A-Level CIE Mathematics: Exponentials and Logarithms – Key Points Revision | A-Level CIE 数学：指数与对数考点精讲

📚 A-Level CIE Mathematics: Exponentials and Logarithms – Key Points Revision | A-Level CIE 数学：指数与对数考点精讲

Exponentials and logarithms are central to CIE A-Level Mathematics, underpinning topics from algebraic manipulation to calculus and real-world modelling. This revision guide covers every key concept, law and technique you need to master, with clear explanations and exam-ready examples. It is designed for both Pure Mathematics 2 and 3 (P2/P3) students who must handle exponential equations, logarithmic graphs, natural logarithms and their differentiation with confidence.

指数与对数是 CIE A-Level 数学的核心内容，从代数运算到微积分再到实际建模都离不开它们。本考点精讲涵盖每一个关键概念、法则和解题技巧，并配有清晰的解释和贴近考题的示例，帮助 P2/P3 考生扎实掌握指数方程、对数图像、自然对数及其微分。

1. Laws of Indices | 指数法则

When multiplying powers with the same base, add the exponents: aᵐ × aⁿ = aᵐ⁺ⁿ.

底数相同幂相乘，指数相加：aᵐ × aⁿ = aᵐ⁺ⁿ。

When dividing powers with the same base, subtract the exponents: aᵐ ÷ aⁿ = aᵐ⁻ⁿ.

底数相同幂相除，指数相减：aᵐ ÷ aⁿ = aᵐ⁻ⁿ。

Raising a power to another power means multiplying the exponents: (aᵐ)ⁿ = aᵐⁿ.

幂的乘方等于指数相乘：(aᵐ)ⁿ = aᵐⁿ。

A negative exponent indicates a reciprocal: a⁻ⁿ = 1 / aⁿ.

负指数表示倒数：a⁻ⁿ = 1 / aⁿ。

Any non-zero base raised to the power 0 equals 1: a⁰ = 1 (a ≠ 0).

任何非零底数的 0 次幂都等于 1：a⁰ = 1（a ≠ 0）。

Fractional exponents correspond to roots: a^(1/n) = ⁿ√a, and a^(m/n) = ⁿ√(aᵐ) = (ⁿ√a)ᵐ.

分数指数对应根式：a^(1/n) = ⁿ√a，且 a^(m/n) = ⁿ√(aᵐ) = (ⁿ√a)ᵐ。

These rules are valid for all real exponents and are the foundation for simplifying exponential expressions and solving equations.

这些法则对所有实数指数都成立，是化简指数表达式和求解方程的基础。

2. Introduction to Logarithms | 对数入门

The logarithm logₐ b answers the question: ‘To what power must the base a be raised to obtain b?’ That is, aʸ = b ⇔ y = logₐ b.

对数 logₐ b 回答这样一个问题：“底数 a 的多少次幂等于 b？”即 aʸ = b ⇔ y = logₐ b。

Logarithms are the inverses of exponentials. The statement 2³ = 8 is equivalent to log₂ 8 = 3.

对数是指数的逆运算。2³ = 8 等价于 log₂ 8 = 3。

Key values to remember: logₐ 1 = 0 because a⁰ = 1; logₐ a = 1 because a¹ = a.

务必牢记的关键值：logₐ 1 = 0 因为 a⁰ = 1；logₐ a = 1 因为 a¹ = a。

In CIE, you will work with logarithms of any positive base a > 0, a ≠ 1, and also with base 10 (common log, written as log x or log₁₀ x) and base e (natural log, ln x).

CIE 考试中你既会用到任意正底数 a > 0, a ≠ 1 的对数，也会用到底数 10（常用对数，写作 log x 或 log₁₀ x）和底数 e（自然对数，ln x）。

Understanding the equivalence between exponential and logarithmic forms is the first step in manipulating equations.

理解指数形式与对数形式的等价关系是处理方程的第一步。

3. Laws of Logarithms | 对数法则

The laws of logarithms mirror the laws of indices and are essential for simplifying logarithmic expressions.

对数法则与指数法则相对应，是化简对数式的必要工具。

Product rule: logₐ (MN) = logₐ M + logₐ N.

乘法法则：logₐ (MN) = logₐ M + logₐ N。

Quotient rule: logₐ (M / N) = logₐ M – logₐ N.

除法法则：logₐ (M / N) = logₐ M – logₐ N。

Power rule: logₐ (Mᵏ) = k logₐ M.

