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Or take a quick quiz: 10 questions on this subtopic in AS Maths (Edexcel), marked as you tap.
The laws of indices for all rational powers, including zero, negative and fractional indices. They are tested directly in short 2 to 4 mark questions (often 'write in the form \(ax^n\)' or 'solve by writing both sides as powers of 2') and they are needed all the time in calculus and exponentials.
\((ab)^n = a^n b^n\) and \(\left(\dfrac{a}{b}\right)^n = \dfrac{a^n}{b^n}\)
\(a^0 = 1\) for any \(a \ne 0\)
The first three laws only combine powers of the same base: \(2^3 \times 3^2\) cannot be written as a single power.
Negative and fractional indices
\(a^{-n} = \dfrac{1}{a^n}\): a negative power means 'one over'. It does not make the answer negative.
\(a^{\frac{1}{n}} = \sqrt[n]{a}\) and \(a^{\frac{m}{n}} = (\sqrt[n]{a})^m\). Take the root first to keep numbers small: \(32^{\frac{3}{5}} = 2^3 = 8\)
Deal with the minus sign, then the root, then the power: \[\begin{aligned}&\left(\frac{4}{9}\right)^{-\frac{3}{2}} \\ &= \left(\frac{9}{4}\right)^{\frac{3}{2}} \\ &= \left(\frac{3}{2}\right)^3 \\ &= \frac{27}{8}\end{aligned}\]
Thinking a negative power gives a negative number: \(2^{-2} = \dfrac{1}{4}\), not \(-4\)
Multiplying the bases: \(2^3 \times 2^4 = 2^7\), not \(4^7\)
Moving the coefficient into the power: \(\dfrac{3}{x^2} = 3x^{-2}\), not \((3x)^{-2}\)
Not dividing every term when splitting a fraction: \(\dfrac{x^2 + 1}{x} = x + x^{-1}\)
\((x^3)^2 = x^6\), not \(x^5\) or \(x^9\)
More tips
Memory tricks
Negative power: flip. Fractional power: the bottom is the root, the top is the power.
Know powers of 2 up to \(2^{10} = 1024\), powers of 3 up to \(3^5 = 243\), and \(4^3 = 64\),\(5^3 = 125\), so you can spot common bases.
Check a simplification by putting a value such as \(x = 4\) into both forms on your calculator.
Recognise \(x\sqrt{x} = x^{\frac{3}{2}}\) and \(\dfrac{1}{x\sqrt{x}} = x^{-\frac{3}{2}}\): they come up constantly in calculus.
Exam technique
All papers allow a calculator, but index questions often say 'show all stages of your working': write each step, e.g. for \(\left(\dfrac{27}{8}\right)^{\frac{2}{3}}\) write \(\left(\dfrac{27}{8}\right)^{\frac{1}{3}} = \dfrac{3}{2}\) before squaring.
Give \(a\) and \(n\) in 'the form \(ax^n\)' as exact numbers (fractions, not rounded decimals).
Rewriting in powers of \(x\) is often part (a) of a calculus question; check it carefully because part (b) depends on it.
What each grade needs
What you need to be able to do, from the first marks up to the top grade.
Grade E
Use the multiplication, division and power laws\(a^m \times a^n = a^{m+n}\), \(a^m \div a^n = a^{m-n}\) and \((a^m)^n = a^{mn}\)
Grade E
Evaluate zero and negative powers\(a^0 = 1\) and \(a^{-n} = \dfrac{1}{a^n}\), e.g. \(2^{-3} = \dfrac{1}{8}\)
Grade D
Evaluate fractional powers without a calculator\(a^{\frac{m}{n}} = (\sqrt[n]{a})^m\), e.g. \(27^{\frac{2}{3}} = 3^2 = 9\)
Grade C
Write expressions as powers of xe.g. \(\dfrac{3}{\sqrt{x}} = 3x^{-\frac{1}{2}}\), ready to differentiate or integrate.
Grade C
Split a fraction into separate powerse.g. \(\dfrac{x^2 + 4}{2\sqrt{x}} = \dfrac{1}{2}x^{\frac{3}{2}} + 2x^{-\frac{1}{2}}\)
Grade B
Solve equations using a common basee.g. \(8^x = 4^{x + 1}\) gives \(2^{3x} = 2^{2x + 2}\), so \(x = 2\)
Grade B
Solve equations with fractional or negative powerse.g. \(x^{-\frac{3}{2}} = \dfrac{1}{8}\) gives \(x^{\frac{3}{2}} = 8\), so \(x = 8^{\frac{2}{3}} = 4\)
Grade A
Simplify expressions with several different basese.g. \[\begin{aligned}\frac{2^{x + 1} \times 4^{x}}{8^{x - 1}} &= 2^{x + 1 + 2x - 3x + 3} \\ &= 2^4 \\ &= 16\end{aligned}\]
Sample questions
Written for this site in the style of Edexcel exam questions. They are not taken from real past papers.
Question 1Easy3 marks
In this question you must show all stages of your working. Solutions relying entirely on calculator technology are not acceptable.
(a) Write down the value of \(256^{\tfrac{1}{4}}\)[1]
(b) Find the value of \(256^{-\tfrac{3}{4}}\)[2]
Show the answer and mark scheme
(a)Answer: \(4\)
B1 for 4
Worked solution: \(256^{\tfrac{1}{4}} = \sqrt[4]{256} = 4\), since \(4^4 = 256\).
(b)Answer: \(\frac{1}{64}\)
M1 for a correct first step, e.g. \(\frac{1}{256^{\tfrac{3}{4}}}\) or \(\left(256^{\tfrac{1}{4}}\right)^{3} = 64\) or \(4^{-3}\)
A1 for \(\frac{1}{64}\) oe (accept 0.015625)
Worked solution: \(256^{-\tfrac{3}{4}} = \frac{1}{\left(256^{\tfrac{1}{4}}\right)^{3}} = \frac{1}{4^{3}} = \frac{1}{64}\).
Question 2Medium6 marks
In this question you must show all stages of your working. Solutions relying entirely on calculator technology are not acceptable.
(a) Given that \(\frac{\left(x\sqrt{x}\right)^{2}}{\sqrt[5]{x}} = x^{n}\), where \(x \gt 0\), find the exact value of \(n\).[3]
(b) Solve the equation \(4x^{-\tfrac{5}{2}} = \frac{4}{243}\)[3]
Show the answer and mark scheme
(a)Answer: \(n = \frac{14}{5}\)
M1 for writing the roots as fractional powers, e.g. \(\sqrt{x} = x^{\tfrac{1}{2}}\) and \(\sqrt[5]{x} = x^{\tfrac{1}{5}}\)
M1 for a correct use of the power law on the numerator: \(x^{2 \times \frac{3}{2}} = x^{3}\)
A1 for \(n = \frac{14}{5}\)
Worked solution: \(x\sqrt{x} = x^{\tfrac{3}{2}}\), so the numerator is \(x^{3}\). Dividing by \(x^{\tfrac{1}{5}}\): \(n = 3 - \frac{1}{5} = \frac{14}{5}\).
(b)Answer: \(x = 9\)
M1 for isolating the power: \(x^{-\tfrac{5}{2}} = \frac{1}{243}\)
M1 for raising both sides to the power \(-\frac{2}{5}\) (or an equivalent two-step method)