Fractional Indices
A fractional index stands for a root: \(a^{\frac{1}{n}} = \sqrt[n]{a}\) and \(a^{\frac{m}{n}} = (\sqrt[n]{a})^m\), so \(27^{\frac{2}{3}} = 9\). This NSW Year 9 Advanced maths subtopic connects surds and indices (NESA outcome MA5‑IND‑P‑02).
You will switch between surd form and index form, evaluate powers such as \(\left(\frac{8}{27}\right)^{-\frac{2}{3}}\) without a calculator, and simplify expressions like \((8a^6b^3)^{\frac{2}{3}}\) and \((16t^8)^{-\frac{3}{4}} \times 4t^7\) with the index laws. You will solve \(8^x = 32\), write a mix of roots as one power of \(x\), and use \(S = 6V^{\frac{2}{3}}\) for a cube's surface area. It is part of Year 9 Advanced Indices and Scientific Notation.
Theory
A fractional index is a root: \(a^{1/n} = \sqrt[n]{a}\), and \(a^{m/n} = \left(\sqrt[n]{a}\right)^m\). This Year 9 Advanced maths page covers evaluating without a calculator, simplifying with the index laws, writing surds as powers and solving index equations.
\(a^{1/2}\) is the number that, multiplied by itself, gives \(a\). By the multiplication law \(a^{1/2} \times a^{1/2} = a^1\), so \(a^{1/2} = \sqrt{a}\). In the same way \(a^{1/3} = \sqrt[3]{a}\) and, in general, \(a^{1/n} = \sqrt[n]{a}\), the \(n\)th root of \(a\).
In \(a^{m/n}\), the denominator \(n\) is the root and the numerator \(m\) is the power. They can be done in either order, \(\left(\sqrt[n]{a}\right)^m\) or \(\sqrt[n]{a^m}\), but taking the root first keeps the numbers small.
A root written with the radical sign is a surd form. Changing between surd form and index form lets you use all the index laws on roots.
All the index laws work with fractional indices. Add, subtract and multiply the fractions with the usual fraction rules. A negative fractional index means a reciprocal as well as a root.
It is part of the preparation for Year 11 Advanced. Questions usually say the pronumerals are positive, so the roots are defined.
| Surd form | Index form |
|---|---|
| \(\sqrt{w}\) | \(w^{1/2}\) |
| \(\sqrt[5]{w}\) | \(w^{1/5}\) |
| \(\sqrt[3]{w^4}\) | \(w^{4/3}\) |
| \(w^2\sqrt{w}\) | \(w^{5/2}\) |
| \(\dfrac{1}{\sqrt[4]{w^3}}\) | \(w^{-3/4}\) |
How to evaluate \(a^{m/n}\)
- Negative index? Take the reciprocal first (flip a fraction).
- Root: take the \(n\)th root of the base.
- Power: raise the result to the power \(m\).
How to simplify
- Surds to indices: write every root as a fractional index.
- Brackets: raise each factor. Multiply each index by the fraction, and evaluate the number part by root then power.
- Multiply or divide: add or subtract the indices using a common denominator.
- Positive indices: move factors with negative indices across the fraction line. Convert back to surd form only if asked.
Equations and formulas
- Same base: write both sides as powers of one prime, including any surd, then equate the indices (Example 4).
- Formulas with fractional powers: substitute and evaluate by root then power. To work backwards, isolate the power and raise both sides to the reciprocal index.
(i) \(32^{3/5}\)
(ii) \(\left(\dfrac{16}{81}\right)^{-3/4}\).
Take the root (the denominator) first, then the power (the numerator).
(i)
| \(32^{3/5}\) | \(=\) | \(\left(\sqrt[5]{32}\right)^3\) |
| \(\) | \(=\) | \(2^3 = \mathbf{8}\) |
(ii) The negative index flips the fraction first.
| \(\left(\dfrac{16}{81}\right)^{-3/4}\) | \(=\) | \(\left(\dfrac{81}{16}\right)^{3/4}\) |
| \(\) | \(=\) | \(\left(\sqrt[4]{\dfrac{81}{16}}\right)^{3}\) |
| \(\) | \(=\) | \(\left(\dfrac{3}{2}\right)^{3} = \mathbf{\dfrac{27}{8}}\) |
\(2^5 = 32\), \(3^4 = 81\) and \(2^4 = 16\). Knowing small powers makes these quick.