幂法则：logₐ (Mᵏ) = k logₐ M。

These three rules allow you to combine or separate logarithmic terms. In CIE exam questions, you often need to use them together to reduce an equation to a form that can be solved.

利用这三条法则可以合并或拆开对数项。CIE 试题中常常需要综合运用它们，将方程化简为可解的形式。

Be careful with the domain: the argument of a logarithm must be strictly positive. For example, logₐ (x–2) requires x > 2.

注意定义域：对数的真数必须严格为正。例如 logₐ (x–2) 要求 x > 2。

4. Change of Base Formula | 换底公式

Sometimes you need to evaluate or compare logarithms with different bases. The change-of-base formula provides the tool:

logₐ b = log꜀ b / log꜀ a

有时需要计算或比较不同底数的对数，换底公式正是这一工具：

logₐ b = log꜀ b / log꜀ a

Common choices for the new base c are 10 (using log on the calculator) or e (using ln). Therefore:

logₐ b = log b / log a = ln b / ln a

新底数 c 通常选 10（计算器上的 log）或 e（用 ln）。因此：

logₐ b = log b / log a = ln b / ln a

This formula is extremely useful when an equation involves several different bases or when an exact algebraic answer is required without a calculator.

当方程涉及多个不同的底数，或在无计算器情况下需要给出精确代数解时，该公式极其实用。

In CIE questions, you may be asked to prove the formula starting from the definition: let y = logₐ b, then aʸ = b, take log꜀ of both sides and apply the power rule.

CIE 试题可能要求你从定义出发证明该公式：设 y = logₐ b，则 aʸ = b，两边取 log꜀ 并运用幂法则即可。

5. Solving Exponential Equations | 解指数方程

Exponential equations have the unknown in the exponent, such as 2ˣ = 7 or e²ˣ⁻¹ = 5. Three approaches are common in CIE exams.

指数方程中未知数出现在指数位置上，如 2ˣ = 7 或 e²ˣ⁻¹ = 5。CIE 考试中常用三种方法。

Method 1 – Equalising bases: If both sides can be written as powers of the same base, equate the exponents. Example: 4ˣ = 8 ⇒ (2²)ˣ = 2³ ⇒ 2²ˣ = 2³ ⇒ 2x = 3 ⇒ x = 1.5.

方法一：统一底数。若两边可写成同底数的幂，则令指数相等。例：4ˣ = 8 ⇒ (2²)ˣ = 2³ ⇒ 2²ˣ = 2³ ⇒ 2x = 3 ⇒ x = 1.5。

Method 2 – Taking logarithms: When bases cannot easily be made the same, take logarithms of both sides (usually ln or log). Example: 3ˣ = 20 ⇒ ln(3ˣ) = ln 20 ⇒ x ln 3 = ln 20 ⇒ x = ln 20 / ln 3.

方法二：取对数。当无法轻易统一底数时，可在方程两边同时取对数（通常用 ln 或 log）。例：3ˣ = 20 ⇒ ln(3ˣ) = ln 20 ⇒ x ln 3 = ln 20 ⇒ x = ln 20 / ln 3。

Method 3 – Substitution and disguised quadratics: Equations like e²ˣ – 5eˣ + 6 = 0 can be solved by letting y = eˣ, giving y² – 5y + 6 = 0, then back-substituting to find x.

方法三：换元与隐藏二次方程。像 e²ˣ – 5eˣ + 6 = 0 这样的方程可设 y = eˣ，化为 y² – 5y + 6 = 0，然后再代回求 x。

Always check that your answers lie in the valid domain, and remember to give exact answers in terms of logarithms unless a decimal approximation is requested.

始终检查所得解是否落在有效定义域内，除非题目明确要求近似值，否则应保留用对数表示的精确解。

6. Solving Logarithmic Equations | 解对数方程

Logarithmic equations contain log terms with the unknown inside the argument. Step-by-step simplification using the laws of logarithms is key.

对数方程含有以未知数为真数的对数项，关键是运用对数法则逐步化简。

Strategy: First, use the laws of logarithms to combine terms into a single logarithm on one or both sides. Example: log₂ (x) + log₂ (x – 2) = 3 ⇒ log₂ [x(x – 2)] = 3.

策略：首先利用对数法则将各项合并成一边或两边的单个对数。例：log₂ (x) + log₂ (x – 2) = 3 ⇒ log₂ [x(x – 2)] = 3。

Then rewrite in exponential form: x(x – 2) = 2³ = 8 ⇒ x² – 2x – 8 = 0 ⇒ (x – 4)(x + 2) = 0.

