Exponential & Logarithmic Functions — SL 1.8 Functions and equations involving exponents/logarithms (links to exponential models) - IB Mathematics Analysis and Approaches

SECTION 1

Historical Context & Motivation

Long before calculators existed, astronomers and navigators faced an exhausting problem: multiplying enormous numbers by hand. In the early 1600s, a Scottish mathematician named John Napier realized that if you could turn multiplication into addition, calculations would become dramatically faster. His insight was that powers of a fixed base follow simple additive patterns — if you add exponents, you multiply the corresponding values. This idea gave birth to logarithms, one of the most powerful tools in mathematics.

Over the following centuries, mathematicians discovered that exponential functions appear everywhere in nature — from population growth to radioactive decay, from compound interest to the spread of diseases. Understanding exponentials and their inverses, logarithms, became essential not just for pure mathematics but for modeling the real world.

1614 • Napier Publishes Logarithm Tables

John Napier published Mirifici Logarithmorum Canonis Descriptio, introducing logarithms as a computational tool that reduced multiplication to addition.

1668 • Natural Logarithm Emerges

Nicolaus Mercator published work on the natural logarithm (ln), connecting logarithms to the area under the hyperbola y = 1/x and laying groundwork for the constant e.

1748 • Euler Formalizes e

Leonhard Euler introduced the notation e ≈ 2.718 as the base of natural logarithms and established the exponential function eˣ as central to analysis.

1798 • Malthus and Exponential Growth

Thomas Malthus used exponential models to describe population growth, demonstrating the real-world power of these functions in economics and social science.

1896 • Radioactive Decay Modeled

Henri Becquerel's discovery of radioactivity led to exponential decay models, showing that the same mathematics governs both growth and decay processes.

The central question of this topic is straightforward: how do we work with functions where the variable appears in the exponent, and how do we "undo" exponentiation to solve equations? This is exactly the role that logarithms play — they are the inverse of exponential functions, and together they form a toolkit for modeling change in the real world.

SECTION 2

Core Principles & Definitions

Before diving into equations and models, you need a solid understanding of the foundational ideas that connect exponential and logarithmic functions. These concepts build on your knowledge of exponents from Algebra 1, extending them into a framework for solving sophisticated equations and modeling real phenomena.

Exponential Function

A function of the form f(x) = aˣ where a > 0 and a ≠ 1. The variable is in the exponent, not the base. This creates rapid growth (a > 1) or decay (0 < a < 1).

Logarithmic Function

The inverse of an exponential: if aˣ = b, then log_a(b) = x. A logarithm answers the question: "What power of a gives b?"

The Number e

Euler's number, e ≈ 2.718, is the base of natural logarithms (ln). It arises naturally in continuous growth/decay and is the most common base in IB models.

Laws of Exponents

The rules aᵐ × aⁿ = aᵐ⁺ⁿ, (aᵐ)ⁿ = aᵐⁿ, and a⁰ = 1 are the algebraic backbone. Every logarithm law is derived directly from these exponent laws.

Inverse Relationship

Exponential and logarithmic functions are perfect inverses: log_a(aˣ) = x and a^{$\log_a$ (x)} = x. Their graphs are reflections across the line y = x.

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KEY TAKEAWAY

Think of logarithms like a "volume dial" for exponents. If an exponential function cranks the output up (or down) at an accelerating rate, the logarithm is the control that tells you exactly which setting (exponent) produced a given output. Asking "what is log₂(8)?" is the same as asking "what power do I put on 2 to get 8?" — the answer is 3, because 2³ = 8.

SECTION 3

Visual Explanation — Graphs of Exponential & Logarithmic Functions

The best way to grasp the relationship between exponential and logarithmic functions is to see them on the same coordinate plane. The following diagram plots y = 2ˣ (an exponential growth curve) and y = log₂(x) (its logarithmic inverse) together, along with the mirror line y = x. Notice how every point (a, b) on one curve corresponds to (b, a) on the other.