(i) \((125x^9y^{-6})^{2/3}\)
(ii) \(10a^{5/6} \div 4a^{1/3}\).
(i) Raise every factor: the number by the root-then-power rule, the pronumerals by multiplying indices.
| \(125^{2/3}\) | \(=\) | \((\sqrt[3]{125})^2 = 5^2 = 25\) |
| \((x^9)^{2/3}\) | \(=\) | \(x^{6}\) |
| \((y^{-6})^{2/3}\) | \(=\) | \(y^{-4}\) |
| \((125x^9y^{-6})^{2/3}\) | \(=\) | \(\boldsymbol{\dfrac{25x^6}{y^4}}\) |
(ii) Divide the coefficients and subtract the fractional indices with a common denominator.
| \(\dfrac{5}{6} - \dfrac{1}{3}\) | \(=\) | \(\dfrac{5}{6} - \dfrac{2}{6} = \dfrac{1}{2}\) |
| \(10a^{5/6} \div 4a^{1/3}\) | \(=\) | \(\boldsymbol{\dfrac{5}{2}a^{1/2}}\) |
\(\dfrac{5}{2}a^{1/2}\) can also be written \(\dfrac{5\sqrt{a}}{2}\).
Change every root into a fractional index, then add and subtract with a common denominator of \(6\).
| \(\sqrt[3]{x}\) | \(=\) | \(x^{1/3}\) |
| \(\sqrt{x^5}\) | \(=\) | \(x^{5/2}\) |
| \(2 + \dfrac{1}{3} - \dfrac{5}{2}\) | \(=\) | \(\dfrac{12}{6} + \dfrac{2}{6} - \dfrac{15}{6} = -\dfrac{1}{6}\) |
| \(\dfrac{x^2\sqrt[3]{x}}{\sqrt{x^5}}\) | \(=\) | \(\boldsymbol{x^{-1/6}}\) |
With a positive index this is \(\dfrac{1}{x^{1/6}}\) \(= \dfrac{1}{\sqrt[6]{x}}\).
(i) \(27^{x} = 9\sqrt{3}\)
(ii) \(16^{x} = \dfrac{1}{8}\).
Write both sides as powers of the same prime, then equate the indices. The answer may be a fraction.
(i) Everything is a power of \(3\):
| \(27^x\) | \(=\) | \((3^3)^x = 3^{3x}\) |
| \(9\sqrt{3}\) | \(=\) | \(3^2 \times 3^{1/2} = 3^{5/2}\) |
| \(3x\) | \(=\) | \(\dfrac{5}{2}\) |
| \(x\) | \(=\) | \(\mathbf{\dfrac{5}{6}}\) |
(ii) Everything is a power of \(2\):
| \((2^4)^x\) | \(=\) | \(2^{-3}\) |
| \(4x\) | \(=\) | \(-3\) |
| \(x\) | \(=\) | \(\mathbf{-\dfrac{3}{4}}\) |
Check (ii): \(16^{-3/4}\) \(= \dfrac{1}{(\sqrt[4]{16})^3}\) \(= \dfrac{1}{2^3}\) \(= \dfrac{1}{8}\).
Common pitfalls
Frequently asked questions
What does a fractional index mean?
The denominator is a root and the numerator is a power: \(a^{m/n} = \left(\sqrt[n]{a}\right)^m\). For example \(8^{2/3} = 2^2 = 4\).
Why is x to the power one half the same as the square root of x?
\(x^{1/2} \times x^{1/2}\) \(= x^{1/2 + 1/2}\) \(= x^1\). A number that multiplies by itself to give \(x\) is \(\sqrt{x}\).
Should I take the root or the power first?
Either order gives the same answer, but the root first keeps numbers small: \(81^{3/4} = 3^3 = 27\).
How do you write a surd with a fractional index?
The root becomes the denominator and any power inside becomes the numerator: \(\sqrt[3]{w^4} = w^{4/3}\).
What does a negative fractional index mean?
Take the reciprocal and apply the root and power: \(25^{-1/2}\) \(= \dfrac{1}{\sqrt{25}}\) \(= \dfrac{1}{5}\).