Index laws
The index laws are shortcuts for combining powers of the same base without working out their values. In NSW Year 8 maths you use the multiplication law \(a^m \times a^n = a^{m+n}\) and the division law \(a^m \div a^n = a^{m-n}\) with numerical bases (NESA outcome MA4‑IND‑C‑01).
You will first write powers in expanded form to see why the laws work, then simplify products and quotients such as \(3^6 \times 3^2 \div 3^4\), find a missing index and rewrite bases like 16 and 64 as powers of 2. You will also check claims made by other students. Photo file sizes sent from a phone and computer memory are counted in powers of 2. It is part of Year 8 Working with Numbers.
Every question with a fully worked solution.
- Index laws - Video - Laws of Indices Watch
Theory
When multiplying numbers with the same base, keep the base and add the indices: \(a^m \times a^n = a^{m+n}\). When dividing, keep the base and subtract the indices: \(a^m \div a^n = a^{m-n}\). This Year 8 maths page shows why the laws work and how to use them.
In \(8^5\) the base is \(8\) and the index is \(5\), so \(8\) is used as a factor \(5\) times.
Multiplication law. Same base: keep the base and add the indices. \(7^4 \times 7^2 = (7 \times 7 \times 7 \times 7) \times (7 \times 7) = 7^6\).
Division law. Same base: keep the base and subtract the indices. \(7^4 \div 7^2 = \dfrac{7 \times 7 \times 7 \times 7}{7 \times 7} = 7^2\). A fraction is a division, so \(\dfrac{a^m}{a^n} = a^{m-n}\) too.
The hidden index. A number on its own has index \(1\): \(7 = 7^1\), so \(7^4 \times 7 = 7^5\).
Order. Do brackets first, then work from left to right through a string of \(\times\) and \(\div\).
Different bases? The laws only work once the bases match. Match them when one base is a power of the other: \(8 = 2^3\), so \(8^2 = 2^6\) and \(8^2 \times 2^3 = 2^9 = 512\). \(2^4 \times 3^2\) cannot be one power: work out \(16 \times 9 = 144\).
Negative bases. Brackets make the whole negative number the base. Use the laws as usual; then an even index gives a positive answer and an odd index a negative one.
The first two index laws (same base a):
How to simplify with index laws
- Check the bases are the same (match them if one is a power of the other).
- Give any lone number its index of \(1\).
- Add indices for \(\times\), subtract for \(\div\), working left to right.
- Evaluate if asked, deciding the sign for a negative base.
Two facts about the indices
- Write each fact as an equation in the indices (multiplying adds them, dividing subtracts them).
- Add the two equations to find one index, then substitute to find the other.
| \(7^4 \times 7^2\) | \(=\) | \(7^{4+2}\) |
| \(\) | \(=\) | \(7^6\) |
\(7^6\), since \((7 \times 7 \times 7 \times 7) \times (7 \times 7)\) is six \(7\)s.
| \(7^4 \div 7^2\) | \(=\) | \(7^{4-2}\) |
| \(\) | \(=\) | \(7^2\) |
\(7^2\), since \(\dfrac{7 \times 7 \times 7 \times 7}{7 \times 7} = 7 \times 7\).
(i) \(3^5 \times 3^2\)
(ii) \(6^4 \times 6 \times 6^3\)
| \(3^5 \times 3^2\) | \(=\) | \(3^{5+2} = 3^7\) |
| \(6^4 \times 6^1 \times 6^3\) | \(=\) | \(6^{4+1+3} = 6^8\) |
(i) \(3^7\). Check: \(243 \times 9 = 2187 = 3^7\).
(ii) \(6^8\) (the lone \(6\) has index \(1\))
(ii) Write \(\dfrac{10^9}{10^5}\) as an ordinary number.
| \(4^{12} \div 4^8\) | \(=\) | \(4^4 = 256\) |
| \(\dfrac{10^9}{10^5}\) | \(=\) | \(10^4 = 10\,000\) |
(i) \(4^4 = 256\)
(ii) \(10\,000\): dividing by \(10^5\) removes five zeros.
(ii) Simplify \((-3)^2 \times (-3)^3\), then evaluate.
| \(3^4 \times 3^5 \div 3^2\) | \(=\) | \(3^{4+5-2} = 3^7\) |
| \((-3)^2 \times (-3)^3\) | \(=\) | \((-3)^5 = -243\) |
(i) \(81 = 3^4\), so \(3^7\)
(ii) \(5\) is odd, so \(-243\)
(ii) A warehouse holds \(5^6\) boxes in \(5^2\) equal rows. How many boxes are in each row?
| \(6 + {?}\) | \(=\) | \(10\) |
| \(5^6 \div 5^2\) | \(=\) | \(5^4 = 625\) |
(i) \(4\)
(ii) \(5^4 = 625\) boxes. Check: \(25 \times 625 = 15\,625 = 5^6\).
Same base, so the indices obey \(a + b = 9\) and \(a - b = 3\).
Adding the two facts: \(2a = 12\), so \(a = 6\), and then \(b = 9 - 6 = 3\).
Check: \(2^6 \times 2^3 = 2^9\) and \(2^6 \div 2^3 = 2^3\).
\(a = 6\), \(b = 3\)
Common pitfalls
Frequently asked questions
What is the index law for multiplying?
Same base: add the indices. \(a^m \times a^n = a^{m+n}\), for example \(5^2 \times 5^4 = 5^6\).
What is the index law for dividing?
Same base: subtract the indices. \(a^m \div a^n = a^{m-n}\), for example \(5^6 \div 5^2 = 5^4\).
What index does a number on its own have?
An index of \(1\): \(9 = 9^1\).
Can you use index laws with different bases?
Only if you can rewrite one base as a power of the other, such as \(4 = 2^2\).
Why do you add the indices when multiplying?
Because you are counting factors: \(3^3 \times 3^2\) is three \(3\)s then two more, five \(3\)s in all.