Index Law 3 and Zero Power
Theory
Index law 3, the power of a power law, says that when a power is raised to another power, you multiply the indices. The zero power rule says any base except \(0\) to the power \(0\) is \(1\). This Year 9 maths page shows why both rules work, how to raise every factor in a bracket, how to handle negatives and fractions, and how to work backwards to an unknown.
A power of a power is a power inside a bracket with another index outside, like \((x^3)^4\). The rule for it is often called index law 3.
To simplify it, multiply the indices: \((x^3)^4 = x^{12}\). The reason: \((x^3)^4\) is four lots of \(x^3\) multiplied, and \(3 + 3 + 3 + 3 = 12\).
The outside index applies to every factor inside the bracket. That includes the number in front (the coefficient). So \((2y^3)^4 = 2^4 y^{12} = 16y^{12}\).
With a negative inside, the outside index decides the sign. An even power gives a positive answer. An odd power keeps the negative.
A fraction inside the bracket: raise the top and the bottom.
The zero power: any number except \(0\) to the power \(0\) equals \(1\). So \(13^0 = 1\), \(x^0 = 1\) and \((4y)^0 = 1\). The index touches only what it is written on: in \(4y^0\) that is the \(y\) alone, so \(4y^0 = 4 \times 1 = 4\).
Index law 3, the power of a power law:
Every factor in the bracket takes the power:
| Expression | What happens | Answer |
|---|---|---|
| \((x^5)^2\) | \(5 \times 2\) | \(x^{10}\) |
| \((3x^5)^2\) | \(3^2\) and \(x^{10}\) | \(9x^{10}\) |
| \((-3x^5)^2\) | even power: positive | \(9x^{10}\) |
| \((-3x^5)^3\) | odd power: negative | \(-27x^{15}\) |
The zero power, for any base \(a\) except \(0\):
How to simplify a power of a power
- List every factor inside the bracket: the number, each letter, and any minus sign.
- Raise each factor to the outside power. For a letter with an index, multiply the indices.
- Decide the sign: even power, positive; odd power, negative.
- Replace any part with a zero index by \(1\). A number outside that bracket is left alone and keeps its sign.
- Tidy up: work out the number unless index notation is asked for, then put the factors together.
(i) \((m^4)^5\)
(ii) \((5t^3)^2\)
(i) Multiply the indices.
| \((m^4)^5\) | \(=\) | \(m^{4 \times 5}\) |
| \(\) | \(=\) | \(\boldsymbol{m^{20}}\) |
Written out, that is five lots of \(m^4\), and \(4 + 4 + 4 + 4 + 4 = 20\).
(ii) Raise both factors in the bracket, the \(5\) and the \(t^3\).
| \((5t^3)^2\) | \(=\) | \(5^2 \times (t^3)^2\) |
| \(\) | \(=\) | \(25 \times t^{3 \times 2}\) |
| \(\) | \(=\) | \(\boldsymbol{25t^6}\) |
Leaving the \(5\) alone would give \(5t^6\), which is wrong.
(i) \((-2k^4)^3\)
(ii) \((-a^3)^2 \times (-a^2)^3\)
(i) The outside index \(3\) is odd, so the answer stays negative.
| \((-2k^4)^3\) | \(=\) | \((-2)^3 \times k^{4 \times 3}\) |
| \(\) | \(=\) | \(\boldsymbol{-8k^{12}}\) |
(ii) Do each bracket first. An even power makes it positive. An odd power keeps the minus.
| \((-a^3)^2\) | \(=\) | \(a^6\) |
| \((-a^2)^3\) | \(=\) | \(-a^6\) |
| \(a^6 \times (-a^6)\) | \(=\) | \(-a^{6+6}\) |
| \(\) | \(=\) | \(\boldsymbol{-a^{12}}\) |
After the brackets, the two terms multiply, so add the indices.
(i) \(\left(\dfrac{2}{3}\right)^3\)
(ii) \(\left(\dfrac{3a}{b^2}\right)^3\)
(i) Raise the top and the bottom to the power \(3\).
| \(\left(\dfrac{2}{3}\right)^3\) | \(=\) | \(\dfrac{2^3}{3^3}\) |
| \(\) | \(=\) | \(\mathbf{\dfrac{8}{27}}\) |
(ii) Do the same with letters. Every factor on top and bottom is cubed.
| \(\left(\dfrac{3a}{b^2}\right)^3\) | \(=\) | \(\dfrac{3^3 a^3}{b^{2 \times 3}}\) |
| \(\) | \(=\) | \(\boldsymbol{\dfrac{27a^3}{b^6}}\) |
(i) \((6k)^0 + 6k^0\)
(ii) \((5n^0)^3\).
(i) Decide what each zero index touches. In \((6k)^0\) it is the whole bracket; in \(6k^0\) it is only the \(k\).
| \((6k)^0\) | \(=\) | \(1\) |
| \(6k^0\) | \(=\) | \(6 \times 1 = 6\) |
| \((6k)^0 + 6k^0\) | \(=\) | \(1 + 6 = \mathbf{7}\) |
(ii) Work inside the bracket first. The zero index touches only the \(n\).
| \(5n^0\) | \(=\) | \(5 \times 1 = 5\) |
| \((5n^0)^3\) | \(=\) | \(5^3 = \mathbf{125}\) |
The other way gives the same answer: \((5n^0)^3 = 5^3 n^{0 \times 3} = 125n^0 = 125\).
Common pitfalls
Frequently asked questions
What is index law 3?
Index law 3 is the power of a power law. When a power is raised to another power, multiply the indices. For example, \((x^3)^4 = x^{12}\).
Do you add or multiply indices?
Add them when you multiply two powers with the same base: \(x^3 \times x^4 = x^7\). Multiply them when a power is raised to a power: \((x^3)^4 = x^{12}\).
Does the number in front get raised to the power too?
Yes, if it is inside the bracket. \((2y^3)^4 = 2^4 y^{12} = 16y^{12}\).
Is a negative number to an even power positive?
Yes. A negative to an even power is positive, and to an odd power it is negative: \((-2)^4 = 16\) and \((-2)^3 = -8\).
How do you raise a fraction to a power?
Raise the top and the bottom separately: \(\left(\dfrac{2}{3}\right)^3 = \dfrac{2^3}{3^3} = \dfrac{8}{27}\).
What is any number to the power of 0?
Any number except \(0\) raised to the power \(0\) equals \(1\). For example, \(13^0 = 1\) and \(x^0 = 1\).
What is the difference between 4y to the power 0 and (4y) to the power 0?
In \(4y^0\) the index touches only the \(y\), so it is \(4 \times 1 = 4\). In \((4y)^0\) the bracket puts the whole \(4y\) under the index, so it is \(1\).