Fractional Indices
A fractional index means a root: \(a^{\frac{1}{n}}=\sqrt[n]{a}\), and \(a^{\frac{m}{n}}\) is the \(n\)th root raised to the power \(m\). So \(243^{\frac{2}{5}}=3^2=9\). In NSW Year 10 Advanced maths you move between surd form and index form, and use all the index laws with fractional and negative indices (NESA outcome MA5‑IND‑P‑02).
You will evaluate expressions such as \((0.027)^{-\frac{1}{3}}\) without a calculator, write roots as powers, simplify algebraic expressions with fractional indices, expand brackets and solve equations like \(x^{\frac{3}{4}}=27\). Real-world formulas link a planet's orbit time to its distance from the Sun and an animal's energy use to its mass. Harder questions write nested roots as a single power and prove identities. It is part of Year 10 Advanced Indices.
Every question with a fully worked solution.
- Fractional Indices - Video - Simplifying Expressions with Rational Exponents Watch
Theory
A fractional index is a root: \(a^{\genfrac{}{}{}{1}{1}{n}} = \sqrt[n]{a}\), and \(a^{\genfrac{}{}{}{1}{m}{n}} = \left(\sqrt[n]{a}\right)^m\). This Year 10 Advanced maths page covers content used throughout Advanced maths in Years 11 and 12: evaluating without a calculator, simplifying algebraic terms, switching between surd form and index form, and solving equations.
The index laws decide what a fractional index means. Since \(\left(a^{\genfrac{}{}{}{1}{1}{2}}\right)^2 = a^{\genfrac{}{}{}{1}{1}{2} \times 2} = a\), the number \(a^{\genfrac{}{}{}{1}{1}{2}}\) squares to \(a\): it is \(\sqrt{a}\). In general \(a^{\genfrac{}{}{}{1}{1}{n}} = \sqrt[n]{a}\).
For \(a^{\genfrac{}{}{}{1}{m}{n}}\), the denominator \(n\) is the root and the numerator \(m\) is the power. You may do them in either order, but taking the root first keeps the numbers small: \(32^{\genfrac{}{}{}{1}{2}{5}} = \left(\sqrt[5]{32}\right)^2 = 2^2 = 4\).
A negative fractional index still means a reciprocal: \(a^{-\genfrac{}{}{}{1}{1}{2}} = \dfrac{1}{\sqrt{a}}\).
All the index laws still apply. With even roots, take \(a \gt 0\) so the expressions are defined.
| Index form | Surd form | Example |
|---|---|---|
| \(a^{\genfrac{}{}{}{1}{1}{2}}\) | \(\sqrt{a}\) | \(49^{\genfrac{}{}{}{1}{1}{2}} = 7\) |
| \(a^{\genfrac{}{}{}{1}{1}{n}}\) | \(\sqrt[n]{a}\) | \(81^{\genfrac{}{}{}{1}{1}{4}} = 3\) |
| \(a^{\genfrac{}{}{}{1}{m}{n}}\) | \(\left(\sqrt[n]{a}\right)^m = \sqrt[n]{a^m}\) | \(27^{\genfrac{}{}{}{1}{4}{3}} = 3^4 = 81\) |
| \(a^{-\genfrac{}{}{}{1}{m}{n}}\) | \(\dfrac{1}{\left(\sqrt[n]{a}\right)^m}\) | \(4^{-\genfrac{}{}{}{1}{5}{2}} = \dfrac{1}{32}\) |
How to evaluate or simplify
- Decimals become fractions; a negative index flips the base.
- Root using the denominator, then power using the numerator.
- Algebraic terms: multiply every index in the bracket by the fractional power, and evaluate the number part.
- Write the answer with positive indices unless asked otherwise.
Other questions
- Roots of fractions: \(\sqrt[3]{\dfrac{p^6}{27}} = \dfrac{p^2}{3}\), taking the root of each part.
- Nested roots: work from the inside out, turning each root into an index.
- Expanding: \(x^{\genfrac{}{}{}{1}{1}{3}}\left(x^{\genfrac{}{}{}{1}{5}{3}} + x^{-\genfrac{}{}{}{1}{1}{3}}\right) = x^2 + 1\); for squares of sums, use \((p + q)^2 = p^2 + 2pq + q^2\).
- Factorising: take out the lowest power, such as \(x^{\genfrac{}{}{}{1}{1}{2}}\); \(x - 9\) is a difference of squares, \(\left(x^{\genfrac{}{}{}{1}{1}{2}}\right)^2 - 3^2\).
- Formulas: substitute, then root and power. To make the base the subject, use the reciprocal power.
- Proofs with a substitution: let \(u = a^{\genfrac{}{}{}{1}{1}{3}}\); then \(u^3 = a\), which can tidy long expansions.
(i) \(64^{\genfrac{}{}{}{1}{2}{3}} + 16^{-\genfrac{}{}{}{1}{3}{4}}\)
(ii) \((0.04)^{-\genfrac{}{}{}{1}{3}{2}}\).
