Master the Graphs & Properties of Logarithmic Functions | 对数函数的图像与性质

📚 Master the Graphs & Properties of Logarithmic Functions | 对数函数的图像与性质

Logarithmic functions are among the most frequently tested topics in IB Mathematics, appearing in both Analysis & Approaches (AA) and Applications & Interpretation (AI). A strong command of their graphs and algebraic properties is essential for solving exponential equations, modelling real-world growth and decay, and understanding inverse relationships with exponential functions. This guide consolidates everything you need to know — from the fundamental definition to advanced graph transformations — with clear examples tailored to IB examination style.

对数函数是 IB 数学中考查频率极高的内容，既出现在 Analysis & Approaches (AA) 中，也出现在 Applications & Interpretation (AI) 中。熟练掌握对数函数的图像与代数性质，是求解指数方程、建立现实世界增长与衰减模型，以及理解指数函数与对数函数互为反函数关系的关键。本指南整合了你需要掌握的全部核心知识——从基本定义到复杂的图像变换——并配有贴合 IB 考试风格的典型例题。

1. Definition and Basic Form | 定义与基本形式

For any positive real number a ≠ 1 and any positive real number x, the logarithmic function is defined as the inverse of the exponential function: if y = logₐx, then aʸ = x. The base a must always satisfy the condition a > 0 and a ≠ 1, and the argument x must be strictly positive.

对于任意正实数 a（且 a ≠ 1）和任意正实数 x，对数函数被定义为指数函数的反函数：若 y = logₐx，则 aʸ = x。底数 a 必须始终满足 a > 0 且 a ≠ 1，真数 x 必须严格大于零。

y = logₐx ⇔ x = aʸ (a > 0, a ≠ 1, x > 0)

Two special bases deserve particular attention in IB: base 10, written as log x or log₁₀x, and base e (Euler’s number ≈ 2.718), written as ln x or logₑx. The natural logarithm ln x is especially important because its derivative is 1/x, and it appears in continuous growth models, integration techniques, and differential equations throughout the IB syllabus.

在 IB 课程中有两个特殊的底数需要特别注意：以 10 为底，记作 log x 或 log₁₀x；以及以 e（欧拉常数 ≈ 2.718）为底，记作 ln x 或 logₑx。自然对数 ln x 尤为重要，因为其导数为 1/x，并且它广泛出现在 IB 教学大纲中的连续增长模型、积分技巧和微分方程等内容中。

2. Key Properties of Logarithms | 对数的核心运算法则

The algebraic properties of logarithms derive directly from the laws of exponents. Memorising these rules is non-negotiable for IB success — they are the foundation for solving logarithmic and exponential equations, simplifying expressions, and evaluating limits.

对数的代数性质直接源自指数运算法则。对于 IB 考试而言，牢记这些规则是硬性要求——它们是求解对数与指数方程、化简表达式以及计算极限的基础。

Product Rule | 积的对数: logₐ(MN) = logₐM + logₐN
Quotient Rule | 商的对数: logₐ(M/N) = logₐM − logₐN
Power Rule | 幂的对数: logₐ(Mⁿ) = n · logₐM
Change of Base | 换底公式: logₐb = logₓb / logₓa = ln b / ln a
Special Values | 特殊值: logₐ1 = 0; logₐa = 1

These properties are valid only when M and N are positive, a > 0, a ≠ 1. A common mistake observed in IB examinations is applying the product rule to logₐ(M + N) — no such simplification exists. The logarithm of a sum is not the sum of the logarithms.

这些运算法则仅在 M 和 N 为正数、a > 0、a ≠ 1 时成立。在 IB 考试中常见的错误是将积的对数法则错误地套用于 logₐ(M + N)——这样的化简是不存在的。和的对数不等于对数的和。

3. The Graph of y = logₐx | 对数函数的图像特征

The graph of a logarithmic function is the reflection of the exponential graph y = aˣ across the line y = x. This symmetry is a direct consequence of the inverse relationship between the two functions. The graph passes through a distinctive shape that must be committed to memory for both sketch-only and transformation-style questions.

对数函数的图像是指数函数 y = aˣ 的图像关于直线 y = x 对称得到的。这种对称性是两个函数互为反函数的直接结果。对数函数的图像具有独特的形状特征，无论是纯作图题还是图像变换题，你都必须牢记这些特征。

y = logₐx 与 y = aˣ 关于直线 y = x 对称 (a > 0, a ≠ 1)

Key characteristics of the graph y = logₐx for a > 1:

当 a > 1 时，y = logₐx 的图像关键特征如下：

The curve passes through (1, 0) — the x-intercept.
The y-axis (x = 0) is a vertical asymptote; the curve approaches it as x → 0⁺.
The function is strictly increasing on its entire domain (0, ∞).
As x → ∞, y grows without bound (slowly, since logarithmic growth is very slow).
The domain is x ∈ (0, ∞); the range is y ∈ ℝ.

