AP Precalculus • Exponential and Logarithmic Functions

Logarithmic Functions

The inverse of exponential growth, logarithms transform multiplicative relationships into additive ones.

A function curve with a triangle and circular angle measure.
Inside this lesson 9 sections 5 questions 10 figures

SECTION 1

Historical Context & Motivation

Before electronic calculators existed, astronomers, navigators, and engineers faced an enormous practical burden: multiplying and dividing very large numbers by hand was slow and error-prone. The key insight that would eventually resolve this problem came from recognizing that exponents convert multiplication into addition. If one could build a systematic table mapping every number to its corresponding exponent—its logarithm—then the laborious task of multiplication would reduce to simple addition. This idea, developed over centuries, gave rise to one of the most versatile functions in all of mathematics.

1614 • Napier's Logarithms

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

1624 • Briggs' Common Logarithms

Henry Briggs refines Napier's work and publishes tables of base-10 (common) logarithms, establishing the standard used for centuries of hand computation.

1668 • Natural Logarithm Emerges

Nicolaus Mercator publishes a series expansion for ln(1 + x), connecting logarithms to the area under the hyperbola y = 1/x and laying groundwork for the natural logarithm.

1748 • Euler's Synthesis

Leonhard Euler formally defines the natural logarithm as the inverse of eˣ in his Introductio in analysin infinitorum, unifying logarithmic and exponential theory.

1935 • Richter Scale & Modern Applications

Charles Richter introduces a base-10 logarithmic scale for earthquake magnitude, exemplifying how logarithms compress vast numerical ranges into manageable scales.

The central question that logarithms answer is deceptively simple: given a base b and a result y, what exponent x satisfies bˣ = y? This inverse relationship to exponential functions makes logarithms indispensable in modeling phenomena that span many orders of magnitude—from sound intensity to radioactive decay to population dynamics.

SECTION 2

Core Principles & Definitions

A logarithmic function is defined as the inverse of an exponential function. If f(x) = bˣ where b > 0 and b ≠ 1, then its inverse is f⁻¹(x) = logb(x). The expression logb(x) = y means precisely that by = x. Understanding this equivalence is the foundation for all logarithmic reasoning in AP Precalculus.

1

Inverse Relationship

logb(x) = y if and only if by = x. The logarithm asks: "To what power must I raise b to obtain x?"
2

Domain & Range

The domain of y = logb(x) is (0, ∞) and the range is (−∞, ∞). Logarithms are undefined for zero and negative inputs.
3

Key Anchor Point

Every logarithmic function passes through (1, 0) because b⁰ = 1 for any valid base. The point (b, 1) is also always on the graph since logb(b) = 1.
4

Vertical Asymptote

The y-axis (x = 0) is a vertical asymptote. As x → 0⁺, logb(x) → −∞. The function never touches or crosses the y-axis.
5

Monotonic Behavior

When b > 1, the function is strictly increasing. When 0 < b < 1, it is strictly decreasing. Either way, logarithmic functions are always one-to-one.

SECTION 3

Visual Explanation — Graphs of Logarithmic Functions

01−1−2212345xy(1, 0)y = ln(x)y = log₂(x)y = log₁₀(x)x = 0 (V.A.)Logarithmic Functions — Three Common Bases
The graph shows three logarithmic functions: y = ln(x) (base e), y = log₂(x), and y = log₁₀(x). All three pass through (1, 0), increase without bound, and have the vertical asymptote x = 0 (shown in red). Larger bases produce flatter curves because larger powers are needed to reach the same output.

Several features of the graph deserve close attention. First, note that all three curves share the anchor point (1, 0) because any base raised to the zero power equals one. Second, the curves are concave down throughout their domains—the rate of increase slows as x grows, which is the hallmark of logarithmic growth. Third, observe how the base affects steepness: log₁₀(x) grows most slowly because base 10 is the largest, meaning x must increase by a factor of 10 for the output to rise by just 1 unit. Conversely, ln(x) (base e ≈ 2.718) grows more quickly than log₁₀(x) but more slowly than log₂(x). The vertical asymptote at x = 0 reflects the fact that no power of a positive base can produce zero or a negative number.