(i) Root first, then the power. A negative index means a reciprocal.
| \(64^{\genfrac{}{}{}{1}{2}{3}}\) | \(=\) | \(\left(\sqrt[3]{64}\right)^2 = 4^2 = 16\) |
| \(16^{-\genfrac{}{}{}{1}{3}{4}}\) | \(=\) | \(\dfrac{1}{\left(\sqrt[4]{16}\right)^3} = \dfrac{1}{2^3} = \dfrac{1}{8}\) |
| \(\text{sum}\) | \(=\) | \(\boldsymbol{16\dfrac{1}{8}}\) |
(ii) Write the decimal as a fraction, then flip it:
| \((0.04)^{-\genfrac{}{}{}{1}{3}{2}}\) | \(=\) | \(\left(\dfrac{1}{25}\right)^{-\genfrac{}{}{}{1}{3}{2}} = 25^{\genfrac{}{}{}{1}{3}{2}}\) |
| \(\) | \(=\) | \(\left(\sqrt{25}\right)^3 = \boldsymbol{125}\) |
Raise every factor to the power \(\genfrac{}{}{}{1}{2}{3}\): multiply each index by \(\genfrac{}{}{}{1}{2}{3}\).
| \(8^{\genfrac{}{}{}{1}{2}{3}}\) | \(=\) | \(\left(\sqrt[3]{8}\right)^2 = 4\) |
| \(\left(a^6\right)^{\genfrac{}{}{}{1}{2}{3}}\) | \(=\) | \(a^{4}\) |
| \(\left(b^{-3}\right)^{\genfrac{}{}{}{1}{2}{3}}\) | \(=\) | \(b^{-2}\) |
| \(\left(8a^6b^{-3}\right)^{\genfrac{}{}{}{1}{2}{3}}\) | \(=\) | \(4a^4b^{-2} = \boldsymbol{\dfrac{4a^4}{b^2}}\) |
(i) Write \(5\sqrt[3]{5} \times \sqrt{5}\) as a single power of \(5\).
(ii) Write \(\dfrac{1}{\sqrt[3]{x^2}}\) as a power of \(x\).
(i) Change each root to an index, then add the indices:
| \(5\sqrt[3]{5} \times \sqrt{5}\) | \(=\) | \(5^1 \times 5^{\genfrac{}{}{}{1}{1}{3}} \times 5^{\genfrac{}{}{}{1}{1}{2}}\) |
| \(\) | \(=\) | \(5^{\genfrac{}{}{}{1}{6}{6} + \genfrac{}{}{}{1}{2}{6} + \genfrac{}{}{}{1}{3}{6}} = \boldsymbol{5^{\genfrac{}{}{}{1}{11}{6}}}\) |
(ii) The root gives the denominator; one over gives a negative index:
| \(\dfrac{1}{\sqrt[3]{x^2}}\) | \(=\) | \(\dfrac{1}{x^{\genfrac{}{}{}{1}{2}{3}}} = \boldsymbol{x^{-\genfrac{}{}{}{1}{2}{3}}}\) |
(i) \(x^{\genfrac{}{}{}{1}{3}{2}} = 125\), for \(x \gt 0\)
(ii) \((3x + 1)^{-\genfrac{}{}{}{1}{2}{3}} = \dfrac{1}{16}\).
Raise both sides to the reciprocal of the index, so the index on the left becomes \(1\).
(i)
| \(x\) | \(=\) | \(125^{\genfrac{}{}{}{1}{2}{3}}\) |
| \(\) | \(=\) | \(\left(\sqrt[3]{125}\right)^2 = 5^2 = \boldsymbol{25}\) |
(ii) Flip both sides first to remove the negative:
| \((3x + 1)^{\genfrac{}{}{}{1}{2}{3}}\) | \(=\) | \(16\) |
| \(3x + 1\) | \(=\) | \(16^{\genfrac{}{}{}{1}{3}{2}} = \left(\sqrt{16}\right)^3 = 64\) |
| \(x\) | \(=\) | \(\boldsymbol{21}\) |
Check (ii): \(64^{-\genfrac{}{}{}{1}{2}{3}} = \dfrac{1}{\left(\sqrt[3]{64}\right)^2} = \dfrac{1}{16}\) ✓
Common pitfalls
Frequently asked questions
What does a fractional index mean?
The denominator is a root and the numerator is a power: \(a^{\genfrac{}{}{}{1}{1}{2}} = \sqrt{a}\), \(a^{\genfrac{}{}{}{1}{1}{3}} = \sqrt[3]{a}\), and \(a^{\genfrac{}{}{}{1}{m}{n}} = \left(\sqrt[n]{a}\right)^m\).
How do you work out 8 to the power two thirds?
Take the cube root of \(8\), which is \(2\), then square it: \(8^{\genfrac{}{}{}{1}{2}{3}} = 4\).
How do you change a surd into index form?
The root becomes the denominator of the index and any power inside becomes the numerator: \(\sqrt[3]{x^4} = x^{\genfrac{}{}{}{1}{4}{3}}\).
What does a negative fractional index mean?
Take the reciprocal as well: \(4^{-\genfrac{}{}{}{1}{1}{2}} = \dfrac{1}{\sqrt{4}} = \dfrac{1}{2}\).
How do you solve an equation like x to the power three halves equals 8?
Raise both sides to the reciprocal power, \(\genfrac{}{}{}{1}{2}{3}\): \(x = 8^{\genfrac{}{}{}{1}{2}{3}} = 4\).