil is approximately 9,950 years old.
MGU Library Connection
  • For more on logarithmic scales in chemistry: MGU Chemistry course pages in the Library
  • For more on Newton's Law of Cooling and applications in physics: MGU Physics course pages
  • For financial modeling with exponential functions: MGU Finance and Data Science courses
Sections 9.13 – 9.14
Common Mistakes and Calculus Readiness
Eight Recurring Errors · Derivatives Preview · Bridge to Calculus I
Section 9.13

Common Exponential and Logarithmic Mistakes

Mistake 1 — Applying log properties to sums
log(x + y) = log(x) + log(y)

The product property applies to products, not sums. log(xy) = log(x) + log(y). There is no simple formula for log(x + y).

Mistake 2 — Misapplying the power property
ln(x²) = (ln x)²

Correct: ln(x²) = 2·ln(x). The power property moves the exponent to a coefficient in front — it does not square the logarithm itself.

Mistake 3 — Forgetting extraneous solutions
Keeping x = –2 in log(x) + log(x–3) = 1 because the algebra works out

Always verify that every argument of every logarithm in the original equation is positive after solving. If log(–2) appears, the solution is extraneous.

Mistake 4 — Treating inverse notation as a reciprocal
log⁻¹(x) = 1/log(x)

In the context of inverse functions, f⁻¹ means the inverse function, not the reciprocal. log⁻¹(x) means 10ˣ, the inverse of log base 10 — not 1/log(x).

Mistake 5 — Distributing exponents over addition
e^(x+y) = eˣ + eʸ

Correct: e^(x+y) = eˣ · eʸ. Exponents distribute over multiplication, not addition.

Mistake 6 — Setting eˣ = negative number
Solving e^(2x) + 5 = 0 and writing e^(2x) = –5, then ln(–5) = 2x

The exponential function eˣ is always positive. If an equation requires eˣ = negative number, there is no real solution.

Mistake 7 — Forgetting domain on logarithms
Writing log(–3) or ln(0) as part of a solution

Both are undefined. Any solution that requires evaluating a logarithm of a non-positive number must be discarded as extraneous.

Mistake 8 — Confusing continuous rate r with discrete rate
Using r = 0.08 in both A = P(1+r)ᵗ and A = Pe^(rt) as if they are the same

The continuous rate r in Pe^(rt) is not the same as the annual rate in P(1+r)ᵗ. The continuous rate is r = ln(1 + rate). For a 8% annual rate, the continuous rate is ln(1.08) ≈ 0.0770.

Section 9.14

Preparing for Calculus Applications

Exponential and logarithmic functions are not isolated topics in precalculus. They are the foundation for some of the most important ideas in Calculus I and beyond.

In Calculus I, students will prove that the derivative of eˣ is eˣ itself — no other function has this property. It means that the rate at which eˣ is growing at any point equals the value of the function at that point. This makes eˣ the natural solution to equations of the form "the rate of change of a quantity equals the quantity itself" — which describes population growth, radioactive decay, Newton's law of cooling, electric current, and quantum probability amplitudes.

The natural logarithm has an equally elegant derivative: the derivative of ln x is 1/x. This allows calculus to integrate functions of the form 1/x, which is impossible using polynomial antiderivatives alone.

Exponential functions are also the solutions to separable differential equations of the form dy/dt = ky. Students who understand these functions well will recognize these solution forms immediately when they appear in Calculus II and Differential Equations.

In Calculus II, the number e and the function eˣ appear in Taylor series, where students discover that eˣ can be written as an infinite sum of polynomial terms — one of the most surprising and beautiful results in all of mathematics.

Calculus Connections — Preview

d/dx (eˣ) = eˣ    The natural exponential function is its own derivative.

d/dx (ln x) = 1/x    The natural logarithm solves the integration problem for 1/x.

dy/dt = ky → y = Ce^(kt)    Exponential functions solve separable growth/decay equations.

Readiness for Calculus I

Before Calculus I, students should be fluent with: the shape and features of eˣ and ln x; the cancellation laws e^(ln x) = x and ln(eˣ) = x; solving exponential and logarithmic equations exactly; building exponential models from growth rate and initial value; and using the continuous growth formula A = A₀e^(rt).

