Mastering Index Laws Revision Notes | GCSE AQA Higher Further Maths

1. Why Index Laws Matter

Indices appear throughout GCSE algebra — in simplifying expressions, solving equations, and working with surds and standard form. Mastering the rules here means you can handle all of those topics with confidence.

Think of index notation as a shorthand for repeated multiplication. Writing $2^5$ is far neater than writing $2 \times 2 \times 2 \times 2 \times 2$, and the laws of indices let you manipulate that shorthand algebraically without expanding everything out.

2. The Five Core Index Laws

There are five rules you need to know. Each one follows from the definition of an index as repeated multiplication — so rather than memorising them blindly, notice how each rule makes sense when you expand a small example.

Before applying these to harder problems, check the multiplication and division laws make sense. Consider $x^3 \times x^4$: expanding gives $(x \times x \times x) \times (x \times x \times x \times x) = x^7$, which matches adding the indices. The same logic underpins every law in the table above.

3. Applying the Laws: Simplifying Expressions

The multiplication, division, and power laws are your main tools for simplifying algebraic expressions. The key habit is to check whether the bases match before applying any rule — you cannot add indices across different bases.

Worked example: Simplifying Using the Multiplication and Power Laws

Simplify $\dfrac{(x^3)^4 \times x^2}{x^5}$

1. $(x^3)^4 = x^{3 \times 4} = x^{12}$ — Power law: multiply the indices. [M1]
2. $x^{12} \times x^2 = x^{12+2} = x^{14}$ — Same base, multiplying: add the indices. [M1]
3. $x^{14} \div x^5 = x^{14-5} = x^9$ — Same base, dividing: subtract the indices. [A1]

Answer: $x^9$

Notice the order of operations: deal with brackets (power law) first, then multiplication, then division. This mirrors standard BIDMAS and keeps the working clean.

4. Negative and Zero Indices

The zero index result ($a^0 = 1$) is not a separate rule to memorise in isolation — it falls straight out of the division law. If you divide $a^3$ by $a^3$, you get 1 (any number divided by itself), and the division law gives $a^{3-3} = a^0$. So $a^0 = 1$.

Negative indices work the same way. Dividing $a^2$ by $a^5$ gives $a^{2-5} = a^{-3}$, and expanding the division as a fraction gives $\frac{a^2}{a^5} = \frac{1}{a^3}$. That confirms $a^{-3} = \frac{1}{a^3}$.

Worked example: Evaluating a Negative Index

Evaluate $4^{-2}$

1. $4^{-2} = \dfrac{1}{4^2}$ — A negative index means take the reciprocal of the positive-index form. [M1]
2. $4^2 = 4 \times 4 = 16$ — Expand the square explicitly to avoid errors.
3. $4^{-2} = \dfrac{1}{16}$ — The answer is a fraction less than 1. [A1]

Answer: $\dfrac{1}{16}$

5. Fractional Indices: Roots and Powers Combined

Fractional indices are where many students lose marks, but the rule is logical once you see where it comes from. Consider $a^{\frac{1}{2}} \times a^{\frac{1}{2}} = a^{\frac{1}{2}+\frac{1}{2}} = a^1 = a$. So $a^{\frac{1}{2}}$ is a number that, when multiplied by itself, gives $a$ — that is exactly the square root. Hence $a^{\frac{1}{2}} = \sqrt{a}$.

Extending this: $a^{\frac{1}{n}} = \sqrt[n]{a}$ (the $n$th root), and $a^{\frac{m}{n}} = \left(\sqrt[n]{a}\right)^m$. The denominator controls the root; the numerator controls the power. A reliable method is to take the root first, then apply the power — this keeps the numbers smaller and easier to handle.

Worked example: Evaluating a Fractional Index

Evaluate $27^{\frac{2}{3}}$

1. $27^{\frac{2}{3}}$: denominator is 3 (cube root), numerator is 2 (square) — Denominator = root, numerator = power. [M1]
2. $\sqrt[3]{27} = 3$ because $3 \times 3 \times 3 = 27$ — Root first keeps numbers manageable.
3. $3^2 = 3 \times 3 = 9$ — Square the result of the root. [A1]

Answer: $27^{\frac{2}{3}} = 9$

Worked example: Combining a Negative and Fractional Index

Evaluate $16^{-\frac{3}{4}}$

1. $16^{-\frac{3}{4}} = \dfrac{1}{16^{\frac{3}{4}}}$ — Negative index means reciprocal. Deal with this first. [M1]
2. $\sqrt[4]{16} = 2$ because $2 \times 2 \times 2 \times 2 = 16$ — Denominator 4 means fourth root.
3. $2^3 = 2 \times 2 \times 2 = 8$ — Numerator 3 means cube the result.
4. $16^{-\frac{3}{4}} = \dfrac{1}{8}$ — Return to the reciprocal from Step 1. [A1]

Answer: $16^{-\frac{3}{4}} = \dfrac{1}{8}$

6. Solving Equations Using Index Laws

Index laws also appear in equations where you need to find an unknown. Two common types come up at GCSE: equations where the unknown is the base (e.g. $x^3 = 125$) and equations where you must rewrite both sides with the same base to compare indices (e.g. $2^x = 32$).

Worked example: Solving an Equation with an Unknown Base

Solve $x^{\frac{3}{2}} = 8$

1. $\left(x^{\frac{3}{2}}\right)^{\frac{2}{3}} = 8^{\frac{2}{3}}$ — The reciprocal of $\frac{3}{2}$ is $\frac{2}{3}$. Raising to the reciprocal isolates $x$. [M1]
2. $x^{\frac{3}{2} \times \frac{2}{3}} = x^1 = x$ — Power law: multiply indices. $\frac{3}{2} \times \frac{2}{3} = 1$.
3. $8^{\frac{2}{3}} = \left(\sqrt[3]{8}\right)^2 = 2^2 = 4$ — $\sqrt[3]{8} = 2$ because $2 \times 2 \times 2 = 8$; then $2^2 = 2 \times 2 = 4$. [A1]

Answer: $x = 4$

Worked example: Solving an Equation by Matching Bases

Solve $3^{2x-1} = 27$

1. $27 = 3^3$ — Express the right side with the same base as the left. [M1]
2. $2x - 1 = 3$ — If $3^{2x-1} = 3^3$, the indices must be equal. [M1]
3. $2x = 4 \Rightarrow x = 2$ — Add 1 to both sides, then divide by 2. [A1]

Answer: $x = 2$

7. Exam Technique and Common Mistakes

The most reliable way to pick up every mark on an indices question is to work in a clear sequence: identify the law needed, apply it to one pair of terms at a time, and write each step on its own line. Examiners award method marks for correct intermediate steps even if the final answer contains an arithmetic slip.

For fractional index questions, always state the root and the power separately before combining them. Writing $27^{\frac{2}{3}} = (\sqrt[3]{27})^2 = 3^2 = 9$ earns full marks and shows clear method. Jumping straight to a number without working risks losing method marks if the answer is wrong.