📚 Index Laws: Complete A-Level Maths Guide | 指数法则：A-Level 数学完整指南
Index laws, also called the laws of exponents, are the algebraic rules that control how powers behave under multiplication, division, raising to a power and taking roots. They are the foundation of almost every later topic in A-Level Mathematics: surds, logarithms, differentiation, sequences, binomial expansion and exponential modelling all assume that you can handle indices fluently. This guide covers every rule you need for Edexcel, AQA, OCR and Cambridge A-Level Maths, with worked examples, exam-style questions and the mistakes that examiners see most often.
指数法则（也叫幂运算法则）是支配幂在乘法、除法、乘方和开方运算中如何变化的代数规则。它是 A-Level 数学后续几乎所有专题的基础：根式、对数、求导、数列、二项式展开以及指数模型，都默认你已能熟练处理指数。本文覆盖 Edexcel、AQA、OCR 与剑桥 A-Level 数学所需的全部法则，配有例题精讲、考试风格题目以及考官最常看到的错误。
e are multiplied together. So a⁵ means a × a × a × a × a, and x³y⁴ means x × x × x × y × y × y × y.
像 aⁿ 这样的幂由两部分组成。底数 a 是被相乘的数或表达式，指数 n 表示底数自乘的次数。因此 a⁵ 表示 a × a × a × a × a，而 x³y⁴ 表示 x × x × x × y × y × y × y。
These laws are valid for real exponents, but in the early A-Level chapters the focus is on integer and rational exponents, which is exactly where the exam marks sit. The table below is the summary you should be able to reproduce from memory.
这些法则对实数指数都成立，但在 A-Level 初期章节中，重点是指数为整数和有理数的情况，而这正是考试得分所在。下表是要求你能凭记忆写出的总结。
| Law | 法则 | Formula | 公式 |
| Multiplication | 同底相乘 | aᵐ × aⁿ = aᵐ⁺ⁿ |
| Division | 同底相除 | aᵐ ÷ aⁿ = aᵐ⁻ⁿ |
| Power of a power | 幂的乘方 | (aᵐ)ⁿ = aᵐⁿ |
| Power of a product | 积的乘方 | (ab)ⁿ = aⁿbⁿ |
| Power of a quotient | 商的乘方 | (a ÷ b)ⁿ = aⁿ ÷ bⁿ, b ≠ 0 |
| Zero index | 零指数 | a⁰ = 1, a ≠ 0 |
| Negative index | 负指数 | a⁻ⁿ = 1 ÷ aⁿ, a ≠ 0 |
| Fractional index | 分数指数 | a1/n = ⁿ√a, a ≥ 0 for even n |
1. What Are Index Laws? | 什么是指数法则？
Index laws are the algebraic rules that govern how powers behave under multiplication, division, raising to another power and taking roots. They exist because exponents are a shorthand for repeated multiplication, and every rule can be proved by simply counting factors. Once you understand the counting argument, you never need to memorise blindly again.
指数法则是支配幂在乘法、除法、乘方和开方运算中如何变化的代数规则。它们之所以成立，是因为指数本身就是重复乘法的简写，每一条法则都可以通过"数因子"来证明。一旦理解了数因子的思路，就不需要死记硬背。
t copies of x, so the answer is x⁸. The indices 3 and 5 were added because the factors were simply collected together.
例如，把 x³ × x⁵ 完全展开是 (x × x × x) × (x × x × x × x × x)，也就是 8 个 x 相乘，所以答案是 x⁸。指数 3 和 5 相加，是因为因子被简单地合并在一起。
must also know the domain restrictions: a⁰ and a⁻ⁿ require a ≠ 0, and even roots such as a1/2 or a1/4 require a ≥ 0 when working with real numbers.