Every calculation in this chapter was preparation for that work. Students who leave Chapter 9 fluent in exponential and logarithmic algebra, graphs, and models are genuinely ready for the next step.

Chapter 9 — Back Matter

Chapter Summary

Key Terms Review

exponential function
A function of the form f(x) = bˣ, where b > 0 and b ≠ 1. The variable is in the exponent.
base
The fixed positive number b raised to the variable exponent in an exponential function.
growth factor
The factor b > 1 by which an exponential function multiplies per unit step.
decay factor
The factor b where 0 < b < 1 by which an exponential decay function multiplies per unit step.
Euler's number e
The irrational number ≈ 2.71828, the base of the natural exponential function; arises from continuous compounding.
continuous growth model
A(t) = A₀·e^(rt), where r is the continuous growth rate and t is time.
logarithm
The exponent to which a base must be raised to produce a given value. log_b(x) = y means b^y = x.
common logarithm
log(x) = log₁₀(x). Uses base 10. On a calculator, the log button.
natural logarithm
ln(x) = log_e(x). Uses base e. On a calculator, the ln button.
product property
log_b(MN) = log_b(M) + log_b(N). The log of a product equals the sum of the logs.
quotient property
log_b(M/N) = log_b(M) – log_b(N). The log of a quotient equals the difference of the logs.
power property
log_b(Mᵖ) = p·log_b(M). An exponent inside a logarithm moves to a coefficient outside.
change of base
log_b(x) = ln(x)/ln(b) = log(x)/log(b). Rewrites any logarithm using base e or base 10.
half-life
The time H required for an exponential decay quantity to halve. Related to rate by H = ln(2)/r.
doubling time
The time T required for a growing quantity to double. Related to rate by T = ln(2)/r.
extraneous solution
An algebraically derived solution that fails to satisfy the original equation due to a domain violation.
logarithmic scale
A scale in which equal distances represent equal multiplicative factors, compressing large data ranges.
pH
pH = –log([H⁺]). A logarithmic measure of hydrogen ion concentration. Neutral = 7; acidic < 7; basic > 7.
decibel
β = 10·log(I/I₀). A logarithmic unit for measuring sound intensity relative to the threshold of hearing.
Richter scale
M = log(A/A₀). A logarithmic scale for earthquake magnitude; each unit increase represents 10× the amplitude.

Concept Review Questions

  1. Explain in words what log_b(x) = y means. What question does it answer?
  2. Why is the domain of a logarithmic function restricted to positive inputs?
  3. How are the graphs of f(x) = 2ˣ and f(x) = log₂(x) related geometrically? What line do they reflect across?
  4. What is the difference between discrete and continuous compound interest? Which gives a larger final amount?
  5. Why does the number e appear in models of continuous growth and decay?
  6. A quantity grows by 8% per year. Write an exponential model using a growth factor and also as a continuous model. Which constant is larger, 0.08 or ln(1.08)? Explain.
  7. Explain why log(x + y) ≠ log(x) + log(y).
  8. What does the power property of logarithms allow you to do, and why is this useful for solving exponential equations?
  9. Why must students check for extraneous solutions when solving logarithmic equations but not when solving exponential equations?
  10. Name two ways the number e and the function eˣ appear in Calculus I.

Exponential Practice

Part AEvaluating and Modeling
  1. Evaluate f(x) = 5ˣ at x = 0, 1, 2, –1, –2, ½.
  2. For g(x) = (⅓)ˣ, evaluate g(0), g(1), g(–1), g(3).
  3. An investment grows from $2,000 at 6% annual interest compounded monthly. Evaluate after 5 years.
  4. Bacteria start at 500 and double every 4 hours. Write a model and find the count after 12 hours.
  5. A 400 g substance has a half-life of 10 years. How much remains after 35 years?
  6. Use continuous compounding to find how much $3,000 grows to in 15 years at r = 5.5%.
  7. Sketch f(x) = 3ˣ and g(x) = 3⁻ˣ on the same axes. Describe their relationship.
  8. Describe the transformation from y = eˣ to y = e^(x+3) – 2. Identify the new asymptote.

