Exponents & Logarithms — SL 1.5 Exponents and logarithms (laws; base 10 and e; ln) - IB Mathematics Analysis and Approaches

SECTION 1

Historical Context & Motivation

Before calculators existed, multiplying large numbers was one of the most time-consuming tasks in science and navigation. Astronomers tracking planetary orbits, merchants calculating compound interest, and navigators charting courses across oceans all needed a way to simplify these enormous calculations. The invention of logarithms transformed multiplication into addition, effectively cutting the workload in half and changing mathematics forever.

Meanwhile, the concept of exponents evolved from simple shorthand for repeated multiplication into a powerful framework for describing growth, decay, and scaling in nature. Together, exponents and logarithms form one of the most important pairs of inverse operations in all of mathematics, much like multiplication and division or addition and subtraction.

1614 • Napier Publishes Logarithms

Scottish mathematician John Napier published Mirifici Logarithmorum Canonis Descriptio, introducing logarithms as a computational tool. His tables allowed astronomers to replace tedious multiplications with simple additions.

1624 • Briggs Creates Common Logarithms

Henry Briggs refined Napier's work by introducing base-10 logarithms (common logarithms), publishing tables of log values that became standard tools for scientists and engineers for over 300 years.

1683 • Jacob Bernoulli Discovers e

While studying compound interest, Jacob Bernoulli discovered that continuously compounded interest converges to a special constant, later named e ≈ 2.71828. This irrational number became the natural base for exponential functions.

1748 • Euler Formalizes eˣ and ln

Leonhard Euler systematically developed the theory of the exponential function eˣ and the natural logarithm (ln), connecting them to calculus, complex numbers, and his famous identity e^iπ + 1 = 0.

1970s • Scientific Calculators Replace Log Tables

With the arrival of electronic calculators, log tables became obsolete as computational tools. However, logarithms and exponents remained essential in modeling real-world phenomena such as population growth, radioactive decay, and sound intensity.

The central question that exponents and logarithms answer is this: if you know a base and a result, what power do you need to raise the base to in order to get that result? This seemingly simple question drives everything from the pH scale in chemistry to the Richter scale for earthquakes and the decibel scale for sound.

SECTION 2

Core Principles & Definitions

At its heart, an exponent tells you how many times to multiply a base by itself. When we write aⁿ, the number a is the base and n is the exponent (or power). A logarithm is the inverse operation: it asks "what exponent do I need?" So if aⁿ = b, then log_a(b) = n. These two operations undo each other, just like squaring and taking a square root.

Exponential Form

In aⁿ = b, the base a is raised to the power n to produce the result b. For example, 2³ = 8.

Logarithmic Form

log_a(b) = n means "the power you raise a to in order to get b." So log₂(8) = 3 because 2³ = 8.

Common Logarithm (log₁₀)

When no base is written, "log" typically means base 10. For example, log(1000) = 3 because 10³ = 1000. This is used in the pH scale, Richter scale, and decibel measurements.

Natural Logarithm (ln)

The natural logarithm uses base e ≈ 2.71828. We write ln(x) instead of log_e(x). It appears naturally in continuous growth, calculus, and many IB problems.

Inverse Relationship

Exponents and logarithms are inverse functions. This means a^{$\log_a$ (x)} = x and log_a(a^x) = x. They "undo" each other.

✦

KEY TAKEAWAY

Think of exponents and logarithms like a lock and key. The exponential function locks a number away by raising a base to a power: 2³ locks the answer as 8. The logarithm is the key that unlocks it: log₂(8) = 3 reveals the hidden exponent. They always come in pairs — you can't have one without the other, just like you can't open a lock without its matching key.

SECTION 3

Visual Explanation

The relationship between exponential and logarithmic functions becomes crystal clear when you see their graphs. The exponential function y = a^x curves upward rapidly, while its inverse y = log_a(x) grows slowly. The two functions are mirror images of each other across the line y = x. This visual symmetry is the hallmark of inverse functions.