The cyan curve shows y = 2ˣ passing through (0, 1) and rising steeply. The violet curve shows y = log₂(x), which passes through (1, 0) and grows slowly. Both curves are perfect reflections of each other across the dashed line y = x.

Several key features are visible in the diagram. The exponential curve y = 2ˣ has a horizontal asymptote at y = 0 — the curve gets closer and closer to the x-axis as x goes to negative infinity, but it never actually touches it. Correspondingly, the logarithmic curve y = log₂(x) has a vertical asymptote at x = 0. The domain of an exponential function is all real numbers, but the domain of a logarithmic function is only positive real numbers (x > 0). Also notice how each labeled point on one curve swaps its coordinates on the other — for example, (1, 2) on the exponential becomes (2, 1) on the logarithm.

SECTION 4

Mathematical Framework

This section lays out the essential formulas and laws you need for SL 1.8. Every logarithm law follows directly from the corresponding exponent law, so keep the connection in mind as you study each one.

EXPONENTIAL-LOGARITHM EQUIVALENCE

aˣ = b ⟺ x = log_a(b)

where a is the base (a > 0, a ≠ 1), x is the exponent, and b is the result (b > 0). This is the fundamental bridge between exponentials and logarithms.

LOGARITHM LAWS

log_a(mn) = log_a(m) + log_a(n) log_a(m/n) = log_a(m) − log_a(n) log_a(mᵖ) = p × log_a(m)

These three laws — the product rule, quotient rule, and power rule — mirror the exponent rules aᵐ × aⁿ = aᵐ⁺ⁿ, aᵐ ÷ aⁿ = aᵐ⁻ⁿ, and (aᵐ)ⁿ = aᵐⁿ.

CHANGE OF BASE FORMULA

log_a(b) = log_c(b) / log_c(a)

This allows you to convert between any bases. Most commonly, you'll use c = 10 (common log) or c = e (natural log, ln) since your calculator has buttons for both. For example, log₅(20) = ln(20) / ln(5).

EXPONENTIAL MODEL (GROWTH & DECAY)

A(t) = A₀ × eᵏᵗ

A₀ = initial amount, k = growth/decay constant (k > 0 for growth, k < 0 for decay), t = time. This is the standard continuous exponential model used in IB applications.

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IB Formula Booklet Reminder

The change of base formula and the three logarithm laws are provided in the IB formula booklet. However, you are expected to use them fluently, so practice applying them in different contexts rather than just memorizing them.

SECTION 5

Detailed Breakdown — Solving Exponential & Logarithmic Equations

Knowing the laws is one thing; knowing when and how to use them is what actually scores marks on an IB exam. The diagram below shows the decision-making process for solving equations that involve exponents or logarithms. Think of it as a flowchart: start with your equation type and follow the branches to the right technique.

This flowchart guides you through the key decision points. For exponential equations, check whether you can rewrite both sides with the same base. If not, take logarithms. For logarithmic equations, convert to exponential form. Always check your answer lies within the valid domain.

Let's illustrate each path with a quick example. For the "same base" path: if you encounter 5²ˣ = 125, recognize that 125 = 5³, so 5²ˣ = 5³ and therefore 2x = 3, giving x = 1.5. For the "take log" path: if you have 3ˣ = 20, no integer power of 3 equals 20, so take ln of both sides to get x × ln(3) = ln(20), giving x = ln(20)/ln(3) ≈ 2.73. For the logarithmic path: if log₃(2x + 1) = 4, convert to exponential form to get 3⁴ = 2x + 1, so 81 = 2x + 1 and x = 40.

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Domain Check — Don't Skip It!

When solving logarithmic equations, you must verify that your answer keeps the argument of every logarithm positive. For instance, if you get x = −5 but the original equation had log(x + 3), then x + 3 = −2 < 0, which is undefined. That solution is extraneous and must be rejected.

SECTION 6

Worked Example — Exponential Decay with Logarithms

A radioactive substance has an initial mass of 200 grams and decays according to the model A(t) = 200e^−0.035t, where t is measured in years. Find the time it takes for the substance to decay to 50 grams.

