Multiplying and Dividing using Indices
When powers with the same base are multiplied, the indices are added, and when they are divided, the indices are subtracted: \(a^m \times a^n = a^{m+n}\) and \(a^m \div a^n = a^{m-n}\). This NSW Year 9 Advanced maths subtopic applies these laws to algebraic terms (NESA outcome MA5‑IND‑C‑01).
You will simplify products and quotients with coefficients and several pronumerals, such as \(\frac{48a^7b^5}{16a^5b^2}\), and correct a student who multiplied the indices. You will find a missing term and the height of a prism from its volume, evaluate \(\frac{6^5 \times 2^3}{4^2 \times 3^4}\) without a calculator, solve \(3^{2x+1} \times 3^{x-4} = 3^{12}\), and simplify powers with \(n\) in the index. It is part of Year 9 Advanced Indices and Scientific Notation.
Theory
To multiply powers of the same base, add the indices; to divide, subtract them. This Year 9 Advanced maths page covers coefficients and signs, answers with the leftover factors underneath, evaluating with prime bases and pronumerals in the index.
In \(a^5\), \(a\) is the base and \(5\) is the index (power, exponent). It means five factors of \(a\) multiplied together. A pronumeral written with no index, such as \(b\), has index \(1\).
A term such as \(7m^6n^2\) has a coefficient (\(7\)) and a pronumeral part. When terms are multiplied or divided, the coefficients are multiplied or divided like ordinary numbers, and each base is handled separately.
Multiplying powers of the same base joins the factors, so the indices are added. Dividing cancels factors in pairs, so the indices are subtracted.
The laws only join powers of the same base. \(x^3 \times y^2\) cannot be simplified, and \(2^3 \times 3^2\) is not \(6^5\). To use the laws on numbers like \(6^5\) or \(10^4\), first write the base as a product of primes.
Indices can be expressions too. \(3^{2x+1}\) follows the same laws, and when two powers of the same base are equal, their indices are equal. That turns an index statement into an ordinary equation.
For any base \(a \neq 0\) and indices \(m\) and \(n\):
| Situation | What to do | Example |
|---|---|---|
| Multiply terms | Multiply coefficients, add indices | \(5p^3 \times 4p^6 = 20p^9\) |
| Divide terms | Divide coefficients, subtract indices | \(\dfrac{21k^8}{7k^3} = 3k^5\) |
| More of a base underneath | Leftover factors stay underneath | \(\dfrac{w^4}{w^7} = \dfrac{1}{w^3}\) |
| Same power top and bottom | The base cancels to \(1\) | \(\dfrac{9t^5}{3t^5} = 3\) |
| Coefficients do not divide exactly | Leave a fraction in simplest form | \(\dfrac{6c^4}{9c} = \dfrac{2c^3}{3}\) |
How to simplify a product or quotient
- Coefficients first: multiply or divide the numbers, keeping track of the sign. Write decimals and fractions as numbers in the working.
- One base at a time: add the indices for each pronumeral that is multiplied; subtract for each that is divided.
- Mixed \(\times\) and \(\div\): collect everything on the top, everything on the bottom, then divide. With a written \(\div\) sign, work left to right.
- Leftovers: if a base has the bigger index underneath, the difference stays underneath. Coefficients that do not divide exactly stay as a fraction.
- Check: substitute a small value such as \(x = 2\) into the question and your answer. They must agree.
Other jobs the same laws do
- Missing factor: if \(4p^2 \times \square = 28p^9\), the missing term is the product divided by the known factor.
- Measurement: for a prism, height \(=\) volume \(\div\) (length \(\times\) width). Divide the terms as above.
- Numbers: regroup \((5 \times 3^4) \times (2 \times 3^2)\) as \((5 \times 2) \times 3^{6}\), or write composite bases as primes (Example 3).
- Substitute after simplifying: simplify first, then substitute. Brackets keep a negative value safe: \((-2)^2 = 4\).
- Index equations: make each side a single power of the same base, then equate the indices (Example 4).
Multiply the numbers with their signs, then deal with one pronumeral at a time. A pronumeral with no index shown has index \(1\).
| \(-6 \times 2 \times (-0.5)\) | \(=\) | \(6\) |
| \(m^3 \times m^1\) | \(=\) | \(m^{3+1} = m^4\) |
| \(n^2 \times n^4 \times n^1\) | \(=\) | \(n^{2+4+1} = n^7\) |
| \(-6m^3n^2 \times 2mn^4 \times (-0.5n)\) | \(=\) | \(\boldsymbol{6m^4n^7}\) |
Two negative factors make the product positive. The decimal \(-0.5\) is just a number: multiply it with the other coefficients.