The cyan curve shows y = 2^x rising sharply, while the violet curve shows y = log₂(x) growing slowly. The dashed line y = x is the mirror axis — every point (a, b) on the exponential curve corresponds to a point (b, a) on the logarithmic curve.

Notice how the exponential function shoots upward while the logarithmic function barely climbs. This asymmetry is fundamental: exponents amplify, while logarithms compress. The exponential curve has a horizontal asymptote at y = 0 (it approaches zero on the left but never touches it), while the logarithmic curve has a vertical asymptote at x = 0. Both functions pass through specific anchor points: the exponential always goes through (0, 1) since a⁰ = 1, and the logarithm always goes through (1, 0) since log_a(1) = 0.

SECTION 4

Mathematical Framework — Laws of Exponents & Logarithms

Laws of Exponents

The exponent laws allow you to simplify expressions involving powers. These rules are essential building blocks for the IB course and appear throughout algebra, calculus, and statistics.

PRODUCT LAW

aᵐ × aⁿ = aᵐ⁺ⁿ

When multiplying powers with the same base, add the exponents. For example, 2³ × 2⁴ = 2⁷ = 128.

QUOTIENT LAW

aᵐ ÷ aⁿ = aᵐ⁻ⁿ

When dividing powers with the same base, subtract the exponents. For example, 5⁶ ÷ 5² = 5⁴ = 625.

POWER LAW

(aᵐ)ⁿ = aᵐⁿ

When raising a power to another power, multiply the exponents. For example, (3²)⁴ = 3⁸ = 6561.

SPECIAL CASES

a⁰ = 1 a⁻ⁿ = 1/aⁿ a^(1/n) = ⁿ√a

Any non-zero number raised to the power 0 equals 1. A negative exponent means "take the reciprocal." A fractional exponent means "take a root." For example, $8^{1/3}$ = ³√8 = 2.

Laws of Logarithms

Since logarithms are the inverses of exponents, each exponent law has a corresponding logarithm law. These laws convert between operations: multiplication becomes addition, division becomes subtraction, and powers become coefficients.

LOG PRODUCT LAW

log_a(xy) = log_a(x) + log_a(y)

The log of a product equals the sum of the logs. This is why Napier invented logarithms — it turns multiplication into addition.

LOG QUOTIENT LAW

log_a(x/y) = log_a(x) − log_a(y)

The log of a quotient equals the difference of the logs. Division becomes subtraction.

LOG POWER LAW

log_a(xⁿ) = n × log_a(x)

An exponent inside a log can be brought out front as a coefficient. This is the most frequently used log law on IB exams.

CHANGE OF BASE FORMULA

log_a(x) = log_b(x) / log_b(a) = ln(x) / ln(a)

This formula lets you convert any logarithm to a different base. Since most calculators only have log (base 10) and ln (base e) buttons, this is essential for evaluating expressions like log₃(20).

💡

IB Formula Booklet Reminder

The IB provides the change of base formula and the three logarithm laws in the SL/HL formula booklet. You do not need to memorize them, but you do need to know when and how to apply them quickly.

SECTION 5

Detailed Breakdown — Base 10, Base e, and ln

While logarithms can have any positive base (other than 1), two bases dominate in mathematics and science. Understanding when to use each one is a key skill for the IB exam.

Side-by-side comparison of common logarithms (log, base 10) and natural logarithms (ln, base e). Both share the same laws — only the base differs. Your calculator has dedicated buttons for both.

The number e might seem like an odd choice for a base, but it arises naturally whenever growth or decay is continuous. Imagine you invest $1 at 100% interest. If compounded once a year, you get $2. Compounded twice, you get $2.25. Compounded infinitely many times, you get exactly $e ≈ $2.718. This limiting behavior makes e the most "natural" base for exponential functions, which is why logarithms with base e are called natural logarithms.