曲线经过点 (1, 0)——即 x 轴截距。
y 轴（即 x = 0）是垂直渐近线；当 x → 0⁺ 时，曲线无限靠近 y 轴。
函数在其整个定义域 (0, ∞) 上严格递增。
当 x → ∞ 时，y 无限增大（且增长速度缓慢，因为对数增长极慢）。
定义域为 x ∈ (0, ∞)；值域为 y ∈ ℝ。

When 0 < a < 1, the graph is the mirror image of the case a > 1 reflected across the x-axis. In this case, the function is strictly decreasing, yet it still passes through (1, 0) and retains the same vertical asymptote and domain restrictions.

当 0 < a < 1 时，图像是 a > 1 情形关于 x 轴对称的镜像。此时函数严格递减，但仍然经过 (1, 0)，并且保持相同的垂直渐近线和定义域限制。

4. Domain, Range, and Asymptote | 定义域、值域与渐近线

For any logarithmic function of the form f(x) = logₐ(g(x)), the argument g(x) must be strictly positive. Setting g(x) > 0 and solving for x gives the domain. This is a routine question type in IB Paper 1, often combined with solving a quadratic inequality.

对于形如 f(x) = logₐ(g(x)) 的任意对数函数，真数 g(x) 必须严格大于零。令 g(x) > 0 并求解 x 即可得到定义域。这是 IB Paper 1 中的常规题型，通常与求解二次不等式结合考查。

Example | 例题: Find the domain of f(x) = log₂(x² − 3x + 2).

Solution | 解答: We require x² − 3x + 2 > 0. Factorising gives (x − 1)(x − 2) > 0, hence x < 1 or x > 2. The domain is therefore x ∈ (−∞, 1) ∪ (2, ∞).

解答：需要满足 x² − 3x + 2 > 0。因式分解得 (x − 1)(x − 2) > 0，因此 x < 1 或 x > 2。所以定义域为 x ∈ (−∞, 1) ∪ (2, ∞)。

The vertical asymptote corresponds to the value(s) of x for which the argument equals zero. For f(x) = logₐ(g(x)), the vertical asymptote is found by solving g(x) = 0. Be mindful that the range of any logarithmic function is always all real numbers, regardless of the base or the complexity of the argument — provided the argument covers all positive values.

垂直渐近线对应于使真数为零的 x 值。对于 f(x) = logₐ(g(x))，通过求解 g(x) = 0 可找到垂直渐近线。请注意，无论底数如何变化、真数多么复杂，任何对数函数的值域始终为全体实数——前提是真数能取遍所有正数。

5. Increasing and Decreasing Behaviour | 增减性分析

The monotonicity of a logarithmic function depends entirely on the base:

对数函数的单调性完全取决于底数：

If a > 1, logₐx is strictly increasing: for 0 < x₁ < x₂, we have logₐx₁ < logₐx₂.
若 a > 1，logₐx 严格递增：当 0 < x₁ < x₂ 时，有 logₐx₁ < logₐx₂。
If 0 < a < 1, logₐx is strictly decreasing: for 0 < x₁ < x₂, we have logₐx₁ > logₐx₂.
若 0 < a < 1，logₐx 严格递减：当 0 < x₁ < x₂ 时，有 logₐx₁ > logₐx₂。

These inequalities are instrumental in solving logarithmic inequalities, a topic that frequently appears in IB. For instance, to solve log₃(x − 1) < 2, first note that the domain requires x > 1. Then rewrite the inequality in exponential form: x − 1 < 3², giving x < 10. Combining with the domain yields the solution 1 < x < 10.

这些不等式在求解对数不等式时至关重要，而对数不等式是 IB 中频繁出现的考点。例如，要求解 log₃(x − 1) < 2，首先注意定义域要求 x > 1。然后将不等式改写为指数形式：x − 1 < 3²，得到 x < 10。结合定义域，最终解为 1 < x < 10。

A crucial subtlety: since log₂x is increasing, the inequality direction is preserved when both sides are raised as powers of 2. However, when the base is between 0 and 1, the inequality direction must be reversed. This is a classic trap in IB Paper 1.

