Mixed Index Laws
Mixed Index Laws brings the multiplication, division, power of a power and zero index laws together in one expression in NSW Year 8 maths (NESA outcome MA4‑IND‑C‑01). The key is to apply one law at a time and write a line for each step, watching for a hidden index of 1.
You will simplify expressions such as \(\frac{(5^4)^2 \times 8^0}{5^6}\) and \((3^4)^3 \div 3^9 \times 3\), handle negative bases and two bases at once, and leave a fraction when the bottom power is larger. You will find hidden indices and finish with pronumeral expressions, a first look at Stage 5 work, like \((4x^3y)^2 \div (2xy)^3\). The volume of a cube-shaped box with edges of \(2^2 \times 5\) cm is one example. It is part of Year 8 Algebra.
Every question with a fully worked solution.
- Mixed Index Laws - Video - Mixed Index Laws - Example 1 Watch
Theory
When an expression uses more than one index law, apply one law at a time and write a line for each step: brackets first, then zero indices, then add or subtract indices from left to right. This Year 8 maths page shows how.
The laws. \(a^m \times a^n = a^{m+n}\); \(a^m \div a^n = a^{m-n}\); \((a^m)^n = a^{mn}\); \((ab)^n = a^nb^n\); \(a^0 = 1\). The base can be a number or a pronumeral.
Adding or multiplying? \(a^m \times a^n\) adds the indices; \((a^m)^n\) multiplies them. Look for the bracket.
A hidden index. A base with no index has index \(1\): \(5^8 \div 5 = 5^7\). A zero index only turns its own power into \(1\): \(5^3 \times 5^0 = 5^3\), but \((5^3)^0 = 1\).
Coefficients. Raise, multiply or divide the numbers in front as ordinary numbers: \((2a^3)^2 = 4a^6\). Two bases: work on each base separately.
When the bottom has more, the leftover power stays on the bottom: \(2^6 \div 2^9 = \dfrac{1}{2^3} = \dfrac{1}{8}\). Negative bases: collect the indices, then an odd index gives a negative value. \(-2^4 = -16\) but \((-2)^4 = 16\).
The index laws together:
| Law | Form |
|---|---|
| Multiplication | \(a^m \times a^n = a^{m+n}\) |
| Division | \(a^m \div a^n = a^{m-n}\) |
| Power of a power | \((a^m)^n = a^{mn}\) |
| Power of a product | \((ab)^n = a^nb^n\) |
| Zero index | \(a^0 = 1\) |
How to simplify a mixed expression
- Expand brackets with powers (raise every factor).
- Replace zero-index powers by \(1\).
- Multiply or divide the coefficients, and add or subtract the indices of each base, left to right.
- Evaluate or leave as a single power.
| \((2a^3)^2\) | \(=\) | \(4a^6\) |
| \(4a^6 \times 3a\) | \(=\) | \(12a^7\) |
| \(12a^7 \div 6a^2\) | \(=\) | \(2a^5\) |
\(2a^5\)
| \(\dfrac{15m^7 \times 1}{3m^2}\) | \(=\) | \((15 \div 3) \times m^{7-2}\) |
| \(\) | \(=\) | \(5m^5\) |
\(5m^5\)
| \((2^4)^3\) | \(=\) | \(2^{12}\) |
| \(2^7 \times 2^1\) | \(=\) | \(2^8\) |
| \(2^{12} \div 2^8\) | \(=\) | \(2^4 = 16\) |
\(16\). Check: \(4096 \div 256 = 16\).
| \(\) | \(=\) | \((-2)^{7+0-4} = (-2)^3\) |
| \(\) | \(=\) | \(-8\) |
\(-8\) (odd index)
| \((3 \times 5^2)^3\) | \(=\) | \(3^3 \times 5^6\) |
| \(3^{3-2} \times 5^{6-4}\) | \(=\) | \(3 \times 25 = 75\) |
\(75\)
(ii) Find the whole number: \((4^{?})^2 \div 4^3 = 4^5\).
| \(\dfrac{2^6}{2^9}\) | \(=\) | \(\dfrac{1}{2^3} = \dfrac{1}{8}\) |
| \(2 \times {?} - 3\) | \(=\) | \(5\) |
(i) \(\dfrac{1}{8}\)
(ii) \(2 \times {?} = 8\), so \({?} = 4\)
| \((3a^2b)^2\) | \(=\) | \(9a^4b^2\) |
| \(9a^4b^2 \times 2ab\) | \(=\) | \(18a^5b^3\) |
| \(18a^5b^3 \div 6a^3b^2\) | \(=\) | \(3a^2b\) |
\(3a^2b\)
Common pitfalls
Frequently asked questions
How do you simplify an expression with several index laws?
Brackets first, then zero indices, then multiply and divide from left to right, one step per line.
When do you add indices and when do you multiply them?
Add for \(a^m \times a^n\); multiply for \((a^m)^n\).
What if the power on the bottom is bigger?
The leftover power stays on the bottom: \(x^2 \div x^5 = \dfrac{1}{x^3}\).
What do you do with the numbers in front?
Work them out as ordinary numbers: \(12a^7 \div 6a^2 = 2a^5\).
Does a zero index make the whole answer 1?
Only the power it sits on: \(4x^0 = 4\).