Logarithm Practice

Part BEvaluating, Converting, and Properties
  1. Evaluate: log₂(64), log₅(125), log₃(1/9), log₇(1).
  2. Convert to exponential form: log₄(16) = 2; log(0.01) = –2; ln(e³) = 3.
  3. Convert to logarithmic form: 2⁸ = 256; 10⁻⁴ = 0.0001; e⁰ = 1.
  4. Use change of base to evaluate log₆(250) to four decimal places.
  5. Expand: log₃(9x²/y³)
  6. Expand: ln(√(x³/e²))
  7. Condense: 3·log₂(x) – log₂(4) + log₂(y)
  8. Condense: 2·ln(x) – ½·ln(x+1) + ln(3)

Equation-Solving Practice

Part CExponential Equations
  1. 4ˣ = 256
  2. 7ˣ = 350
  3. 2^(x–1) = 32
  4. e^(3x) = 12
  5. 5 · 3ˣ = 135
  6. 2^(x+1) = 3ˣ (take ln of both sides)
  7. e^(2x) – 5eˣ + 6 = 0 (let u = eˣ)
  8. 3 · e^(2x+1) = 75
Part DLogarithmic Equations
  1. log₃(x+5) = 2
  2. ln(2x–3) = 1
  3. log(x) + log(x–21) = 2
  4. log₂(x+3) + log₂(x–1) = 5
  5. 2·ln(x) = ln(4x–3)
  6. log(x) – log(x–2) = log(3)
  7. ln(x+4) – ln(x) = 2
  8. log₄(x²–1) = log₄(3x–3)

Modeling Problems

Define your variables, write a model, and answer the specific question. Include units.

  1. A city's population in 2000 was 450,000 and grew at 2.5% per year. Write a model and predict the population in 2030.
  2. A radioactive element has a continuous decay rate of r = 0.0083 per year. Find its half-life.
  3. A $15,000 car depreciates at 18% per year. Write a model and find the value after 6 years.
  4. An account earns 4.2% interest compounded continuously. How long until an investment doubles?
  5. A population of 1,000 bacteria grows continuously at r = 0.35 per hour. How long until there are 50,000 bacteria?
  6. An earthquake of magnitude 6.5 and another of magnitude 8.0 are compared. How many times larger in amplitude is the magnitude 8.0 quake?
  7. A solution has [H⁺] = 3.7 × 10⁻⁹ mol/L. Is it acidic, basic, or neutral?
  8. A forensic scientist finds a body has cooled from 37°C to 25°C in a room at 18°C, with k = 0.0411. How long has the person been dead?

Applications

  1. Finance: An investor wants her account to grow from $8,000 to $20,000 in 10 years using continuous compounding. What annual rate is required?
  2. Medicine: A drug is metabolized continuously at 12% per hour. A patient receives a 250 mg dose. When does the amount fall below 20 mg?
  3. Sound: A sound has intensity I = 3.2 × 10⁻⁵ W/m². Find its decibel level using I₀ = 10⁻¹² W/m².
  4. Chemistry: Vinegar has [H⁺] ≈ 10⁻²·⁹ mol/L. What is the pH of vinegar?
  5. Archaeology: A bone fragment contains 18% of its original carbon-14. Using a half-life of 5,730 years, estimate its age.
  6. Ecology: A lake's fish population satisfies P(t) = 4000·e^(–0.06t) due to overfishing. When does the population fall to 500?

Error Analysis

Identify and correct the error in each of the following.

Problem 1
log₂(8 + 4) = log₂(8) + log₂(4) = 3 + 2 = 5

What is the error? Compute the correct value of log₂(12).

Problem 2
Solving ln(x²) = 4: x² = e⁴, so x = e² only

What solution is missing? Why must both be checked?

Problem 3
log₃(27) = 27/3 = 9

What is the correct value? What does log₃(27) actually ask?

Problem 4
Solving log(x–4) + log(x+2) = 2: x = 8 and x = –8, keeping both

Check both solutions in the original equation. Which must be rejected and why?

