Algebra

The index laws with worked examples, including negative and fractional indices

By Mr Gan · O-Level E Maths · Updated August 2026 · 9 min read

Four rules cover almost every index question: multiply same-base terms by adding the indices (aᵐ × aⁿ = aᵐ⁺ⁿ), divide by subtracting them (aᵐ ÷ aⁿ = aᵐ⁻ⁿ), raise a power to a power by multiplying the indices ((aᵐ)ⁿ = aᵐⁿ), and rewrite negative or fractional indices as reciprocals and roots (a⁻ⁿ = 1/aⁿ, a^(1/n) = ⁿ√a) before evaluating. These rules only combine directly when the base is the same on both sides.

Why index questions cost easy marks

The rules themselves are short, but students lose marks in two specific ways. The first is adding indices when the bases are different: 2³ × 3² is not 6⁵, because the rule aᵐ × aⁿ = aᵐ⁺ⁿ only works when the base a is identical on both sides. The second is panicking at a negative index and writing a negative answer: a⁻² means 1/a², a positive fraction, not a negative number.

Both mistakes come from applying a rule without checking it actually applies. The method below fixes that by giving you a fixed order to work through: check the bases, then apply the matching rule, then clean up any negative or fractional index at the end.

The method: the index laws in one place

Product law

aᵐ × aⁿ = aᵐ⁺ⁿ

Quotient law

aᵐ ÷ aⁿ = aᵐ⁻ⁿ

Power law

(aᵐ)ⁿ = aᵐⁿ

Zero index

a⁰ = 1 (a ≠ 0)

Negative index

a⁻ⁿ = 1/aⁿ

Fractional index

a^(1/n) = ⁿ√a, a^(m/n) = ⁿ√(aᵐ)

Every one of these rules needs the same base a throughout. If a question mixes bases, for example and , rewrite one base in terms of the other first (4 = 2², so 4ˣ = 2²ˣ) before applying any law, the same rewriting skill used in HCF and LCM by prime factorisation.

Watch the whole method in about a minute.

Worked example 1, simplify and evaluate

Simplify (3a²b)³ × 2a⁴b² ÷ 6a⁵b³, then find its value when a = 2 and b = 3.

Solution

1
Apply the power law to the bracket first. (3a²b)³ means every factor inside is cubed: (3a²b)³ = 3³ × a²ˣ³ × b¹ˣ³ = 27a⁶b³.
2
Multiply by 2a⁴b² using the product law (add the indices of matching bases):
27a⁶b³ × 2a⁴b² = (27 × 2) × a⁶⁺⁴ × b³⁺² = 54a¹⁰b⁵.
3
Divide by 6a⁵b³ using the quotient law (subtract the indices of matching bases):
54a¹⁰b⁵ ÷ 6a⁵b³ = (54 ÷ 6) × a¹⁰⁻⁵ × b⁵⁻³ = 9a⁵b².
4
Substitute a = 2, b = 3: a⁵ = 32 and b² = 9, so 9a⁵b² = 9 × 32 × 9.
5
Work it out: 9 × 32 = 288, then 288 × 9 = 2592. The simplified expression is 9a⁵b², and its value is 2592.

Quick check: tackle brackets first (power law), then products (product law), then quotients (quotient law), in that order, the same way you handle BODMAS. Doing the steps out of order is where most simplification slips creep in.

Worked example 2, negative and fractional indices

Evaluate 27^(2/3) + 16^(-1/2) − 8⁰ without a calculator.

Solution

1
For 27^(2/3), the denominator 3 means "cube root" and the numerator 2 means "square" (either order works, cube root first keeps the numbers small): 27^(1/3) = 3, since 3 × 3 × 3 = 27. Then 27^(2/3) = 3² = 9.
2
For 16^(-1/2), deal with the negative sign first by flipping to a reciprocal: 16^(-1/2) = 1 ÷ 16^(1/2). Since 16^(1/2) = √16 = 4, this gives 16^(-1/2) = 1/4.
3
For 8⁰, any non-zero base raised to the power 0 equals 1: 8⁰ = 1.
4
Combine: 9 + 1/4 − 1 = 8 + 1/4 = 8¼.

The step students get wrong

Reading a⁻ⁿ as "make the answer negative" instead of "take the reciprocal." 16^(-1/2) is a small positive fraction (1/4), not -4. The negative sign in an index never touches the sign of the final value: it only tells you to flip the base to a reciprocal before you evaluate. Sort the negative sign out first, then handle any fractional part of the index separately.

Fractional indices: which part is the root and which is the power

For a^(m/n), the denominator n is the root and the numerator m is the power: a^(m/n) = (ⁿ√a)ᵐ, which is the same value as ⁿ√(aᵐ). Taking the root first, then the power, almost always gives smaller, easier numbers, as in worked example 2 above (cube-rooting 27 to 3, then squaring, is far easier than cubing 27² first). Powers of a fixed base like this also form a pattern worth recognising, the same skill tested in special sequences questions.


Frequently asked questions

Can I add the indices if the bases are different, like 2³ × 3²?

No. The product and quotient laws only apply when the base is exactly the same on both sides. 2³ × 3² must be worked out separately: 2³ = 8 and 3² = 9, so 2³ × 3² = 8 × 9 = 72, not 6⁵.

Why does a⁰ equal 1 and not 0?

Follow the quotient law: aⁿ ÷ aⁿ = aⁿ⁻ⁿ = a⁰. But any non-zero number divided by itself is 1, so a⁰ must equal 1 to keep the rule consistent. This holds for every non-zero base, including large or fractional ones.

Is a negative index the same as a negative number?

No, and this is the single most common index mistake. A negative index means "take the reciprocal," not "make it negative." 2⁻³ = 1/2³ = 1/8, which is a small positive fraction.

How do index laws connect to standard form?

Standard form writes numbers as A × 10ⁿ, and multiplying or dividing two numbers in standard form uses exactly the same product and quotient laws covered here, applied to the power of 10. See our standard form guide for the worked examples.

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