Finding the Time for Radioactive Decay

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Step 1 — Set Up the Equation

We want A(t) = 50, so we set up: 50 = 200e^−0.035t. Our goal is to isolate t.
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Step 2 — Isolate the Exponential

Divide both sides by 200: 50/200 = e^−0.035t, which simplifies to 0.25 = e^−0.035t.

0.25 = e^−0.035t
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Step 3 — Take the Natural Logarithm of Both Sides

Apply ln to both sides: ln(0.25) = ln(e^−0.035t). Since ln and e are inverses, the right side simplifies: ln(0.25) = −0.035t.

ln(0.25) = −0.035t
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Step 4 — Solve for t

Divide both sides by −0.035: t = ln(0.25) / (−0.035). Using a calculator, ln(0.25) ≈ −1.3863.

t = (−1.3863) / (−0.035) ≈ 39.6 years
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Step 5 — Interpret the Result

The substance decays from 200 g to 50 g in approximately 39.6 years. Notice that 50 is one-quarter of 200, so this is two half-lives. We can verify: the half-life is ln(2)/0.035 ≈ 19.8 years, and 2 × 19.8 = 39.6 ✓.

t ≈ 39.6 years

SECTION 7

Exponential vs. Logarithmic — Strengths & Limitations

Exponential and logarithmic functions each have distinct advantages depending on the context. Understanding when to use each form — and their limitations — will help you model problems efficiently and avoid common mistakes.

Feature	Exponential Form	Logarithmic Form
Best for	Predicting future values ("How much will there be after 10 years?")	Finding time or rate ("When will it reach 500?")
Domain	All real numbers (−∞, +∞)	Positive reals only (0, +∞)
Range	Positive reals only (0, +∞)	All real numbers (−∞, +∞)
Growth behavior	Increases/decreases at an accelerating rate	Increases/decreases at a decelerating rate
Common pitfall	Forgetting that aˣ is always positive; it never equals zero or a negative number	Forgetting to check that the argument is positive; extraneous solutions are common
Asymptote	Horizontal: y = 0	Vertical: x = 0

Comparing exponential and logarithmic functions side by side

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KEY TAKEAWAY

Think of exponentials and logarithms as two sides of the same coin. Exponential form is your "forward gear" — plug in time and get an amount. Logarithmic form is your "reverse gear" — plug in an amount and find the time. In IB exam questions, you'll often need to switch between gears mid-problem, so practice converting fluently in both directions.

SECTION 8

Connections to Advanced Theory & Other Topics

The exponential and logarithmic functions you've learned in SL 1.8 form the foundation for several more advanced topics, both within the IB course and beyond. This section gives you a preview of where these ideas lead.

SL 1.8 Concept	Where It Leads	Connection
Exponential growth/decay model A₀eᵏᵗ	SL 2.10: Exponential models in context	Real-world applications — population, finance, temperature cooling — all use this same structure.
Logarithm laws	HL: Calculus of logarithmic functions	The derivative of ln(x) is 1/x, and the integral of 1/x is ln\|x\| + C — logarithms become central to calculus.
Solving aˣ = b	SL 5.6: Differential equations	The equation dy/dx = ky has the solution y = Aeᵏˣ — the same exponential model, now derived from calculus.
Change of base formula	Logarithmic scales (pH, decibels, Richter)	Understanding different bases lets you interpret scientific scales that compress enormous ranges into manageable numbers.

How SL 1.8 connects to other IB topics and beyond

If you continue to HL Mathematics, you'll discover that the function eˣ is extraordinary — it's the only function that is its own derivative. This property makes it the backbone of differential equations, which model everything from electrical circuits to epidemics. The tools you're building now with logarithmic and exponential manipulation will be used constantly in those advanced contexts.

SECTION 9

Practice Problems

Question 1: CONCEPTUAL

Explain why log_a(1) = 0 for any valid base a. Use the relationship between exponential and logarithmic forms in your reasoning.