SECTION 4

Mathematical Framework

Fundamental Definition

LOGARITHMIC DEFINITION

y=logb(x)⟺by=x,whereb>0,b≠1,x>0y = log_b(x) ⟺ b^y = x, where b > 0, b ≠ 1, x > 0

b = base of the logarithm; x = argument (input); y = exponent (output). The biconditional arrow (⟺) emphasizes that the logarithmic and exponential forms are completely equivalent statements.

Properties of Logarithms

PRODUCT RULE

logb(MN)=logb(M)+logb(N)log_b(MN) = log_b(M) + log_b(N)

The logarithm of a product equals the sum of the logarithms. This is the property that originally made logarithms so useful for computation—multiplication becomes addition.

QUOTIENT RULE

logb(M/N)=logb(M)−logb(N)log_b(M / N) = log_b(M) − log_b(N)

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

POWER RULE

logb(Mk)=k⋅logb(M)log_b(M^k) = k · log_b(M)

An exponent on the argument can be brought out front as a coefficient. This property is essential for solving exponential equations.

CHANGE OF BASE FORMULA

logb(x)=loga(x)/loga(b)=ln(x)/ln(b)log_b(x) = log_a(x) / log_a(b) = ln(x) / ln(b)

Converts between bases. On the AP exam, this formula is critical because most calculators only compute log₁₀ and ln. Setting a = 10 or a = e lets you evaluate any logarithm numerically.

Inverse Composition Identities

Because the logarithmic and exponential functions are inverses, composing them in either order returns the original input. The identity log⁡b\log_b(bˣ) = x holds for all real x, and blogb(x)b^{log_b(x)} = x holds for all x > 0. These identities are the algebraic expression of the graphical reflection symmetry between y = bˣ and y = logb(x) across the line y = x. They are frequently tested on the AP exam in both directions: simplifying nested expressions and verifying that two functions are inverses.

SECTION 5

Transformations & End Behavior

Just as with other function families, transformations of the parent logarithmic function y = logb(x) follow the standard rules for shifts, reflections, stretches, and compressions. The general transformed form is y = a · logb(x − h) + k, where h produces a horizontal shift, k a vertical shift, and a a vertical stretch or reflection. The critical feature to track through any transformation is the vertical asymptote, which shifts to x = h, and the anchor point, which moves from (1, 0) to (1 + h, k).

Transformations of y = log₂(x)xy0123−1−212−1−2y = log₂(x)y = log₂(x − 1)y = log₂(x) + 1y = −log₂(x)(1, 0)(2, 0)(1, 1)
Four variants of y = log₂(x) are shown. The horizontal shift (cyan) moves the asymptote from x = 0 to x = 1. The vertical shift (amber) raises the curve by 1 unit. The reflection (pink) flips the graph over the x-axis while keeping the same asymptote and x-intercept.

End Behavior and Limits

Conditionb > 10 < b < 1
As x → ∞logb(x) → ∞logb(x) → −∞
As x → 0⁺logb(x) → −∞logb(x) → ∞
Increasing / DecreasingAlways increasingAlways decreasing

End behavior comparison based on base value

The end behavior reversal when 0 < b < 1 can be understood through the change-of-base formula: logb(x) = ln(x) / ln(b), and when b < 1, ln(b) is negative. Dividing by a negative constant flips the sign of every output, which is equivalent to reflecting the graph of ln(x) over the x-axis. On the AP exam, remember that the concavity of a logarithmic function with b > 1 is always concave down, and its rate of change (though always positive) is always decreasing.

SECTION 6

Worked Example

Worked example

  1. 1

    Step 1 — State the Problem

    Solve for x: log₃(x + 6) − log₃(x) = 2. State the domain restrictions.
  2. 2

    Step 2 — Apply the Quotient Rule

    Combine the left side using the quotient property: log₃((x + 6) / x) = 2.
    log₃((x + 6) / x) = 2
  3. 3

    Step 3 — Convert to Exponential Form

    Rewrite the equation using the definition of logarithm: if log₃(A) = 2, then 3² = A. Therefore (x + 6) / x = 9.
    (x + 6) / x = 9
  4. 4

    Step 4 — Solve the Rational Equation

    Multiply both sides by x: x + 6 = 9x. Subtract x: 6 = 8x. Divide: x = 6/8 = 3/4.
    x = 3/4
  5. 5