Expression	Value	Why?
`log(10)`	1	10¹ = 10
`log(1000)`	3	10³ = 1000
`log(1)`	0	10⁰ = 1
`ln(e)`	1	e¹ = e
`ln(e⁵)`	5	log and exponent cancel
`ln(1)`	0	e⁰ = 1

Quick reference values for log and ln

SECTION 6

Worked Example

Let's work through a multi-step problem that combines several logarithm laws — exactly the type you'll encounter on an IB exam.

Simplify and solve: 2 log(5) + log(8) − log(2)

1
Step 1 — Apply the Power Law

The coefficient 2 in front of log(5) can be moved inside as an exponent using the power law: log(xⁿ) = n × log(x). So 2 log(5) becomes log(5²) = log(25).

2 log(5) = log(25)
2
Step 2 — Apply the Product Law

Now combine log(25) + log(8) using the product law: log(x) + log(y) = log(xy). We get log(25 × 8) = log(200).

log(25) + log(8) = log(200)
3
Step 3 — Apply the Quotient Law

Finally, subtract log(2) using the quotient law: log(x) − log(y) = log(x/y). We get log(200/2) = log(100).

log(200) − log(2) = log(100)
4
Step 4 — Evaluate

Since log means base 10, we need the power of 10 that gives 100. Because 10² = 100, we know log(100) = 2.

2 log(5) + log(8) − log(2) = 2

Solve for x: eˣ⁺¹ = 15

1
Step 1 — Take the Natural Logarithm of Both Sides

To isolate the exponent, take ln of both sides. Since ln is the inverse of e, we get: ln(e^x+1) = ln(15). By the inverse property, ln(e^anything) = anything.

x + 1 = ln(15)
2
Step 2 — Isolate x

Subtract 1 from both sides to solve for x.

x = ln(15) − 1 ≈ 2.708 − 1 = 1.71 (3 s.f.)

SECTION 7

Common Mistakes & Exam Tips

Logarithm questions are among the most error-prone on IB exams. The following table highlights the traps that students frequently fall into and how to avoid them.

Common Mistake ✗	Correct Version ✓	Why It Matters
`log(a + b) = log(a) + log(b)`	`log(ab) = log(a) + log(b)`	The product law applies to products inside the log, not sums. There is no shortcut for log(a + b).
`log(a − b) = log(a) − log(b)`	`log(a/b) = log(a) − log(b)`	The quotient law applies to quotients inside the log, not differences.
`log(a) × log(b) = log(ab)`	No simplification exists	Multiplying two logs does not simplify. The product law adds logs, not multiplies them.
`ln(0) = 0`	ln(0) is undefined	You cannot take the log of zero or any negative number. The domain of all log functions is x > 0.
Forgetting to check domain	Always verify x > 0 in solutions	When solving log equations, extraneous solutions can arise. Always check your answer in the original equation.

Five most common logarithm errors on IB exams

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EXAM STRATEGY

When you see a log expression on an IB exam, pause and ask: "Am I dealing with a product, quotient, or power inside the log?" The three log laws only apply to these three situations. If the argument contains addition or subtraction, the laws do not apply. This single check will prevent the most common errors.

SECTION 8

Connection to Advanced Topics

The exponent and logarithm concepts from SL 1.5 are foundational to many later topics in the IB course and in university mathematics. Understanding where these ideas lead can help you see why mastering them now is so important.

SL 1.5 Foundation	Where It Leads	Topic / Course
Exponent laws	Exponential functions and their graphs (SL 2.5)	Functions
Solving exponential equations with logs	Exponential growth/decay models (SL 2.10)	Modeling
ln(x) as inverse of eˣ	Derivative of eˣ and ln(x) in calculus (SL 5.6)	Calculus
Log laws for simplification	Logarithmic scales in statistics (HL)	Statistics
Change of base formula	Complex exponentials and Euler's formula (HL)	Further Math

How SL 1.5 connects to later IB topics

In calculus, you'll discover that the function e^x is special because it is its own derivative — the rate at which it grows equals its current value. This unique property is why e appears in models of population growth, radioactive decay, and financial modeling. For now, focus on the algebraic tools — they are the gateway to all of these applications.