然后改写为指数形式：x(x – 2) = 2³ = 8 ⇒ x² – 2x – 8 = 0 ⇒ (x – 4)(x + 2) = 0。

Potential solutions are x = 4 and x = –2. Since the argument of a logarithm must be positive, x = –2 is rejected because log₂(x–2) would involve a negative argument. Final answer: x = 4.

可能的解为 x = 4 和 x = –2。由于对数真数必须为正，x = –2 会使得 log₂(x–2) 的真数为负，应舍弃。最终解：x = 4。

When two logarithms with the same base are equated, simply equate their arguments: logₐ M = logₐ N ⇒ M = N, provided M, N > 0.

当两个同底对数相等时，可直接令真数相等：logₐ M = logₐ N ⇒ M = N，并确保 M, N > 0。

Stay alert for extraneous solutions, which frequently appear after removing logs, and always verify against the original equation.

务必警惕去掉对数后产生的增根，一定要代入原方程检验。

7. Exponential Functions and Their Graphs | 指数函数及其图像

The basic exponential function is y = aˣ where a > 0 and a ≠ 1. Its graph passes through (0,1) because a⁰ = 1.

基本指数函数为 y = aˣ，其中 a > 0 且 a ≠ 1。图像经过 (0,1)，因为 a⁰ = 1。

For a > 1, the function is strictly increasing and growth gets steeper as x increases. The x‑axis (y = 0) is a horizontal asymptote as x → –∞.

若 a > 1，函数严格递增，随着 x 增大曲线愈加陡峭。x 轴（y = 0）是当 x → –∞ 时的水平渐近线。

For 0 < a < 1, the function is strictly decreasing and the asymptote still occurs as x → +∞. The y‑intercept remains at (0,1).

若 0 < a < 1，函数严格递减，当 x → +∞ 时曲线趋近于 x 轴。y 轴截距仍在 (0,1)。

Transformations of y = aˣ follow the usual patterns: y = aˣ⁺ᵏ shifts horizontally, y = aˣ + c shifts vertically, and y = b·aˣ stretches vertically by factor b.

y = aˣ 的图像变换遵循常见规律：y = aˣ⁺ᵏ 水平平移，y = aˣ + c 垂直平移，y = b·aˣ 垂直拉伸 b 倍。

CIE often asks you to sketch these graphs, label the intercept and asymptote, and describe the effect of parameter changes.

CIE 常要求你画出这类图像的草图，标出截距和渐近线，并描述参数变化带来的影响。

8. Logarithmic Functions and Their Graphs | 对数函数及其图像

The logarithmic function y = logₐ x is the inverse of y = aˣ. Its domain is x > 0 and its range is all real numbers.

对数函数 y = logₐ x 是 y = aˣ 的反函数。定义域为 x > 0，值域为全体实数。

The graph passes through (1,0) because logₐ 1 = 0, and through (a,1) because logₐ a = 1.

图像经过 (1,0)（因为 logₐ 1 = 0）和 (a,1)（因为 logₐ a = 1）。

For a > 1, the function is increasing and has a vertical asymptote at x = 0 (the y‑axis). As x → 0⁺, y → –∞.

若 a > 1，函数递增，在 x = 0 处有一条垂直渐近线（即 y 轴）。当 x → 0⁺ 时，y → –∞。

Because exponential and logarithmic functions are inverses, their graphs are reflections of each other in the line y = x.

由于指数函数与对数函数互为反函数，其图像关于直线 y = x 对称。

When solving inequalities or finding domains, remember that you can only take the log of a positive quantity. This restriction is a common source of mark loss.

解不等式或确定定义域时，切记只有正数才能取对数，这一限制条件常是失分点。

9. The Natural Exponential and Natural Logarithm | 自然指数与自然对数

The number e ≈ 2.71828 is a fundamental constant in mathematics. The function y = eˣ is called the natural exponential function.

e ≈ 2.71828 是数学中的一个基本常数。y = eˣ 称为自然指数函数。

Its remarkable property is that the gradient of the graph at any point is equal to the value of the function itself. This leads to d/dx (eˣ) = eˣ.

其非凡性质在于，图像上任一点的斜率都等于该点的函数值，从而有 d/dx (eˣ) = eˣ。

The natural logarithm, written as ln x, is the logarithm to base e: ln x = logₑ x. It inherits all the logarithm laws, so ln (ab) = ln a + ln b, ln (1) = 0, ln e = 1.