    Step 5 — Verify Domain and Check

    Both arguments must be positive: x + 6 > 0 and x > 0. Since x = 3/4 > 0, both conditions hold. Check: log₃(3/4 + 6) − log₃(3/4) = log₃(27/4) − log₃(3/4) = log₃((27/4) ÷ (3/4)) = log₃(9) = 2. ✓
    x = 3/4 is the valid solution
ℹ
EXAM TIP

SECTION 7

Logarithmic vs. Exponential vs. Power Functions

A central theme in AP Precalculus is comparing how different function families grow. Logarithmic, exponential, and power functions can all model increasing behavior, but their long-run growth rates differ dramatically. Understanding these comparisons is essential for selecting appropriate models in context and for analyzing limits at infinity.

FeatureLogarithmic y = log⁡b\log_b(x)Power y = xⁿExponential y = bˣ
Growth rateSlowest — increases without bound but ever more slowlyModerate — polynomial rate depending on nFastest — eventually dominates all polynomials
Domain(0, ∞)Varies by n; often (−∞, ∞) or [0, ∞)(−∞, ∞)
Range(−∞, ∞)Varies by n(0, ∞)
Concavity (x > 0)Concave down (b > 1)Concave up if n > 1; concave down if 0 < n < 1Concave up (b > 1)
InverseExponential functionRoot / rational power functionLogarithmic function

Comparison of three major function families for x > 0 with b > 1

SECTION 8

Connections to Advanced Topics

Logarithmic functions serve as a gateway to several advanced mathematical and scientific concepts. In calculus, the natural logarithm is the antiderivative of 1/x, which means that ln(x) = ∫₁ˣ (1/t) dt. This integral definition not only provides a rigorous foundation for the function but also explains its connection to area under the hyperbola. The derivative d/dx [ln(x)] = 1/x and the more general d/dx [logb(x)] = 1/(x · ln(b)) are among the most important differentiation formulas in all of calculus.

AP Precalculus ConceptAdvanced Extension
logb(x) as inverse of bˣIntegral definition: ln(x) = ∫₁ˣ (1/t) dt (Calculus)
Change-of-base formulaComplex logarithm: ln(z) = ln|z| + i·arg(z) (Complex Analysis)
Logarithmic scales (pH, decibels)Information entropy: H = −Σ p(x) log₂ p(x) (Information Theory)
Solving exponential equationsDifferential equations: dy/dx = ky → y = CektCe^{kt} (Differential Equations)

From AP Precalculus to advanced mathematics

In applied contexts, semi-log and log-log plots use logarithmic scales on one or both axes to linearize exponential and power-law data. When you plot data on a semi-log graph (log scale on the y-axis, linear on x) and the points form a straight line, the data follows an exponential model. Similarly, a straight line on a log-log plot indicates a power function relationship. These graphical techniques, while introduced conceptually in AP Precalculus, are indispensable tools in data science, physics, and engineering.

SECTION 9

Practice Problems

Question 1: CONCEPTUAL

Which of the following statements about the function f(x) = log₅(x − 3) + 2 is true?
  • A. The domain is (3, ∞) and the vertical asymptote is x = 3.
  • B. The domain is (−∞, ∞) and there is no vertical asymptote.
  • C. The domain is (3, ∞) and the vertical asymptote is x = 2.
  • D. The domain is (−3, ∞) and the vertical asymptote is x = −3.
  • E. The range is (2, ∞) and the vertical asymptote is x = 3.
Show answer (A)
A is correct. The argument of the logarithm must be positive: x − 3 > 0 implies x > 3, so the domain is (3, ∞). The vertical asymptote occurs where the argument equals zero, which is x = 3. The +2 shifts the graph vertically but does not affect the domain or the location of the asymptote. B is wrong because logarithmic functions always have a restricted domain; they cannot accept non-positive arguments. C incorrectly places the asymptote at x = 2, confusing the vertical shift k = 2 with the horizontal shift h = 3. D reverses the sign of the shift, treating (x − 3) as (x + 3). E incorrectly restricts the range; the range of any logarithmic function (with nonzero leading coefficient) is all real numbers (−∞, ∞).

