Power of a Power
A power of a power is simplified by multiplying the indices: \((a^m)^n = a^{mn}\), so \((5^2)^4 = 5^8\). This NSW Year 9 Core maths subtopic extends the law to brackets containing several factors, and is Stage 5 Core content under NESA outcome MA5‑IND‑C‑01.
You will raise every factor inside a bracket to the outside power, including the coefficient, so \((2c^5)^3 = 8c^{15}\), and decide the sign when a negative is inside the bracket from whether the index is odd or even. You will raise fractions such as \(\left(\frac{3p}{q^3}\right)^2\) to a power, combine a power of a power with a multiplication, and work backwards to find an unknown index, as in \(4^3 \times 8 = 2^n\). You will also see why \((4y)^2\) is not \(4y^2\). It is part of Year 9 Core Indices and Scientific Notation.
Every question with a fully worked solution.
- Power of a Power - Video - Index Law Watch
Theory
The power of a power law says that when a power is raised to another power, you multiply the indices. This Year 9 maths page shows why, how to raise every factor in a bracket, how to handle negatives and fractions, and how to work backwards to an unknown index.
A power of a power is a power inside a bracket with another index outside, like \((x^3)^4\).
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 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 | \(-27x^{15}\) |
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.
- 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) If \((4x^5)^2 = ax^b\), find \(a\) and \(b\).
(ii) If \(25^3 \times 125 = 5^n\), find \(n\).
(i) Simplify the left side, then match it with \(ax^b\).
| \((4x^5)^2\) | \(=\) | \(4^2 \times x^{5 \times 2}\) |
| \(\) | \(=\) | \(16x^{10}\) |
So \(a = \mathbf{16}\) and \(b = \mathbf{10}\).
(ii) Write \(25\) and \(125\) as powers of \(5\): \(25 = 5^2\) and \(125 = 5^3\).
| \(25^3 \times 125\) | \(=\) | \((5^2)^3 \times 5^3\) |
| \(\) | \(=\) | \(5^6 \times 5^3\) |
| \(\) | \(=\) | \(5^{6+3} = 5^9\) |
The bases match, so the indices match: \(n = \mathbf{9}\).
Common pitfalls
Frequently asked questions
What is the power of a power rule?
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: \(\left(\dfrac{2}{3}\right)^3 = \dfrac{2^3}{3^3} = \dfrac{8}{27}\).