Index Laws 1 and 2
Theory
The index laws make products and quotients of powers quick to simplify: multiplying powers of the same base adds the indices (index law 1), and dividing them subtracts the indices (index law 2). This Year 9 maths page shows why, then applies the laws to numbers, coefficients and terms with several pronumerals.
In \(a^5\), \(a\) is the base and \(5\) is the index (or power). It means five \(a\)s multiplied: \(a \times a \times a \times a \times a\). A pronumeral with no index written has index \(1\): \(a = a^1\).
When two powers of the same base are multiplied, the indices are added (index law 1). When they are divided, the indices are subtracted (index law 2).
The number in front of a term is its coefficient. Coefficients are multiplied or divided in the ordinary way; the index laws apply only to the bases.
Powers of different bases, such as \(p^2\) and \(q^3\), cannot be combined.
Index law 1 and index law 2:
With coefficients:
| Expression | What happens | Answer |
|---|---|---|
| \(a^6 \times a^3\) | \(6 + 3\) | \(a^9\) |
| \(a^6 \div a^3\) | \(6 - 3\) | \(a^3\) |
| \(3a^6 \times 2a^3\) | \(3 \times 2\), then \(6 + 3\) | \(6a^9\) |
| \(a^6 \times b^3\) | bases differ | \(a^6b^3\) |
How to simplify with the index laws
- Write any missing index as \(1\).
- Work out the coefficients: multiply or divide them, keeping the sign. With several negative factors, an even number of them gives a positive answer and an odd number a negative one.
- Combine each base on its own: add the indices to multiply, subtract to divide.
- Write the answer as the coefficient followed by each base in order. Leftover factors from the bottom stay on the bottom.
(i) Simplify \(5^4 \times 5^3\), leaving the answer in index form.
(ii) Simplify \(3^2 \times 3^2\) with an index law, then evaluate.
The bases match, so add the indices:
| \(5^4 \times 5^3\) | \(=\) | \(5^{4+3}\) |
| \(\) | \(=\) | \(\mathbf{5^7}\) |
| \(3^2 \times 3^2\) | \(=\) | \(3^{4}\) |
| \(\) | \(=\) | \(3 \times 3 \times 3 \times 3 = \mathbf{81}\) |
Check (ii) with the values: \(9 \times 9 = 81\). The base stays \(3\); it does not become \(9\).
Do the numbers, then each letter on its own. A letter with no index has index \(1\).
| \(-3 \times 5\) | \(=\) | \(-15\) |
| \(a^4 \times a^1\) | \(=\) | \(a^5\) |
| \(b^1 \times b^3\) | \(=\) | \(b^4\) |
| \(-3a^4b \times 5ab^3\) | \(=\) | \(\boldsymbol{-15a^5b^4}\) |
The coefficients multiply; only the indices add.
Divide the numbers, then subtract the indices:
| \(14 \div 4\) | \(=\) | \(\dfrac{14}{4} = \dfrac{7}{2}\) |
| \(m^8 \div m^3\) | \(=\) | \(m^{8-3} = m^5\) |
| \(14m^8 \div 4m^3\) | \(=\) | \(\boldsymbol{\dfrac{7m^5}{2}}\) |
When the numbers do not divide exactly, simplify them as a fraction.
A fraction is a division. Take it in three pieces:
| \(9 \div 3\) | \(=\) | \(3\) |
| \(\dfrac{p^6}{p^2}\) | \(=\) | \(p^4 \text{ (more on top)}\) |
| \(\dfrac{q^2}{q^5}\) | \(=\) | \(\dfrac{1}{q^3} \text{ (more underneath)}\) |
| \(\dfrac{9p^6q^2}{3p^2q^5}\) | \(=\) | \(\boldsymbol{\dfrac{3p^4}{q^3}}\) |
When the larger power is on the bottom, the leftover factors stay on the bottom.
Common pitfalls
Frequently asked questions
What happens to the indices when you multiply powers?
If the bases are the same, add the indices. For example \(a^3 \times a^4 = a^7\).
What happens to the indices when you divide powers?
If the bases are the same, subtract the indices. For example \(a^9 \div a^2 = a^7\).
What do you do with the numbers in front?
Multiply or divide the coefficients as ordinary numbers. The index laws only apply to the bases.
Can you add indices with different bases?
No. The index laws only work when the bases are the same. \(p^2 \times q^3\) stays as \(p^2q^3\).