这些法则对实数指数都成立，但在 A-Level 初期章节中，重点是指数为整数和有理数的情况。你还必须掌握定义域限制：a⁰ 与 a⁻ⁿ 要求 a ≠ 0；而在实数范围内，偶次根如 a1/2 或 a1/4 要求 a ≥ 0。
- Integer indices: 2³, x⁵, a⁻⁴ | 整数指数：2³、x⁵、a⁻⁴
- Rational indices: 91/2 = 3, 82/3 = 4 | 有理指数：91/2 = 3、82/3 = 4
- Algebraic bases: (2x)⁴, (3x²y)³ | 代数底数：(2x)⁴、(3x²y)³
2. Multiplying Powers with the Same Base | 同底数幂相乘
When multiplying two powers with the same base, keep the base and add the exponents. This is the single most used index law in A-Level algebra, and it appears in nearly every simplification question.
当两个同底数的幂相乘时，底数不变，指数相加。这是 A-Level 代数中使用频率最高的指数法则，几乎出现在每一道化简题中。
aᵐ × aⁿ = aᵐ⁺ⁿ
and 2³ × 2⁴ = 2⁷ = 128. The rule also works with more than two factors: x² × x³ × x⁴ = x⁹.
例如，x³ × x⁵ = x⁸，以及 2³ × 2⁴ = 2⁷ = 128。该法则对多于两个因子的情况同样成立：x² × x³ × x⁴ = x⁹。
rst, then apply the index law to the letters: 3x² × 4x⁵ = 12x⁷, and 2x³y² × 5x⁴y = 10x⁷y³.
考试题中底数常常带有系数。先把数字相乘，再对字母使用指数法则：3x² × 4x⁵ = 12x⁷，以及 2x³y² × 5x⁴y = 10x⁷y³。
x² × 2³ either, because one base is x and the other is 2. Leave such products as they are.
底数必须完全相同。你不能把 x² × y³ 合并成单个幂，因为底数不同；你也不能合并 x² × 2³，因为一个底数是 x，另一个是 2。这类乘积只能保持原样。
- x⁴ × x⁶ = x¹⁰ | x⁴ × x⁶ = x¹⁰
- 5a² × 3a⁷ = 15a⁹ | 5a² × 3a⁷ = 15a⁹
- x²y³ × x⁵y⁴ = x⁷y⁷ | x²y³ × x⁵y⁴ = x⁷y⁷
- x² × y³ stays as x²y³ | x² × y³ 保持为 x²y³
3. Dividing Powers with the Same Base and the Zero Index | 同底数幂相除与零指数
When dividing powers with the same base, subtract the index of the denominator from the index of the numerator. This is the mirror image of the multiplication rule.
当同底数的幂相除时，用分子的指数减去分母的指数。这是乘法法则的镜像版本。
aᵐ ÷ aⁿ = aᵐ⁻ⁿ
divide the numbers and subtract the indices: 12x⁹ ÷ 4x³ = 3x⁶, and 15x⁵y⁴ ÷ 5x²y = 3x³y³.
例如，x⁷ ÷ x² = x⁵，以及 5⁶ ÷ 5⁴ = 5² = 25。带系数时，先除数字再减指数：12x⁹ ÷ 4x³ = 3x⁶，以及 15x⁵y⁴ ÷ 5x²y = 3x³y³。
zero quantity divided by itself equals 1, so a⁰ must equal 1. This is the standard proof expected in a "show that" question.
这条法则也解释了为什么 a⁰ = 1。若 m = n，则 aᵐ ÷ aᵐ = aᵐ⁻ᵐ = a⁰。但任何非零量除以自身都等于 1，所以 a⁰ 必须等于 1。这是"证明题"中要求写出的标准证明。
index: x³ ÷ x⁷ = x³⁻⁷ = x⁻⁴, which equals 1 ÷ x⁴. Do not stop at x⁻⁴ if the question says "write with positive indices".
