Index Laws
Theory
The index laws are shortcuts for powers of the same base. To multiply, add the indices; to divide, subtract them. This Year 10A maths page applies the laws to algebraic terms with coefficients and several pronumerals, to algebraic fractions, to indices that contain a pronumeral, to evaluating with prime powers, and to finding an unknown index or a missing term.
In \(a^n\), \(a\) is the base and \(n\) is the index (or power, or exponent). \(a^n\) means \(n\) factors of \(a\) multiplied together: \(a^4 = a \times a \times a \times a\). A pronumeral on its own has index \(1\): \(a = a^1\).
The multiplication law: when multiplying powers of the same base, keep the base and add the indices, because you are counting all the factors.
The division law: when dividing powers of the same base, keep the base and subtract the indices, because the matching factors cancel.
With coefficients (the numbers in front), multiply or divide the numbers as usual, then apply the laws to each pronumeral separately. A coefficient that does not divide exactly stays as a fraction.
The laws only work for the same base. \(2^3 \times 5^2\) cannot be combined into a single power.
For any base \(a\) and positive integers \(m\) and \(n\):
With coefficients:
How to simplify with the index laws
- Check the bases are the same. Write any lone pronumeral as power \(1\).
- Deal with the numbers: multiply them, or divide them as a fraction in simplest form.
- Add the indices of each pronumeral when multiplying; subtract when dividing.
- Write the answer with the number first, then the pronumerals in alphabetical order.
Finding a missing index
- Simplify one side to a single power of the base.
- Match the indices on both sides and solve the equation.
Other questions
- Division written with \(\div\): turn it into a fraction, or multiply by the reciprocal.
- Missing term: if \(\dfrac{P}{\square} = Q\), then \(\square = \dfrac{P}{Q}\).
- Numbers with different bases (\(25\), \(10\), \(12\)): write them as products of prime powers first.
- Unknown index: simplify one side to a single power of the base, then equate the indices.
- Given \(b^k\): split \(b^{k+2} = b^k \times b^2\) and \(b^{k-1} = b^k \div b\).
- Measurements: area of a rectangle or triangle, or how many small pieces fit, are products and quotients of terms.
- Spot the error: a common mistake is to divide the indices (\(8 \div 4\)) instead of subtracting them.
Multiply the numbers, then add the indices of each pronumeral. \(a\) means \(a^1\).
| \(-4a^3b \times 5ab^2\) | \(=\) | \((-4 \times 5) \times a^{3 + 1} \times b^{1 + 2}\) |
| \(\) | \(=\) | \(\boldsymbol{-20a^4b^3}\) |
(ii) Find \(n\) if \(\dfrac{x^{3n} \times x^{2}}{x^{n-1}} = x^{11}\).
(i) Add, then subtract the whole index in brackets:
| \(\dfrac{2^{n+3} \times 2^{n}}{2^{2n+1}}\) | \(=\) | \(2^{(n+3)+n-(2n+1)}\) |
| \(\) | \(=\) | \(2^{2}\) |
| \(\) | \(=\) | \(\boldsymbol{4}\) |
The \(n\) terms cancel, so the value is \(4\) for every \(n\).
(ii) Simplify the left side, then equate the indices:
| \(x^{3n+2-(n-1)}\) | \(=\) | \(x^{11}\) |
| \(2n + 3\) | \(=\) | \(11\) |
| \(n\) | \(=\) | \(\boldsymbol{4}\) |
Write each base as a product of primes, then use the index laws on each prime.
| \(6^4\) | \(=\) | \((2 \times 3)^4 = 2^4 \times 3^4\) |
| \(12^3\) | \(=\) | \((2^2 \times 3)^3 = 2^6 \times 3^3\) |
| \(\dfrac{2^4 \times 3^4 \times 2^3}{2^6 \times 3^3}\) | \(=\) | \(2^{4+3-6} \times 3^{4-3}\) |
| \(\) | \(=\) | \(2 \times 3 = \boldsymbol{6}\) |
Check: \(1296 \times 8 \div 1728 = 6\) ✓
(ii) Find the missing term: \(8y^3 \times \square = 56y^{10}\).
(i) The bases match, so the indices must match: \(n + 4 = 10\).
| \(n + 4\) | \(=\) | \(10\) |
| \(\therefore\ n\) | \(=\) | \(\boldsymbol{6}\) |
(ii) The missing term is the answer divided by the known term.
| \(\square\) | \(=\) | \(\dfrac{56y^{10}}{8y^3}\) |
| \(\) | \(=\) | \(\boldsymbol{7y^7}\) |
Check: \(8y^3 \times 7y^7 = 56y^{10}\).
Common pitfalls
Frequently asked questions
What are the index laws for multiplying and dividing?
To multiply powers of the same base, add the indices: \(a^m \times a^n = a^{m + n}\). To divide, subtract them: \(a^m \div a^n = a^{m - n}\).
Why do you add indices when multiplying?
Because an index counts factors. \(a^3 \times a^2\) is three factors of \(a\) times two more, which is five factors of \(a\) altogether: \(a^5\).
What do you do with the numbers in front?
Multiply or divide them as normal numbers. The index laws only apply to the indices. For example, \(-4a^3 \times 5a = -20a^4\).
Can you use the index laws with different bases?
Not directly. \(2^3 \times 3^2\) cannot be written as one power. The bases must be the same before you add or subtract the indices.
What is x to the power 1?
It is just \(x\). Any pronumeral written without an index has index \(1\), which matters when you add indices: \(x^2 \times x = x^3\).
How do you find an unknown index?
Simplify each side to a single power of the same base, then set the indices equal and solve the equation.