Logarithmic Functions — SL 2.5 Logarithmic functions and models - IB Mathematics Analysis and Approaches

SECTION 1

Historical Context & Motivation

Before calculators and computers existed, mathematicians and scientists needed a practical way to handle extremely large and extremely small numbers. Multiplying two ten-digit numbers by hand is tedious and error-prone. In the late sixteenth century, astronomers spent enormous amounts of time performing these calculations to track the motion of planets and stars. The logarithm was invented precisely to solve this problem — it transforms multiplication into addition, making calculations dramatically faster and more reliable.

1614 • Napier Publishes Logarithms

Scottish mathematician John Napier published Mirifici Logarithmorum Canonis Descriptio, introducing logarithms to simplify astronomical calculations. His tables reduced multiplication to addition.

1624 • Briggs Creates Common Logarithms

Henry Briggs refined Napier's work and published tables of base-10 logarithms, which became the standard for scientific computation for over three centuries.

1668 • Natural Logarithms Emerge

Mathematicians including Nicolaus Mercator explored logarithms with base e ≈ 2.718, connecting them to areas under the hyperbola y = 1/x and laying groundwork for calculus.

1935 • Richter Scale Introduced

Charles Richter developed the Richter magnitude scale for earthquakes using base-10 logarithms, demonstrating how logarithmic models handle quantities spanning many orders of magnitude.

Modern • Logarithms in Data Science

Today logarithmic functions appear in pH chemistry, decibel scales, information theory, machine learning algorithms, and financial modeling — anywhere exponential growth or vast ranges of data occur.

The central question that logarithms answer is deceptively simple: if we know a base and a result, what exponent was used? For example, 2 raised to what power gives 32? The logarithm provides the machinery to answer this question systematically, and as we will see, the resulting function has a beautifully distinctive shape with powerful modeling applications.

SECTION 2

Core Principles & Definitions

A logarithm is fundamentally the inverse of an exponential function. If you are comfortable with the idea that 2³ = 8, then you already understand the seed of logarithmic thinking: log₂(8) = 3. The logarithm asks, "what exponent do I need?" Let's formalize the core ideas that make logarithmic functions work.

Definition of a Logarithm

If b^x = a, then log_b(a) = x. The base b must be positive and not equal to 1, and a must be positive.

Common & Natural Logs

log(x) with no base written means log₁₀(x) (the common logarithm). ln(x) means log_e(x) (the natural logarithm), where e ≈ 2.718.

Domain & Range

The domain of y = log_b(x) is x > 0 (only positive inputs). The range is all real numbers. The graph passes through (1, 0) and has a vertical asymptote at x = 0.

Inverse Relationship

The functions y = b^x and y = log_b(x) are reflections of each other across the line y = x. They "undo" each other: log_b(b^x) = x.

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KEY TAKEAWAY

Think of logarithms like a detective's tool. If exponentiation is the process of building a skyscraper — starting from a base and stacking floors — the logarithm is the act of looking at the finished building and counting how many floors it has. You know the base (floor height) and the total height (result), and the logarithm tells you the number of floors (the exponent).

SECTION 3

Visual Explanation — The Logarithmic Curve

The graph of a logarithmic function has a distinctive shape that reflects its nature as the inverse of an exponential. It rises steeply for small positive values of x, then gradually flattens out as x increases. This "diminishing returns" behavior is central to why logarithmic models are used in science. The following diagram shows y = log₂(x) alongside its inverse y = 2^x and the mirror line y = x.

The cyan curve shows y = log₂(x), which passes through (1, 0), (2, 1), and (4, 2). The violet curve shows y = 2ˣ. Notice they are mirror images across the dashed line y = x. The red dashed line at x = 0 is the vertical asymptote — the log function approaches but never touches or crosses the y-axis.

Notice the key features visible in the diagram. First, the log curve passes through (1, 0) regardless of the base, because any number raised to the power 0 equals 1. Second, the curve increases without bound as x grows, but it does so at a decreasing rate — each time x doubles, y only increases by 1. Third, as x approaches 0 from the right, the curve plunges downward toward negative infinity, creating the vertical asymptote at x = 0. These features are what make logarithmic functions so useful for compressing wide-ranging data into manageable scales.

SECTION 4

Mathematical Framework — Laws of Logarithms

Just as exponents follow predictable rules (like aᵐ × aⁿ = aᵐ⁺ⁿ), logarithms have their own set of laws. These laws are direct consequences of exponent rules, and they are essential for simplifying expressions, solving equations, and transforming between logarithmic and exponential forms. The IB syllabus expects you to know and apply all of the following.

PRODUCT LAW

logb(M × N) = logb(M) + logb(N)

The logarithm of a product equals the sum of the logarithms. This is the property that originally made logarithms so powerful — it turns multiplication into addition.

QUOTIENT LAW

logb(M ÷ N) = logb(M) − logb(N)

The logarithm of a quotient equals the difference of the logarithms. Division becomes subtraction.

POWER LAW

logb(Mⁿ) = n × logb(M)

The logarithm of a power brings the exponent down as a multiplier. This is especially useful for solving exponential equations where the unknown is in the exponent.