Collect the top first, then divide one part at a time.
| \(18a^2b^7 \times 2a^5b\) | \(=\) | \(36a^{2+5}b^{7+1} = 36a^7b^8\) |
Now divide. For \(a\) there are more factors underneath (\(9 > 7\)), so the leftover \(a\)s stay underneath.
| \(\dfrac{36}{12}\) | \(=\) | \(3\) |
| \(\dfrac{a^7}{a^9}\) | \(=\) | \(\dfrac{1}{a^{9-7}} = \dfrac{1}{a^2}\) |
| \(\dfrac{b^8}{b^3}\) | \(=\) | \(b^{8-3} = b^5\) |
| \(\therefore\ \dfrac{18a^2b^7 \times 2a^5b}{12a^9b^3}\) | \(=\) | \(\boldsymbol{\dfrac{3b^5}{a^2}}\) |
Check with \(a = b = 1\): the question gives \(\dfrac{36}{12} = 3\), and so does the answer.
Index laws only join powers of the same base. Rewrite every base using the primes \(2\), \(3\) and \(5\).
| \(10^4\) | \(=\) | \((2 \times 5)^4 = 2^4 \times 5^4\) |
| \(6^3\) | \(=\) | \((2 \times 3)^3 = 2^3 \times 3^3\) |
| \(\dfrac{2^4 \times 5^4 \times 3^5}{2^3 \times 3^3 \times 5^2}\) | \(=\) | \(2^{4-3} \times 3^{5-3} \times 5^{4-2}\) |
| \(\) | \(=\) | \(2 \times 9 \times 25\) |
| \(\) | \(=\) | \(\mathbf{450}\) |
Check: \(\dfrac{10\,000 \times 243}{216 \times 25}\) \(= \dfrac{2\,430\,000}{5400}\) \(= 450\).
(i) Solve \(2^{3x-1} \times 2^{x+5} = 2^{20}\).
(ii) Simplify \(\dfrac{a^{2n+1} \times a^{n+3}}{a^{n-2}}\).
The laws work the same way when an index is an expression. Put each index in brackets so every term is added or subtracted.
(i)
| \(2^{3x-1} \times 2^{x+5}\) | \(=\) | \(2^{(3x-1)+(x+5)} = 2^{4x+4}\) |
| \(2^{4x+4}\) | \(=\) | \(2^{20}\) |
| \(4x + 4\) | \(=\) | \(20\) |
| \(x\) | \(=\) | \(\boldsymbol{4}\) |
The bases match, so the indices must be equal. Check: \(2^{11} \times 2^{9} = 2^{20}\).
(ii)
| \(a^{2n+1} \times a^{n+3}\) | \(=\) | \(a^{3n+4}\) |
| \(\dfrac{a^{3n+4}}{a^{n-2}}\) | \(=\) | \(a^{(3n+4)-(n-2)}\) |
| \(\) | \(=\) | \(a^{3n+4-n+2}\) |
| \(\) | \(=\) | \(\boldsymbol{a^{2n+6}}\) |
Subtracting \((n-2)\) means \(-n + 2\). Dropping the bracket gives \(a^{2n+2}\), a common slip.
Common pitfalls
Frequently asked questions
What is the index law for multiplying?
When two powers of the same base are multiplied, add the indices: \(a^m \times a^n = a^{m+n}\). Multiply any coefficients as ordinary numbers.
What is the index law for dividing?
When a power is divided by a power of the same base, subtract the indices: \(a^m \div a^n = a^{m-n}\). Divide the coefficients as a fraction in simplest form.
What happens if the bigger power is on the bottom?
The leftover factors stay on the bottom. For example \(\dfrac{y^2}{y^5} = \dfrac{1}{y^3}\).
Can you use index laws with different bases?
Not directly. \(2^3 \times 3^2\) cannot be written as one power. If the bases share prime factors, such as \(6\) and \(2\), rewrite them as products of primes first.
How do you solve an equation with the unknown in the index?
Simplify each side to a single power of the same base. When the bases are equal, the indices are equal, so set the two indices equal and solve the ordinary equation.