Show answer

By the exponential-logarithm equivalence, log_a(1) = 0 means the same thing as a⁰ = 1. Since any non-zero number raised to the power of zero equals 1 (this is a fundamental exponent law), the equation a⁰ = 1 is always true regardless of the base. Therefore, the logarithm of 1 is always 0, no matter what valid base you choose. This makes sense graphically too: every logarithmic curve passes through the point (1, 0).

Question 2: BASIC CALCULATION

Solve for x: 2^3x−1 = 32.

Show answer

Rewrite 32 as a power of 2: 32 = 2⁵. Now the equation becomes 2^3x−1 = 2⁵. Since the bases are the same, the exponents must be equal: 3x − 1 = 5. Solving: 3x = 6, so x = 2. You can verify: 2³⁽²⁾⁻¹ = 2⁵ = 32 ✓.

Question 3: INTERMEDIATE

Solve for x: log₃(x + 4) + log₃(x − 2) = 3.

Show answer

Using the product rule for logarithms: log₃[(x + 4)(x − 2)] = 3. Convert to exponential form: (x + 4)(x − 2) = 3³ = 27. Expanding: x² + 2x − 8 = 27, so x² + 2x − 35 = 0. Factoring: (x + 7)(x − 5) = 0, giving x = −7 or x = 5. Now check the domain: for x = −7, the argument x − 2 = −9 < 0, which is invalid for a logarithm. For x = 5, both x + 4 = 9 > 0 and x − 2 = 3 > 0 ✓. Therefore x = 5 is the only valid solution.

Question 4: APPLIED

A bacteria population starts at 500 and grows continuously at a rate of 4.6% per hour, modeled by P(t) = 500e^0.046t. How many hours will it take for the population to reach 3000?

Show answer

Set P(t) = 3000: 3000 = 500e^0.046t. Divide both sides by 500: 6 = e^0.046t. Take the natural log of both sides: ln(6) = 0.046t. Since ln(6) ≈ 1.7918, we get t = 1.7918 / 0.046 ≈ 38.95. Therefore it takes approximately 39.0 hours for the population to reach 3000. This makes sense because the population must increase by a factor of 6, which requires multiple doubling periods.

Question 5: CRITICAL THINKING

Show that the equation 2ˣ + 2^−x = 5 can be transformed into a quadratic equation by substituting u = 2ˣ. Hence find the exact values of x.

Show answer

Let u = 2ˣ. Then 2^−x = 1/2ˣ = 1/u. The equation becomes u + 1/u = 5. Multiply through by u: u² + 1 = 5u, or u² − 5u + 1 = 0. Using the quadratic formula: u = (5 ± √(25 − 4))/2 = (5 ± √21)/2. Since u = 2ˣ > 0, both roots are valid (both are positive). Now solve for x: 2ˣ = (5 + √21)/2 gives x = log₂((5 + √21)/2) = ln((5 + √21)/2)/ln(2), and 2ˣ = (5 − √21)/2 gives x = log₂((5 − √21)/2) = ln((5 − √21)/2)/ln(2). These are the two exact solutions. Numerically, x ≈ 2.26 or x ≈ −2.26. Notice the symmetry: since the original equation is unchanged when x is replaced by −x, the solutions are negatives of each other.

SUMMARY

Lesson Summary

Summary

In this lesson, you learned that exponential functions have the form f(x) = aˣ, where the variable sits in the exponent, producing rapid growth or decay. Their inverses, logarithmic functions, answer the question "what exponent gives this value?" using the equivalence aˣ = b ⟺ x = log_a(b). The three logarithm laws — product, quotient, and power rules — mirror the laws of exponents and are essential tools for simplifying and solving equations. The change of base formula lets you evaluate logarithms in any base using your calculator's ln or log buttons.

To solve exponential equations, either rewrite both sides with the same base and equate exponents, or take logarithms of both sides and use the power rule. To solve logarithmic equations, convert to exponential form and solve algebraically — always checking for extraneous solutions. The continuous exponential model A(t) = A₀eᵏᵗ connects these functions to real-world applications including population growth, radioactive decay, and financial modeling. Mastering the interplay between exponential and logarithmic forms is the central skill of SL 1.8 and a foundation for advanced IB topics.