CHANGE OF BASE

logb(a) = logc(a) ÷ logc(b)

This formula lets you convert between any two bases. On your calculator, you can evaluate log₅(20) as ln(20) ÷ ln(5) or log(20) ÷ log(5). Both give the same result.

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IB Exam Tip

The IB formula booklet includes the change of base rule, but the product, quotient, and power laws are not provided. You must memorize them. A helpful mnemonic: logs turn operations "one level down" — multiplication becomes addition, division becomes subtraction, and exponentiation becomes multiplication.

SECTION 5

Transformations & Key Features of Logarithmic Graphs

Just like other functions you've studied, logarithmic functions can be shifted, stretched, and reflected. Understanding these transformations is crucial because IB questions frequently present logarithmic functions in the general form y = a × log_b(x − h) + k and ask you to identify features like asymptotes, intercepts, and domain.

Four transformations of the parent function y = log₂(x) (shown as a dashed curve). Amber: vertical shift up by 2. Cyan: horizontal shift right by 3 (asymptote moves to x = 3). Pink: reflection in the x-axis. Green: vertical stretch by factor 3.

Transformation	Equation Form	Effect on Graph	Asymptote
Vertical shift	y = log_b(x) + k	Shifts graph up (k > 0) or down (k < 0)	x = 0 (unchanged)
Horizontal shift	y = log_b(x − h)	Shifts graph right (h > 0) or left (h < 0)	x = h
Vertical stretch/compress	y = a × log_b(x)	Steeper (\|a\| > 1) or flatter (\|a\| < 1)	x = 0 (unchanged)
Reflection in x-axis	y = −log_b(x)	Flips graph upside down	x = 0 (unchanged)

Summary of common transformations for logarithmic functions

SECTION 6

Worked Example — Solving a Logarithmic Equation

Let's work through a typical IB-style problem that combines logarithmic laws with equation solving. This example demonstrates how to use the product and power laws, convert between logarithmic and exponential form, and verify your answer.

Solve: log₃(x) + log₃(x − 6) = 3

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Step 1 — Apply the Product Law

The left side contains two logarithms of the same base being added. By the product law, log₃(x) + log₃(x − 6) = log₃(x(x − 6)). So the equation becomes:

log₃(x² − 6x) = 3
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Step 2 — Convert to Exponential Form

Using the definition of logarithm (if log_b(a) = c then b^c = a), we convert: 3³ = x² − 6x.

x² − 6x = 27
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Step 3 — Solve the Quadratic

Rearranging gives x² − 6x − 27 = 0. We can factor this by finding two numbers that multiply to −27 and add to −6. Those numbers are −9 and 3.

(x − 9)(x + 3) = 0, so x = 9 or x = −3
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Step 4 — Check Domain Restrictions

Remember that logarithms are only defined for positive arguments. We need x > 0 and x − 6 > 0, which means x > 6. Testing x = −3: log₃(−3) is undefined. Testing x = 9: log₃(9) + log₃(3) = 2 + 1 = 3 ✓

x = 9

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Common Mistake Alert

Always check your solutions against the domain restrictions of the original equation. When combining logarithms, each individual logarithmic term must have a positive argument. Extraneous solutions are very common in logarithmic equations, and the IB frequently tests this.

SECTION 7

Logarithmic vs. Exponential vs. Power Models

In the IB course, you encounter several types of function models. Knowing when to choose a logarithmic model instead of an exponential or power model is a key skill. Each model suits a different type of growth or decay behavior. The table below compares these three model types across several criteria.

Feature	Exponential: y = abˣ	Power: y = axⁿ	Logarithmic: y = a + b ln(x)
Growth behavior	Increases faster and faster (or decays)	Increases at a rate dependent on n	Rapid initial growth, then levels off
Typical real-world use	Population growth, radioactive decay, compound interest	Area vs. length, gravitational force	Decibel scale, pH, Richter scale, learning curves
Key question it answers	How much is there after time t?	How does output scale with input?	How big is this on a compressed scale?
Asymptote	Horizontal (y = 0 for decay)	None in general	Vertical (x = 0 or x = h)
Shape on scatter plot	J-curve (growth) or decay curve	Curve through origin	Steep rise then flattening

Comparison of three function model types commonly tested in IB SL

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WHEN TO USE A LOG MODEL

Choose a logarithmic model when your data shows diminishing returns — large changes in the input produce smaller and smaller changes in the output. Think of volume on a speaker: turning the dial from 1 to 2 feels like a huge jump, but turning from 8 to 9 barely changes what you hear. That's logarithmic behavior in action.

SECTION 8

Connections to Advanced Topics

The logarithmic functions you learn in SL 2.5 are not just a self-contained topic — they form a bridge to several more advanced mathematical ideas. Understanding where logarithms lead will help you see their importance in the broader curriculum and in HL Mathematics if you choose to continue.

