Laws of Exponents - Standard Maths

The three index laws

An exponent or index (plural indices) tells you how many times to multiply the base by itself, so \(2^4 = 2\times2\times2\times2 = 16\). Three laws let you combine powers without expanding them out:

\[ a^m \times a^n = a^{m+n} \qquad \frac{a^m}{a^n} = a^{m-n} \qquad (a^m)^n = a^{mn} \]

In words: multiplying powers of the same base, you add the indices; dividing, you subtract; a power of a power, you multiply. Every one of them relies on the bases being the same, the laws say nothing about \(2^3 \times 5^2\).

Worked example

Simplify \(a^5 \times a^2 \div a^3\).

Multiplying the first two adds the indices: \(a^5 \times a^2 = a^{5+2} = a^7\).

Dividing by \(a^3\) subtracts: \(a^{7} \div a^{3} = a^{7-3}\).

\(a^4\)

Zero and negative indices

Two special cases fall straight out of the division law. If you divide \(a^m\) by itself you get 1, but the law says you get \(a^{m-m}=a^0\). So:

\[ a^0 = 1 \qquad\qquad a^{-n} = \frac{1}{a^{n}} \]

Anything (except 0) to the power zero is 1. A negative index means "one over" the positive power: \(2^{-3} = \dfrac{1}{2^3} = \dfrac{1}{8}\). The minus sign does not make the answer negative; it flips the term into a fraction.

A negative power is not a negative answer

\(5^{-2}\) is \(\tfrac{1}{25}\), not \(-25\) and not \(-10\). The negative index sends the term to the denominator, it never touches the sign of the value.

Using several laws at once

Most questions ask you to combine the laws. Deal with any brackets first (power of a power), then multiply and divide by adding and subtracting the indices.

Worked example

Simplify \((2x^3)^2 \times x^{-4}\).

Clear the bracket first, the power hits both the 2 and the \(x^3\): \((2x^3)^2 = 2^2 x^{6} = 4x^6\).

Now multiply by \(x^{-4}\), adding the indices: \(4x^{6} \times x^{-4} = 4x^{6+(-4)}\).

\(4x^{2}\)

Where this is assessed

The index laws are pure Criterion A (knowing which rule applies) and Criterion C (writing the powers correctly). When a bracket has a number and a letter inside, the outer power must reach the number too: \((2x^3)^2\) is \(4x^6\), not \(2x^6\).

Check yourself

1. Simplify \(y^6 \times y^2\). +

Multiplying the same base adds the indices: \(y^{6+2} = \mathbf{y^{8}}\).

2. Evaluate \(5^0 + 2^{-2}\). +

\(5^0 = 1\) and \(2^{-2} = \frac{1}{2^2} = \frac{1}{4}\). Adding: \(1 + \frac{1}{4} = \mathbf{\frac{5}{4}}\) (or \(1.25\)).

3. Simplify \((2x^3)^2\). +

Square both parts: \(2^2 (x^3)^2 = \mathbf{4x^{6}}\).

Part of the Standard Maths library. Spotted an error or want a topic added? That feedback makes the notes better.