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Year 11 Methods (Unit 1 & 2) Exponential Functions And Logarithms

The Index Laws

20 practice questions 1 video lesson Theory + worked examples

Master the index laws for Queensland Year 11 Mathematical Methods (QCAA). Indices, or powers, are a shorthand for repeated multiplication, and the index laws are the rules that let you multiply, divide and raise powers.

You will learn to apply the product, quotient and power laws, simplify powers of products and quotients, and handle the zero and negative index rules — algebra that prepares you for surds, exponentials and logarithms.

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Theory

In Year 11 Mathematical Methods (QCAA, Unit 2), the index laws are the rules for multiplying, dividing and raising powers with the same base. This page covers the product, quotient and power laws, powers of products and quotients, and the zero and negative index rules, with fully worked simplifications.

A power such as \(a^{n}\) is written with a base \(a\) and an index (or exponent) \(n\); it means \(a\) multiplied by itself \(n\) times. The index laws are shortcuts that follow directly from this meaning.

The product law adds indices when powers of the same base are multiplied, and the quotient law subtracts them when they are divided. The power law multiplies indices when a power is raised to another power. A base to the zero index equals \(1\), and a negative index means a reciprocal, \(a^{-n}=\dfrac{1}{a^{n}}\).

Same base is the key. The product and quotient laws only apply when the bases match. Deal with the numbers, then each variable, one law at a time.
Product law shown as repeated factors Three factors of a times two factors of a give five factors of a, showing a cubed times a squared equals a to the fifth. a·a·a × a·a = a·a·a·a·a a⁵ add the indices: 3 + 2 = 5
Product law: \(a^{3}\times a^{2}=a^{5}\) — multiplying powers adds the indices.
Powers of two decreasing to zero and negative indices Each step down divides by two: two cubed is eight, two squared is four, two to the one is two, two to the zero is one, two to the minus one is one half. 2³ = 8 2² = 4 2¹ = 2 2⁰ = 1 2⁻¹ = ½ ÷ 2 ÷ 2 ÷ 2 ÷ 2 dividing past 2¹ forces 2⁰=1 and negative indices
Dividing by the base steps the index down, giving \(a^{0}=1\) and \(a^{-n}=\dfrac{1}{a^{n}}\).

For any base \(a\neq 0\) and integer indices \(m,\,n\):

\[a^{m}\times a^{n}=a^{m+n}\qquad \dfrac{a^{m}}{a^{n}}=a^{m-n}\qquad (a^{m})^{n}=a^{mn}\]
am×an=am+n
\[(ab)^{n}=a^{n}b^{n}\qquad \left(\dfrac{a}{b}\right)^{n}=\dfrac{a^{n}}{b^{n}}\]
(ab)n=anbn
\[a^{0}=1\qquad a^{-n}=\dfrac{1}{a^{n}}\]
a-n=1an
Positive-index form: a tidy answer usually shows every index as a positive whole number, so move any negative-index factor across the fraction bar.

How to simplify an index expression

  1. Numbers first: evaluate or simplify the numerical coefficients on their own.
  2. Group by base: collect the powers of each variable and apply one law at a time — add indices for a product, subtract for a quotient, multiply for a power of a power.
  3. Tidy up: write \(a^{0}=1\) and rewrite any negative index as a reciprocal so every index is a positive whole number.
Example 1 — Product and quotient laws
Simplify \(a^{5}\times a^{3}\div a^{2}\).
Solution

Product law — multiply, so add the indices:

\(a^{5}\times a^{3}\)\(=\)\(a^{5+3}\)
\(=\)\(a^{8}\)

Quotient law — divide, so subtract the index:

\(a^{8}\div a^{2}\)\(=\)\(a^{8-2}\)
\(=\)\(a^{6}\)

\(a^{5}\times a^{3}\div a^{2}=a^{6}\).

a6
Example 2 — Power of a product
Simplify \((2x^{3})^{4}\).
Solution

Raise each factor to the power \(4\):

\((2x^{3})^{4}\)\(=\)\(2^{4}\times (x^{3})^{4}\)

Power law on \(x\) — multiply the indices:

\(=\)\(2^{4}\times x^{3\times 4}\)
\(=\)\(16\,x^{12}\)

\((2x^{3})^{4}=16x^{12}\).