Question 2: BASIC CALCULATION

What is the value of log₈(32)?
  • A. 3/5
  • B. 4
  • C. 5/3
  • D. 5/2
  • E. 3/2
Show answer (C)
C is correct. Express both 8 and 32 as powers of 2: 8 = 2³ and 32 = 2⁵. Then log₈(32) = log₂₃(2⁵). Using the change-of-base approach or the property that log_(bⁿ)(bᵐ) = m/n, we get 5/3. A (3/5) inverts the fraction, confusing the exponents. B (4) is the value of log₂(16), not log₈(32). D (5/2) would be correct if the base were 4 (since 4 = 2²), not 8. E (3/2) is log₄(8), not log₈(32).

Question 3: INTERMEDIATE

If log₂(x) + log₂(x − 2) = 3, what is the value of x?
  • A. x = 4
  • B. x = −2
  • C. x = 4 and x = −2
  • D. x = 8
  • E. x = 3
Show answer (A)
A is correct. Apply the product rule: log₂(x(x − 2)) = 3. Convert to exponential form: x(x − 2) = 2³ = 8, giving x² − 2x − 8 = 0. Factor: (x − 4)(x + 2) = 0, so x = 4 or x = −2. Now check domains: log₂(x) requires x > 0 and log₂(x − 2) requires x > 2. Since x = −2 fails both conditions, it is extraneous. Only x = 4 is valid. B is the extraneous solution alone. C includes the extraneous solution. D (x = 8) likely results from a setup error such as treating the sum of logs as log₂(x + x − 2). E (x = 3) does not satisfy the equation: log₂(3) + log₂(1) = log₂(3) ≈ 1.585, not 3.

Question 4: APPLIED

The intensity of sound in decibels is given by D = 10 log₁₀(I / I₀), where I₀ = 10⁻¹² W/m² is the threshold of hearing. A rock concert registers at 115 dB. A library is measured at 40 dB. (a) Find the intensity I of the rock concert in W/m². (b) How many times more intense is the rock concert than the library? (c) If two identical rock concerts play simultaneously in the same space, what is the combined decibel level?
Show answer
(a) From 115 = 10 log₁₀(I/I₀), divide by 10: log₁₀(I/I₀) = 11.5. Convert to exponential form: I/I₀ = 10¹¹·⁵, so I = 10⁻¹² × 10¹¹·⁵ = 10⁻⁰·⁵ ≈ 0.316 W/m². (b) The library intensity is ILI_L = 10⁻¹² × 10⁴ = 10⁻⁸ W/m². The ratio is 10⁻⁰·⁵ / 10⁻⁸ = 10⁷·⁵ ≈ 31,622,777 times more intense. Alternatively, the difference in decibels is 75, so the intensity ratio is 1075/1010^{75/10} = 10⁷·⁵. (c) Doubling the intensity adds 10 log₁₀(2) ≈ 3.01 dB to the original level: 115 + 3.01 ≈ 118 dB. This illustrates a key logarithmic principle: doubling intensity does not double decibels—it adds approximately 3 dB.

Question 5: CRITICAL THINKING

Let f(x) = log₃(9x²). (a) Rewrite f(x) as a sum of logarithmic terms. (b) Determine the domain of f. (c) Explain why f is NOT the same function as g(x) = 2 + 2 log₃(x), even though the algebraic simplifications look identical. Identify the precise difference in their domains and describe how their graphs compare.
Show answer
(a) Using the product rule: f(x) = log₃(9) + log₃(x²) = 2 + log₃(x²). Applying the power rule carefully: log₃(x²) = 2 log₃|x|, so f(x) = 2 + 2 log₃|x|. (b) The argument 9x² must be positive. Since 9 > 0, we need x² > 0, which holds for all x ≠ 0. Domain: (−∞, 0) ∪ (0, ∞). (c) The function g(x) = 2 + 2 log₃(x) requires x > 0, so its domain is (0, ∞). The naive application of the power rule log₃(x²) = 2 log₃(x) is only valid when x > 0 because log₃(x) is undefined for negative x. For negative x, f(x) = 2 + 2 log₃(−x), which means the left branch of f's graph is a reflection of g across the y-axis. The graph of f is symmetric about the y-axis with a vertical asymptote at x = 0 (approached from both sides), while g has only the right branch. This problem illustrates why domain analysis must precede algebraic manipulation of logarithmic expressions.

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