Multiplication and Division Index Laws
The multiplication and division index laws say that powers of the same base are multiplied by adding the indices, \(a^m\times a^n=a^{m+n}\), and divided by subtracting them, \(a^m\div a^n=a^{m-n}\). In NSW Year 10 Advanced maths these laws are applied to algebraic terms with several pronumerals, and to indices that are themselves expressions such as \(2k-3\) (NESA outcome MA5‑IND‑C‑01).
You will simplify products and quotients such as \(\frac{9p^4q^2\times4pq^3}{6p^3q^4}\), find a missing term, and spot the mistake of dividing indices instead of subtracting. You will evaluate expressions like \(\frac{15^3\times3^2}{45^2}\) using prime powers, and find unknown indices. Problems include the area of a triangle, cutting card into small rectangles, photos on a phone and a growing bacteria culture. It is part of Year 10 Advanced Indices.
Theory
To multiply powers of the same base, add the indices: \(a^m \times a^n = a^{m+n}\). To divide, subtract them: \(a^m \div a^n = a^{m-n}\). This Year 10 Advanced maths page covers coefficients and several pronumerals, pronumerals in the index, evaluating with prime powers, and finding an unknown index.
In \(a^n\), \(a\) is the base and \(n\) is the index (power or exponent). \(a^n\) means \(n\) factors of \(a\) multiplied together.
Multiplying \(a^m\) by \(a^n\) gives \(m + n\) factors of \(a\), so the indices add. Dividing cancels \(n\) factors from \(m\), so the indices subtract. These laws only work when the bases are the same.
In an algebraic term such as \(6x^4y^2\), the number \(6\) is the coefficient. Coefficients are multiplied or divided as ordinary numbers; the index laws apply to each pronumeral separately.
A pronumeral on its own has index \(1\): \(x = x^1\). When everything cancels, \(a^0 = 1\).
| Law | In symbols | Example |
|---|---|---|
| Multiplying | \(a^m \times a^n = a^{m+n}\) | \(y^4 \times y^7 = y^{11}\) |
| Dividing | \(a^m \div a^n = a^{m-n}\) | \(\dfrac{y^9}{y^4} = y^5\) |
| Zero index | \(a^0 = 1\ (a \ne 0)\) | \(\dfrac{c^3}{c^3} = c^0 = 1\) |
How to simplify a product or quotient of terms
- Multiply or divide the coefficients as numbers (simplify fractions).
- Group each pronumeral and apply the law: add indices for \(\times\), subtract for \(\div\).
- Handle signs: negative \(\times\) or \(\div\) negative gives positive.
- Tidy up: drop any power \(x^0 = 1\) and write \(x^1\) as \(x\).
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.
Work on the numbers, then each pronumeral separately. Remember \(a = a^1\).
| \(\text{numerator}\) | \(=\) | \(24a^{5+1}b^{2+4} = 24a^6b^6\) |
| \(\dfrac{24a^6b^6}{12a^3b^5}\) | \(=\) | \(\dfrac{24}{12} \times a^{6-3} \times b^{6-5}\) |
| \(\) | \(=\) | \(\boldsymbol{2a^3b}\) |
The coefficients are divided; only the indices are subtracted.
(i) Show that \(\dfrac{2^{n+3} \times 2^{n}}{2^{2n+1}}\) has the same value for every \(n\), and find it.
(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: \(1296 \times 8 \div 1728 = 6\) ✓
(i) \(5^{k+2}\)
(ii) \(5^{k-1}\).
Run the index laws backwards: an index that is a sum splits into a product, and a difference splits into a quotient.
(i)
| \(5^{k+2}\) | \(=\) | \(5^k \times 5^2\) |
| \(\) | \(=\) | \(3 \times 25 = \boldsymbol{75}\) |
(ii)
| \(5^{k-1}\) | \(=\) | \(5^k \div 5^1\) |
| \(\) | \(=\) | \(\boldsymbol{\dfrac{3}{5}}\) |
The same idea shows that a ratio such as \(\dfrac{N(t+3)}{N(t)}\) for \(N(t) = c \times 2^t\) does not depend on \(t\): it is \(2^3 = 8\).
Common pitfalls
Frequently asked questions
What are the multiplication and division index laws?
When the bases are the same, add the indices to multiply and subtract them to divide: \(a^m \times a^n = a^{m+n}\) and \(a^m \div a^n = a^{m-n}\).
What do you do with the numbers in front (coefficients)?
Multiply or divide them as ordinary numbers. Only the indices are added or subtracted: \(12x^5 \div 4x^2 = 3x^3\).
Can you use the index laws with different bases?
Not directly. Write the numbers with a common prime base first, for example 8 as 2 cubed and 4 as 2 squared, then apply the laws.
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.
Why is any number to the power zero equal to 1?
Dividing a power by itself gives \(1\), but the division law gives \(a^{m-m} = a^0\). So \(a^0 = 1\) for any \(a \ne 0\).