一个关键的细微之处：由于 log₂x 是递增的，因此以 2 为底取幂时不等式方向保持不变。然而，当底数在 0 和 1 之间时，不等式方向必须反转。这是 IB Paper 1 中的经典陷阱。

6. Graph Transformations | 图像变换

All standard transformations apply to logarithmic functions. These questions are extremely common and test your ability to sketch and interpret modified log graphs without using a calculator.

所有标准图像变换都适用于对数函数。这类题目非常常见，考查你在不使用计算器的情况下绘制和解读变形式对数函数图像的能力。

f(x) = logₐx 的基本变换规则

Transform \| 变换	Function \| 函数	Effect \| 效果
Vertical translation \| 垂直平移	y = logₐx + k	Shifts up (k > 0) or down (k < 0)
Horizontal translation \| 水平平移	y = logₐ(x − h)	Shifts right (h > 0) or left (h < 0)
Vertical stretch \| 垂直伸缩	y = k · logₐx	Scale factor k away from or toward x-axis
Horizontal stretch \| 水平伸缩	y = logₐ(kx)	Note: logₐ(kx) = logₐk + logₐx — vertical shift equivalent
Reflection in x-axis \| 关于 x 轴对称	y = −logₐx	Flips graph upside down
Reflection in y-axis \| 关于 y 轴对称	y = logₐ(−x)	Reflects across y-axis; domain becomes x < 0

An important observation: replacing x with (−x) reverses the sign of the argument, which restricts the domain to negative numbers. This transformation is often overlooked in sketching questions, but it is a favourite in IB Paper 2 Section A.

一个重要的观察：将 x 替换为 (−x) 会改变真数的符号，从而将定义域限制为负数。这种变换在作图题中常被忽略，但在 IB Paper 2 Section A 中备受青睐。

7. The Inverse Relationship with Exponentiation | 与指数函数的反函数关系

The single most important concept connecting logarithmic and exponential functions is their inverse relationship. Fully grasping this relationship unlocks the ability to solve equations, simplify expressions, and understand why the graphs are reflections of each other.

联系对数函数与指数函数的唯一最重要的概念就是它们的反函数关系。充分理解这一关系，将为你解锁求解方程、化简表达式以及理解二者图像互为镜像的能力。

logₐ(aˣ) = x (for all real x) and a^(logₐx) = x (for x > 0)

These identities are essentially two sides of the same coin. They explain why exponential equations such as 2ˣ = 32 can be solved directly, while more complex equations such as 2ˣ = 30 require the use of logarithms: take ln of both sides to get x ln 2 = ln 30, hence x = ln 30/ln 2.

这些恒等式本质上是同一枚硬币的两面。它们解释了为什么像 2ˣ = 32 这样的指数方程可以直接求解，而诸如 2ˣ = 30 之类的更复杂的方程则需要借助对数：在等式两边取自然对数得 x ln 2 = ln 30，从而 x = ln 30 / ln 2。

In IB examinations, you are expected to recognise when to apply the inverse identities directly versus when to use the change-of-base formula. For example, solve 5^(2x−1) = 20: taking natural log on both sides gives (2x − 1)ln 5 = ln 20, hence 2x − 1 = ln 20 / ln 5, so x = (ln 20 / ln 5 + 1)/2 ≈ 1.43.

在 IB 考试中，你需要能够判断何时直接运用反函数恒等式，何时使用换底公式。例如，求解 5^(2x−1) = 20：两边取自然对数得 (2x − 1)ln 5 = ln 20，从而 2x − 1 = ln 20 / ln 5，故 x = (ln 20 / ln 5 + 1) / 2 ≈ 1.43。

8. Solving Logarithmic Equations | 求解对数方程

Logarithmic equations appear in both Paper 1 (without GDC) and Paper 2 (with GDC). The standard strategy involves two stages: first, use the properties of logarithms to combine terms on each side into a single logarithm; second, use the one-to-one property to equate arguments.

对数方程在 Paper 1（不使用图形计算器）和 Paper 2（允许使用图形计算器）中均有出现。标准解题策略分两个阶段：首先利用对数运算法则将各侧项合并为单一对数；然后利用一一对应性质令真数相等。

Example | 例题: Solve log₂x + log₂(x − 2) = 3.

Solution | 解答: Apply the product rule: log₂[x(x − 2)] = 3. Convert to exponential form: x(x − 2) = 2³ = 8. Thus x² − 2x − 8 = 0, giving (x − 4)(x + 2) = 0. The solutions are x = 4 and x = −2; however, x = −2 is extraneous because the original equation contains log₂x, which requires x > 0, and log₂(x − 2), which requires x > 2. Hence the only valid solution is x = 4.