SECTION 9

Practice Problems

Question 1: CONCEPTUAL

Explain in your own words why log_a(1) = 0 for any valid base a. What does this tell you about the relationship between logarithms and exponents?

Show answer

By definition, log_a(1) asks: "What power do I raise a to in order to get 1?" Since any non-zero number raised to the power 0 equals 1 (a⁰ = 1), the answer is always 0. This reinforces the idea that logarithms and exponents are inverse operations: the log asks the question that the exponent answers. On a graph, this means every logarithmic function passes through the point (1, 0).

Question 2: BASIC CALCULATION

Simplify the expression log(50) + log(2) without using a calculator.

Show answer

Apply the product law: log(50) + log(2) = log(50 × 2) = log(100). Since 10² = 100, log(100) = 2. The product law transformed an addition of logs into a single log of a product, which then simplified to a neat integer. This is the fundamental technique Napier designed logarithms for — turning multiplication into addition.

Question 3: INTERMEDIATE

Solve for x: 3^2x−1 = 45. Give your answer correct to 3 significant figures.

Show answer

Take log (or ln) of both sides: (2x − 1) × log(3) = log(45). Isolate the bracket: 2x − 1 = log(45)/log(3). Evaluate using a calculator: log(45) ≈ 1.6532, log(3) ≈ 0.4771, so 2x − 1 ≈ 3.465. Add 1: 2x ≈ 4.465. Divide by 2: x ≈ 2.23 (3 s.f.). The key step was applying the change of base approach — taking log of both sides converts the exponential equation into a linear one that's easy to solve.

Question 4: APPLIED

A bacteria culture doubles every 3 hours. Starting with 500 bacteria, the population is modeled by P(t) = 500 × 2^t/3, where t is time in hours. How many hours will it take for the population to reach 16 000?

Show answer

Set up the equation: 16 000 = 500 ×

2^{t/3}

. Divide both sides by 500: 32 =

2^{t/3}

. Recognize that 32 = 2⁵, so 2⁵ =

2^{t/3}

. Since the bases are equal, the exponents must be equal: 5 = t/3, giving t = 15 hours. Alternatively, take log₂ of both sides: log₂(32) = t/3, so 5 = t/3. This problem shows how logarithms solve real exponential growth problems — you're finding the time when a known growth pattern reaches a target value.

Question 5: CRITICAL THINKING

Prove that log_a(b) × log_b(a) = 1 for any valid bases a and b. Then use this result to evaluate log₃(8) × log₈(3).

Show answer

Start with the change of base formula:

\log_a

(b) = ln(b)/ln(a) and

\log_b

(a) = ln(a)/ln(b). Multiply them: [ln(b)/ln(a)] × [ln(a)/ln(b)] = ln(b) × ln(a) / [ln(a) × ln(b)] = 1. The ln terms cancel perfectly. This elegant result shows that swapping the base and argument of a logarithm gives you the reciprocal. For the specific case, log₃(8) × log₈(3) = 1 by the property just proved. This kind of structural reasoning — proving a general rule and then applying it — is exactly what the IB values in Paper 1 questions.

SUMMARY

Lesson Summary

Summary

In this lesson you learned that exponents represent repeated multiplication and follow three core laws: the product law (add exponents when multiplying same bases), the quotient law (subtract exponents when dividing), and the power law (multiply exponents when raising a power to a power). Logarithms are the inverse of exponents: log_a(b) = n means aⁿ = b. The three log laws mirror the exponent laws, converting products into sums, quotients into differences, and powers into coefficients.

The two most important bases are base 10 (common log, written log) and base e ≈ 2.718 (natural log, written ln). The change of base formula allows conversion between any bases using $\log_a$ (x) = ln(x)/ln(a). Remember that logarithms are only defined for positive arguments (x > 0), and always check for extraneous solutions when solving log equations. These tools are your foundation for exponential models, calculus, and real-world applications throughout the IB course.