Rational Indices
Every question with a fully worked solution.
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Theory
A rational index (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 10A maths page covers evaluating without a calculator, simplifying algebraic terms, switching between surd form and index form, and solving equations and formulas.
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: \(x^{\genfrac{}{}{}{1}{1}{3}}\left(x^{\genfrac{}{}{}{1}{2}{3}} + x^{-\genfrac{}{}{}{1}{1}{3}}\right) = x + 1\); 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 64?
Raise both sides to the reciprocal power, \(\genfrac{}{}{}{1}{2}{3}\): \(x = 64^{\genfrac{}{}{}{1}{2}{3}} = \left(\sqrt[3]{64}\right)^2 = 16\).