自然对数记作 ln x，是以 e 为底的对数：ln x = logₑ x。它继承所有对数法则，如 ln (ab) = ln a + ln b，ln 1 = 0，ln e = 1。

In CIE exams, you will frequently see e and ln in calculus, exponential growth/decay, and in solving equations where base e simplifies differentiation.

在 CIE 考试中，你会在微积分、指数增长与衰减以及利用以 e 为底简化导数的方程求解中频繁遇到 e 和 ln。

When converting between forms, recall that y = eˣ ⇔ x = ln y, which is extremely useful for linearising data in modelling questions.

进行形式转换时，记住 y = eˣ ⇔ x = ln y，这在建模题中对数据线性化极为有用。

10. Differentiation of eˣ and ln x | eˣ 与 ln x 的导数

CIE Pure Mathematics 3 requires you to differentiate and integrate exponential and logarithmic functions accurately.

CIE 纯数 3 要求准确地对指数函数和对数函数求导与积分。

d/dx (eˣ) = eˣ

By applying the chain rule, d/dx (eᵏˣ) = k eᵏˣ. More generally, d/dx (eᶠ⁽ˣ⁾) = f'(x) eᶠ⁽ˣ⁾.

结合链式法则，d/dx (eᵏˣ) = k eᵏˣ，更一般地 d/dx (eᶠ⁽ˣ⁾) = f'(x) eᶠ⁽ˣ⁾。

The derivative of the natural logarithm is:

d/dx (ln x) = 1 / x, for x > 0

自然对数的导数为：

d/dx (ln x) = 1 / x，其中 x > 0

Using the chain rule again, d/dx (ln(f(x))) = f'(x) / f(x). This form appears frequently in integration and differential equations.

再次使用链式法则，d/dx (ln(f(x))) = f'(x) / f(x)。这种形式在积分和微分方程中频繁出现。

Integration reverses these results: ∫ eˣ dx = eˣ + C, and ∫ (1/x) dx = ln|x| + C. Students often forget the absolute value inside the logarithm.

积分是逆运算：∫ eˣ dx = eˣ + C，∫ (1/x) dx = ln|x| + C。学生常忘记对数内的绝对值。

These differentiation rules underpin many CIE mark-heavy questions on stationary points, rates of change, and integration by recognition.

这些导数规则是许多 CIE 高分值试题的基础，涉及驻点、变化率以及配型积分法。

11. Exponential Growth and Decay Models | 指数增长与衰减模型

Real-world processes such as population growth, radioactive decay and compound interest are modelled using exponential functions of the form N = N₀ e^(kt) or N = N₀ aˣ.

人口增长、放射性衰变和复利等现实过程通常用 N = N₀ e^(kt) 或 N = N₀ aˣ 这样的指数函数来建模。

The sign of k determines behaviour: if k > 0 the model represents growth; if k < 0 it represents decay. The constant N₀ is the initial value at t = 0.

k 的符号决定模型行为：k > 0 表示增长；k < 0 表示衰减。常数 N₀ 是 t = 0 时的初始值。

To find k from given data, substitute known (t, N) pairs and solve the resulting equation using natural logarithms. For instance, if N = 200 e^(-0.1t), you can find the half-life by setting N = 100 and solving for t.

要从给定数据求 k，可代入已知 (t, N) 数据对，并使用自然对数求解方程。例如若 N = 200 e^(-0.1t)，设 N = 100 求解 t 即得半衰期。

CIE questions often require you to convert between forms, such as rewriting N = a bᵗ as N = a e^(kt) by using b = eᵏ ⇒ k = ln b.

CIE 试题常要求在不同形式间进行转换，例如利用 b = eᵏ ⇒ k = ln b 将 N = a bᵗ 改写为 N = a e^(kt)。

Linearising an exponential model is another favourite: taking ln of both sides yields ln N = ln N₀ + k t, which is a straight line relationship between ln N and t.

将指数模型线性化是另一考点：两边取 ln 得 ln N = ln N₀ + k t，表明 ln N 与 t 成线性关系。

Mastering these modelling steps will help you tackle the structured ‘real-world problem’ questions that appear at the end of CIE papers with confidence.

掌握这些建模步骤，你就能自信地应对 CIE

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A-Level CIE Mathematics: Exponentials and Logarithms – Key Points Revision | A-Level CIE 数学：指数与对数 考点精讲