SL 2.5 Foundation	Advanced Connection
log laws (product, quotient, power)	Simplifying expressions in calculus (e.g., differentiating products via logarithmic differentiation)
Solving exponential equations using logs	Modeling half-life and continuous growth/decay (SL 2.10 and beyond)
The natural logarithm ln(x)	The integral ∫(1/x) dx = ln\|x\| + C, a cornerstone of HL Calculus
Change of base formula	Information theory (log₂ for bits) and entropy in physics and computer science
Logarithmic scales (e.g., Richter, pH)	Log-log and semi-log plots for linearizing data in the IA and science courses

How SL 2.5 concepts connect to advanced mathematics and other subjects

One particularly powerful connection worth previewing is the idea of linearization. If you suspect your data follows an exponential model y = ab^x, you can take the logarithm of both sides to get log(y) = log(a) + x × log(b). This transforms the exponential relationship into a linear one, which is much easier to analyze and graph. This technique is invaluable for the Internal Assessment and for science courses where you need to determine whether data fits an exponential pattern.

SECTION 9

Practice Problems

Question 1: CONCEPTUAL

Explain in your own words why the domain of f(x) = log₃(x) is restricted to x > 0. What would it mean graphically if we tried to input x = 0 or a negative value?

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The domain is restricted because the logarithm asks "3 raised to what power gives x?" Since 3 raised to any real number always produces a positive result, there is no exponent that makes 3ⁿ = 0 or 3ⁿ = negative number. Graphically, the curve approaches the y-axis (x = 0) but never touches it — this is the vertical asymptote. The function heads toward negative infinity as x approaches 0 from the right, reflecting that we'd need an infinitely negative exponent, which is impossible to reach.

Question 2: BASIC CALCULATION

Evaluate without a calculator: (a) log₂(64) (b) log₅(1/25) (c) ln(e³)

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(a) Since 2⁶ = 64, log₂(64) = 6. (b) Since 1/25 = 5⁻², we have log₅(1/25) = −2. Negative exponents correspond to fractions, so negative log values indicate inputs between 0 and 1. (c) By the inverse property, ln(e³) = 3, because the natural log and e are inverses — log base e of e to a power simply returns the exponent.

Question 3: INTERMEDIATE

Use logarithm laws to write the expression 2 log(x) − log(y) + (1/2) log(z) as a single logarithm.

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Apply the power law to each term: 2 log(x) = log(x²), and (1/2) log(z) = log(

z^{1/2}

) = log(√z). Now apply the product and quotient laws: log(x²) − log(y) + log(√z) = log(x² × √z ÷ y). The final answer is log(x²√z / y). This type of condensing is a common IB exam skill — practice moving coefficients up as exponents and then combining using product and quotient laws.

Question 4: APPLIED

The loudness of sound in decibels is modeled by L = 10 log(I / I₀), where I is the intensity and I₀ = 10⁻¹² W/m² is the threshold of hearing. A rock concert has intensity I = 10⁻¹ W/m². (a) Find the loudness in decibels. (b) If the intensity doubles to 2 × 10⁻¹ W/m², by how many decibels does the loudness increase?

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(a) L = 10 log(10⁻¹ / 10⁻¹²) = 10 log(10¹¹) = 10 × 11 = 110 dB. (b) New loudness = 10 log(2 × 10⁻¹ / 10⁻¹²) = 10 log(2 × 10¹¹) = 10[log 2 + log(10¹¹)] = 10[0.301 + 11] = 10 × 11.301 = 113.01 dB. The increase is about 3 dB. This illustrates a key property of logarithmic scales: doubling the intensity always adds approximately 3 dB, regardless of the starting level.

Question 5: CRITICAL THINKING

The function g(x) = a ln(x − h) + k has a vertical asymptote at x = 2, passes through the point (3, 5), and passes through the point (e² + 2, 11). Find the values of a, h, and k.

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Since the vertical asymptote is at x = 2, we know h = 2 (the asymptote of ln(x − h) is x = h). Substituting the point (3, 5): 5 = a ln(3 − 2) + k = a ln(1) + k = a(0) + k, so k = 5. Substituting the point (e² + 2, 11): 11 = a ln(e² + 2 − 2) + 5 = a ln(e²) + 5 = 2a + 5, giving 2a = 6, so a = 3. Therefore g(x) = 3 ln(x − 2) + 5. This problem synthesizes understanding of asymptotes, the defining property of the natural logarithm (ln(1) = 0 and ln(e²) = 2), and simultaneous equations.

SUMMARY

Lesson Summary

Summary

A logarithm is the inverse of an exponential function: if b^x = a, then log_b(a) = x. The domain of y = log_b(x) is x > 0, the range is all real numbers, and the graph has a vertical asymptote at x = 0 and passes through the point (1, 0). The three key laws — product (log of a product = sum of logs), quotient (log of a quotient = difference of logs), and power (log of a power = exponent × log) — allow you to simplify expressions and solve equations. The change of base formula lets you evaluate any logarithm using your calculator's log or ln button.

When solving logarithmic equations, always check solutions against domain restrictions to eliminate extraneous answers. Logarithmic models are ideal for real-world data that grows rapidly at first and then levels off (diminishing returns), and they power familiar scales like the Richter scale, decibels, and pH. Transformations of the parent function follow the same rules as other functions: vertical/horizontal shifts move the curve and its asymptote, while vertical stretches and reflections change its steepness and orientation.