Further Index laws
Further Index Laws adds two more rules in NSW Year 8 maths (NESA outcome MA4‑IND‑C‑01). A power of a power multiplies the indices, \((a^m)^n = a^{mn}\), and any non-zero number to the power zero equals 1, as the pattern \(3^2 = 9\), \(3^1 = 3\), \(3^0 = 1\) shows.
You will simplify expressions like \((7^3)^5\) and \((4^5)^2 \div 4^7\), evaluate zero powers such as \(56 + 56^0\), and use \((6 \times 5)^2 = 6^2 \times 5^2\) as a shortcut. You will find missing indices, write \(9^5\) as a power of 3 and decide whether \(2^{30}\) or \(3^{20}\) is larger without a calculator. The area of a square paddock and the volume of a storage box can be written as powers. It is part of Year 8 Working with Numbers.
Every question with a fully worked solution.
- Further Index laws - Video - Further Index Laws Watch
Theory
When a power is raised to another power, the indices are multiplied: \((a^m)^n = a^{m \times n}\). Any non-zero number to the power \(0\) equals \(1\): \(a^0 = 1\). This Year 8 maths page shows why, and how to use them with the multiplication and division laws.
Power of a power. The index outside the bracket says how many lots of the inside power to multiply: \((5^3)^4 = 5^3 \times 5^3 \times 5^3 \times 5^3 = 5^{3+3+3+3} = 5^{12}\). So the indices multiply.
Zero index. Dividing a number by itself gives \(1\), and the division law gives an index of \(0\): \(7 \div 7 = 1\) and \(7 \div 7 = 7^{1-1} = 7^0\). So \(7^0 = 1\). This is true for any number except \(0\).
A product inside a bracket. \((2 \times 7)^2 = 14^2 = 196\) and \(2^2 \times 7^2 = 4 \times 49 = 196\), so \((a \times b)^n = a^n \times b^n\). Read backwards it is a shortcut: \(5^2 \times 8^2 = 40^2 = 1600\).
Order. Simplify inside a bracket first, then apply the outside index. A zero index belongs only to what it sits on: in \(3^0 \times 3^4\) the first factor is \(1\).
A missing index. Read the law backwards: if \((5^{{?}})^3 = 5^{12}\), then \({?} = 12 \div 3 = 4\).
Rewriting a base. Write one base as a power of another: \(16^3 = (2^4)^3 = 2^{12}\). To compare two powers with the same index, the larger positive base gives the larger number.
The index laws (a and b not zero):
How to simplify with all the index laws
- Simplify inside any bracket first.
- Apply a power of a power: multiply the indices.
- Replace any zero-index power by \(1\).
- Use the multiplication and division laws from left to right.
| \((5^3)^4\) | \(=\) | \(5^{3 \times 4}\) |
| \(\) | \(=\) | \(5^{12}\) |
\(5^{12}\), since \(5^3 \times 5^3 \times 5^3 \times 5^3 = 5^{3+3+3+3}\).
\(7 \div 7 = 1\), and by the division law \(7 \div 7 = 7^{1-1} = 7^0\).
| \(7^0\) | \(=\) | \(1\) |
\(7^0 = 1\)
(i) \((6^4)^3\)
(ii) \(\left((-4)^3\right)^2\)
| \((6^4)^3\) | \(=\) | \(6^{12}\) |
| \(\left((-4)^3\right)^2\) | \(=\) | \((-4)^6 = 4096\) |
(i) \(6^{12}\)
(ii) \(6\) is even, so positive: \(4^6 = 4096\)
(i) \(4 \times 9^0 + 2^0\)
(ii) \((7 \times 3)^0 - 1\)
| \(4 \times 1 + 1\) | \(=\) | \(5\) |
| \(21^0 - 1\) | \(=\) | \(1 - 1 = 0\) |
(i) \(5\)
(ii) \(0\)
(i) \((3^4)^2 \div 3^5 \times 3^0\)
(ii) \((2^3 \times 2^2)^2\)
| \(3^8 \div 3^5 \times 1\) | \(=\) | \(3^3 = 27\) |
| \((2^5)^2\) | \(=\) | \(2^{10} = 1024\) |
(i) \(27\)
(ii) \(1024\)
(ii) Evaluate \(2^5 \times 5^5\) without a calculator.
| \((2^4)^2\) | \(=\) | \(2^8 = 256\) |
| \(2^5 \times 5^5\) | \(=\) | \((2 \times 5)^5 = 10^5\) |
(i) \(2^8 = 256\) m\(^2\)
(ii) \(100\,000\)
Common pitfalls
Frequently asked questions
What is the power of a power law?
Multiply the indices: \((a^m)^n = a^{m \times n}\), so \((2^3)^4 = 2^{12}\).
Why does any number to the power 0 equal 1?
Because \(a^n \div a^n = 1\), and the division law gives \(a^{n-n} = a^0\).
Is 0 to the power 0 equal to 1?
The zero index law is for non-zero bases only.
How do you raise a product to a power?
Raise each factor: \((2 \times 5)^3 = 2^3 \times 5^3 = 1000\).
How do you write 9 to a power as a power of 3?
Since \(9 = 3^2\), \(9^4 = (3^2)^4 = 3^8\).