一个常见的陷阱是分母指数更大的情况，此时结果为负指数：x³ ÷ x⁷ = x³⁻⁷ = x⁻⁴，即 1 ÷ x⁴。如果题目要求"用正指数表示"，就不要停在 x⁻⁴。
- x⁹ ÷ x⁴ = x⁵ | x⁹ ÷ x⁴ = x⁵
- 8x⁶ ÷ 2x² = 4x⁴ | 8x⁶ ÷ 2x² = 4x⁴
- a⁵ ÷ a⁵ = a⁰ = 1 | a⁵ ÷ a⁵ = a⁰ = 1
- x³ ÷ x⁷ = x⁻⁴ = 1 ÷ x⁴ | x³ ÷ x⁷ = x⁻⁴ = 1 ÷ x⁴
4. Power of a Power | 幂的乘方
When a power is raised to another power, multiply the exponents together. The brackets mean that the whole expression aᵐ is used n times as a factor.
当一个幂再被乘方时，指数相乘。括号意味着整个表达式 aᵐ 被当作因子使用了 n 次。
(aᵐ)ⁿ = aᵐⁿ
64. Checking the second example by expansion: 2² = 4 and 4³ = 64, which confirms the rule.
例如，(x³)⁴ = x¹²，以及 (2²)³ = 2⁶ = 64。用展开验证第二例：2² = 4，4³ = 64，与法则一致。
er power as well as multiplying the indices: (2x³)⁴ = 2⁴x¹² = 16x¹², and (3x²y)³ = 27x⁶y³.
若括号内有系数，除了指数相乘，还要把系数取外层幂：(2x³)⁴ = 2⁴x¹² = 16x¹²，以及 (3x²y)³ = 27x⁶y³。
dices; in (aᵐ)ⁿ you multiply them. Writing (x³)⁴ as x⁷ instead of x¹² is one of the most frequent lost marks in A-Level papers.
千万不要混淆两个最常见的指数运算。在 aᵐ × aⁿ 中指数相加；在 (aᵐ)ⁿ 中指数相乘。把 (x³)⁴ 写成 x⁷ 而不是 x¹²，是 A-Level 试卷中最常见的失分之一。
- (x⁵)³ = x¹⁵ | (x⁵)³ = x¹⁵
- (a²)⁶ = a¹² | (a²)⁶ = a¹²
- (2x²)⁵ = 32x¹⁰ | (2x²)⁵ = 32x¹⁰
- (xᵐ)² = x²ᵐ | (xᵐ)² = x²ᵐ
5. Power of a Product and a Quotient | 积的乘方与商的乘方
When a product is raised to a power, every factor inside receives that power. When a quotient is raised to a power, both the numerator and the denominator receive that power.
当一个乘积被乘方时，括号内每个因子都要取该次幂。当一个商被乘方时，分子与分母都要取该次幂。
(ab)ⁿ = aⁿbⁿ and (a ÷ b)ⁿ = aⁿ ÷ bⁿ
³ = 8x³, and (3x²y)² = 9x⁴y². For fractions, (x ÷ 3)⁴ = x⁴ ÷ 81, and (2x ÷ y)³ = 8x³ ÷ y³.
例如，(2x)³ = 2³x³ = 8x³，以及 (3x²y)² = 9x⁴y²。对于分数，(x ÷ 3)⁴ = x⁴ ÷ 81，以及 (2x ÷ y)³ = 8x³ ÷ y³。
This law is what makes the negative index version of a fraction work so neatly. Raising a fraction to a negative power flips the fraction and makes the power positive.