16x12
Example 3 — Negative and zero indices
Simplify \(\dfrac{12a^{5}b^{-3}}{4a^{2}b}\), giving your answer with positive indices.
Solution

Numbers first:

\(\dfrac{12}{4}\)\(=\)\(3\)

Quotient law on each variable — subtract the indices:

\(a\text{-part}\)\(=\)\(a^{5-2}=a^{3}\)
\(b\text{-part}\)\(=\)\(b^{-3-1}=b^{-4}\)

Rewrite the negative index as a reciprocal:

\(=\)\(3a^{3}b^{-4}\)
\(=\)\(\dfrac{3a^{3}}{b^{4}}\)

\(\dfrac{12a^{5}b^{-3}}{4a^{2}b}=\dfrac{3a^{3}}{b^{4}}\).

3a3b4
Example 4 — Combining several laws
Simplify \(\dfrac{(2x^{2}y)^{3}\times x^{-4}}{4xy^{2}}\).
Solution

Expand the bracket with the power law:

\((2x^{2}y)^{3}\)\(=\)\(2^{3}x^{6}y^{3}\)
\(=\)\(8x^{6}y^{3}\)

Multiply the numerator (product law on \(x\)):

\(8x^{6}y^{3}\times x^{-4}\)\(=\)\(8x^{6+(-4)}y^{3}\)
\(=\)\(8x^{2}y^{3}\)

Divide by \(4xy^{2}\) (quotient law, and numbers):

\(\dfrac{8x^{2}y^{3}}{4xy^{2}}\)\(=\)\(2x^{2-1}y^{3-2}\)
\(=\)\(2xy\)

\(\dfrac{(2x^{2}y)^{3}\times x^{-4}}{4xy^{2}}=2xy\).

2xy

Common pitfalls

Multiplying the bases. \(a^{5}\times a^{3}=a^{8}\), not \(a^{15}\). You add the indices; the base stays the same.
Forgetting the coefficient in a bracket. \((2x^{3})^{4}=2^{4}x^{12}=16x^{12}\); the \(2\) is raised to the power too, not just the \(x\).
Thinking a zero index gives zero. Any non-zero base to the power \(0\) is \(1\), so \(5x^{0}=5\), not \(0\).
Sign slips with negative indices. \(a^{-3-1}=a^{-4}=\dfrac{1}{a^{4}}\); a negative index means a reciprocal, not a negative number.

Frequently asked questions

What are the index laws?

They are the rules for powers with the same base: \(a^{m}\times a^{n}=a^{m+n}\), \(\dfrac{a^{m}}{a^{n}}=a^{m-n}\), \((a^{m})^{n}=a^{mn}\), plus \(a^{0}=1\) and \(a^{-n}=\dfrac{1}{a^{n}}\).

When do you add and when do you multiply the indices?

Add the indices when you multiply powers of the same base; multiply the indices when a power is itself raised to a power, as in \((a^{m})^{n}=a^{mn}\).

What does a negative index mean?

A negative index means a reciprocal: \(a^{-n}=\dfrac{1}{a^{n}}\). For example \(2^{-3}=\dfrac{1}{8}\).

Why does anything to the power zero equal one?

Using the quotient law, \(a^{0}=\dfrac{a^{n}}{a^{n}}=1\) for any non-zero base \(a\).

Do the index laws work if the bases are different?

The product and quotient laws only apply when the bases are equal. With different bases, such as \(2^{3}\times 3^{2}\), you evaluate each power separately.