解答：应用积的对数法则：log₂[x(x − 2)] = 3。转换为指数形式：x(x − 2) = 2³ = 8。即 x² − 2x − 8 = 0，因式分解得 (x − 4)(x + 2) = 0。解为 x = 4 和 x = −2；但 x = −2 是增根，因为原方程中包含 log₂x（要求 x > 0）和 log₂(x − 2)（要求 x > 2）。因此唯一有效解为 x = 4。

Always check for extraneous solutions — a hallmark of IB exam questions. The safest approach is to substitute your final answers back into the original equation and verify that all arguments are positive.

务必检查增根——这是 IB 考题的标志性特征。最稳妥的方法是将最终答案代回原方程，验证所有真数均为正数。

9. Applications in Real-World Contexts | 现实应用

Logarithmic functions are a cornerstone of modelling in IB Applications & Interpretation. In this course, you will encounter them in exponential growth and decay models, pH calculations, Richter scale magnitudes, sound intensity (decibels), and compound interest problems.

对数函数是 IB Applications & Interpretation 中建模的基石。在这门课程中，你将在指数增长与衰变模型、pH 值计算、里氏震级、声音强度（分贝）以及复利问题中遇见它们。

In the context of continuous growth, the exponential model is N(t) = N₀e^(kt). Taking the natural logarithm of both sides transforms exponential relationships into linear ones: ln N(t) = ln N₀ + kt. This log-linearisation technique is widely used to estimate growth rates and half-lives from experimental data.

在连续增长的背景下，指数模型为 N(t) = N₀e^(kt)。在等式两边取自然对数，可以将指数关系转化为线性关系：ln N(t) = ln N₀ + kt。这种对数线性化技术被广泛用于从实验数据中估算增长率与半衰期。

Example | 例题: A radioactive substance has a half-life of 5 days. Given N₀ = 100 mg, find the time required for the substance to decay to 25 mg.

Solution | 解答: The decay model is N(t) = 100e^(−kt). Using the half-life condition: 50 = 100e^(−5k), so e^(−5k) = 0.5. Take ln: −5k = ln 0.5, giving k = −ln 0.5/5 ≈ 0.1386. Now solve 25 = 100e^(−0.1386t): e^(−0.1386t) = 0.25, so −0.1386t = ln 0.25, t = ln 0.25 / (−0.1386) = 10 days. Alternatively, recognise that 25 mg is one quarter of 100 mg — exactly two half-lives of 5 days each.

解答：衰变模型为 N(t) = 100e^(−kt)。利用半衰期条件：50 = 100e^(−5k)，即 e^(−5k) = 0.5。取自然对数：−5k = ln 0.5，得 k = −ln 0.5 / 5 ≈ 0.1386。接着解 25 = 100e^(−0.1386t)：e^(−0.1386t) = 0.25，所以 −0.1386t = ln 0.25，t = ln 0.25 / (−0.1386) = 10 天。或者直接理解：25 mg 是 100 mg 的四分之一——正好等于两个 5 天的半衰期。

10. Summary and Key Takeaways | 总结与核心要点

Mastering logarithmic functions requires integrating three layers of understanding: (1) the algebraic properties that allow you to manipulate expressions; (2) the graphical features — domain, range, asymptote, intercept, and monotonicity — that allow you to sketch and interpret graphs; and (3) the inverse relationship with exponentials that underpins equation-solving and real-world modelling.

掌握对数函数需要整合三个层次的理解：（1）代数运算法则，用于化简和变换表达式；（2）图像特征——定义域、值域、渐近线、截距和单调性——用于绘制和解读图像；（3）与指数函数的反函数关系，这是求解方程和建立现实世界模型的基础。

When tackling exam questions, always begin by stating the domain. Check whether the base is greater than or less than 1, as this determines monotonicity and the direction of inequalities. Finally, verify all solutions by substitution to eliminate extraneous roots. With systematic practice, logarithms become one of the most predictable and rewarding topics on the IB mathematics papers.

在做题时，从定义域开始分析。判断底数是大于 1 还是小于 1，因为这决定了单调性和不等号的方向。最后，通过代回验证所有解，以排除增根。通过系统化的练习，对数函数将成为 IB 数学考试中最具规律性和得分收益最高的考点之一。

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Master the Graphs & Properties of Logarithmic Functions | 对数函数的图像与性质