Problem 5
eˣ · eˣ = e^(x²)

What is the correct product? State the exponent rule that applies.

Problem 6
Using P(t) = 1000·e^(0.05t), evaluating P(–10) and concluding the population was negative

Can an exponential function ever produce a negative value? What does P(–10) actually represent in context?

Chapter 9 Checkpoint

Chapter 9 Checkpoint
Complete without assistance — assesses all key skills and concepts from Chapter 9
Part A — Multiple Choice / Short Answer
  1. Which of the following is the inverse of f(x) = 4ˣ?
    (a) f⁻¹(x) = x⁴    (b) f⁻¹(x) = log₄(x)    (c) f⁻¹(x) = log(4x)    (d) f⁻¹(x) = 4⁻ˣ
  2. Evaluate log₅(125).
  3. Which expression is equivalent to ln(x²√y / z³)?
    (a) 2·ln(x) + ½·ln(y) – 3·ln(z)    (b) 2·ln(x)·½·ln(y) / 3·ln(z)    (c) ln(2x) + ln(y/2) – ln(3z)
  4. Solve e^(x–1) = 12. Express the answer exactly and approximately.
  5. Solve log₂(x + 4) = 5.
Part B — Written Problems
  1. A substance decays according to A(t) = 300·e^(–0.045t), where t is in years.
    (a) What is the initial amount?
    (b) What is the continuous decay rate?
    (c) How much remains after 10 years?
    (d) Find the half-life.
  2. An account earns 3.8% annual interest compounded monthly. A deposit of $6,500 is made. How much is in the account after 12 years? How does this compare to continuous compounding at 3.8%?
  3. The pH of black coffee is approximately 5.0. The pH of baking soda solution is approximately 8.5. How many times greater is the hydrogen ion concentration in the coffee than in the baking soda?
  4. Solve log(x + 5) + log(x – 2) = 1. Identify any extraneous solutions.
  5. A city has population 600,000 growing at 1.8% per year. A neighboring city has population 400,000 growing at 3.2% per year. In how many years will they have the same population?
Part C — Extended Response
  1. Carbon-14 dating is used to estimate the age of an ancient manuscript. Analysis shows the manuscript retains 62% of its original carbon-14. The half-life of carbon-14 is 5,730 years.

    (a) Write an exponential decay model for the fraction of carbon-14 remaining as a function of time.
    (b) Using your model, estimate the age of the manuscript.
    (c) If the measurement of 62% has an uncertainty of ±3%, compute the estimated age using 59% and 65% and compare. How sensitive is the age estimate to measurement error?
    (d) Explain in words what the half-life of carbon-14 means and why it makes carbon-14 useful for dating materials thousands of years old but not millions of years old.
Looking Ahead
Chapter 10: Sequences, Series, and Mathematical Induction

Chapter 9 studied functions defined on all real numbers — continuous, smooth, and modeled by algebraic formulas involving exponents and logarithms. Chapter 10 will shift to functions defined on the positive integers: sequences. A sequence is a discrete function whose inputs are whole numbers. This connects to the idea of compounding studied in Chapter 9, where discrete time steps produce discrete outputs. Chapter 10 will also extend to series — accumulated sums of sequences — and introduce mathematical induction as a method for proving statements about all integers. These ideas connect directly to limits, to calculus, and to the foundations of mathematics.

MGU Library Connection
  • Appendix K: Exponential and Logarithmic Rules Reference — this textbook
  • Appendix R: Calculus I Readiness Check — this textbook
  • Chapter 2: Transformations, Inverses, and Function Operations — inverse function background
  • MGU Calculus I: Derivatives of eˣ and ln x, separable differential equations, L'Hôpital's Rule
  • MGU Chemistry: pH, acid-base equilibria, and logarithmic concentration scales
  • MGU Physics: Newton's Law of Cooling, radioactive decay, half-life in nuclear physics
  • MGU Finance: Compound interest, present value, annuities, and the time value of money
  • MGU Data Science: Logarithmic transformations in regression and machine learning
  • MGU Library Dictionary entries: exponential function, logarithm, Euler's number, half-life, pH, Richter scale, decibel