正是这条法则让分数的负指数形式如此简洁。把分数取负指数，等于把分数上下翻转并让指数变正。
(a ÷ b)⁻ⁿ = (b ÷ a)ⁿ
2.25. Students who remember this one line save a great deal of time in algebra questions.
例如，(2 ÷ 3)⁻² = (3 ÷ 2)² = 9 ÷ 4 = 2.25。记住这一行的学生在代数题中能省下大量时间。
- (4x)² = 16x² | (4x)² = 16x²
- (5xy²)³ = 125x³y⁶ | (5xy²)³ = 125x³y⁶
- (x ÷ 2)⁵ = x⁵ ÷ 32 | (x ÷ 2)⁵ = x⁵ ÷ 32
- (3 ÷ x)⁻² = x² ÷ 9 | (3 ÷ x)⁻² = x² ÷ 9
6. Negative Indices | 负指数
A negative index does not make the answer negative. It signals a reciprocal: the power moves from the numerator to the denominator, or the other way round.
负指数并不会让结果变成负数。它表示取倒数：幂从分子移到分母，或从分母移到分子。
a⁻ⁿ = 1 ÷ aⁿ and 1 ÷ a⁻ⁿ = aⁿ
⁵ = 1 ÷ x⁵. With coefficients, the coefficient keeps its sign: 3x⁻² = 3 ÷ x², not 1 ÷ 3x².
例如，2⁻³ = 1 ÷ 8 = 0.125，以及 x⁻⁵ = 1 ÷ x⁵。带系数时，系数保持原样：3x⁻² = 3 ÷ x²，而不是 1 ÷ (3x²)。
or a product. (2 ÷ 5)⁻³ = (5 ÷ 2)³ = 125 ÷ 8 = 15.625, and (2x)⁻² = 1 ÷ (2x)² = 1 ÷ (4x²).
当底数是分数或乘积时，多条法则同时起作用。(2 ÷ 5)⁻³ = (5 ÷ 2)³ = 125 ÷ 8 = 15.625，以及 (2x)⁻² = 1 ÷ (2x)² = 1 ÷ 4x²。
Exam questions frequently ask you to "write without negative indices" or "express as a single fraction". In both cases the goal is the same: move every negative power to the opposite side of the fraction bar and leave the indices positive.
考试题经常要求"用正指数表示"或"化为单个分数"。这两种情况的思路一致：把每个负指数移到分数线的另一侧，使指数全部为正。
- 5⁻² = 1 ÷ 25 | 5⁻² = 1 ÷ 25
- x⁻¹ = 1 ÷ x | x⁻¹ = 1 ÷ x
- (2 ÷ 3)⁻¹ = 3 ÷ 2 | (2 ÷ 3)⁻¹ = 3 ÷ 2
- (4x²)⁻¹ = 1 ÷ (4x²) | (4x²)⁻¹ = 1 ÷ 4x²
7. Fractional Indices and Roots | 分数指数与根式
A fractional index is another way of writing a root. The denominator of the fraction tells you which root to take, and the numerator tells you the power to apply afterwards.
分数指数是根式的另一种写法。分母告诉你开几次方，分子告诉你要取的幂次。
a1/n = ⁿ√a and am/n = (ⁿ√a)ᵐ = ⁿ√aᵐ
For example, 91/2 = √9 = 3, 81/3 = ∛8 = 2, and 161/4 = ∜16 = 2. With a numerator, 82/3 = (∛8)² = 2² = 4, and 163/4 = (∜16)³ = 2³ = 8.
例如，91/2 = √9 = 3，81/3 = ∛8 = 2，以及 161/4 = ∜16 = 2。带分子时，82/3 = (∛8)² = 2² = 4，以及 163/4 = (∜16)³ = 2³ = 8。
Always take the root first and the power second. For 82/3, doing 8² = 64 and then ∛ = 4 gives the same answer, but with larger numbers it is far easier to root first. For 274/3, root first: ∛27 = 3, then 3⁴ = 81.
务必先开方、再取幂。对 82/3，先算 8² = 64 再算 ∛64 = 4 结果相同，但数字越大，先开方越容易。对 274/3，先开方：∛27 = 3，再算 3⁴ = 81。
l and negative indices combine into the hardest-looking but simplest form. 27⁻^(1/3) = 1 ÷ ∛27 =
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