Home / IGCSE Maths 0580 / Algebra and graphs / Indices II: algebraic and fractional indices
Cambridge IGCSE · Topic 2.4
IGCSE Maths 0580 / Topic 2 / IGCSE

Powers that obey rules.

Apply the laws of indices to letters as well as numbers: positive, zero and negative powers, then (Extended) fractional powers, and use matching bases to solve equations such as \(2^{x} = 32\).

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The key idea

Multiply: add the powers. Divide: subtract the powers. Power of a power: multiply the powers. \(x^{0} = 1\), \(x^{-n} = \dfrac{1}{x^{n}}\), and (Extended) \(x^{\frac{m}{n}} = (\sqrt[n]{x})^{m}\). Numbers in front are multiplied or divided as normal.

Section 01

The three laws.

In the lab pick a rule and move the sliders for \(m\) and \(n\). The lab writes out every factor of \(x\) so you can count them and see why the powers add, subtract or multiply. Then try the negative-power rule and watch \(x^{-n}\) turn into a fraction.

RuleLawExample
Multiply\(x^{m} \times x^{n} = x^{m + n}\)\(x^{5} \times x^{3} = x^{8}\)
Divide\(x^{m} \div x^{n} = x^{m - n}\)\(x^{7} \div x^{2} = x^{5}\)
Power of a power\((x^{m})^{n} = x^{mn}\)\((x^{4})^{3} = x^{12}\)
Worked example Power of a product
Simplify \((5x^{3})^{2}\).
  1. Everything inside the bracket is squared: \(5^{2} \times (x^{3})^{2}\).
  2. \(5^{2} = 25\) and \((x^{3})^{2} = x^{6}\).
Answer: \(25x^{6}\)
Worked example Dividing by a negative power
Simplify \(12a^{5} \div 3a^{-2}\).
  1. Numbers: \(12 \div 3 = 4\).
  2. Powers: \(a^{5 - (-2)} = a^{7}\).
Answer: \(4a^{7}\)
Worked example Two letters
Simplify \(6x^{7}y^{4} \times 5x^{-5}y\).
  1. Numbers: \(6 \times 5 = 30\).
  2. \(x\): \(7 + (-5) = 2\). \(y\): \(y = y^{1}\), so \(4 + 1 = 5\).
Answer: \(30x^{2}y^{5}\)
Examiner trap

\((5x^{3})^{2}\) is not \(5x^{6}\): the 5 is inside the bracket, so it is squared too. And the numbers in front are multiplied, never added: \(3x^{2} \times 4x^{3} = 12x^{5}\).

Section 02

Zero and negative powers.

Divide \(x^{3}\) by \(x^{3}\): the answer is 1, and the law gives \(x^{0}\). Divide \(x^{3}\) by \(x^{5}\): the answer is \(\dfrac{1}{x^{2}}\), and the law gives \(x^{-2}\). So a negative power means 'one over'.

Worked example Writing without negative powers
Write (a) \(3x^{-2}\) and (b) \((2x)^{-3}\) without negative indices.
  1. (a) Only \(x\) has the power \(-2\), so the 3 stays on top: \(3 \times \dfrac{1}{x^{2}}\).
  2. (b) The whole bracket has the power \(-3\): \(\dfrac{1}{(2x)^{3}}\).
  3. \((2x)^{3} = 8x^{3}\).
Answer: (a) \(\dfrac{3}{x^{2}}\)   (b) \(\dfrac{1}{8x^{3}}\)
Your turn
Without a calculator, find the value of \(2^{-3} \times 4^{0}\).
Examiner trap

\(3x^{-2}\) is \(\dfrac{3}{x^{2}}\), not \(\dfrac{1}{3x^{2}}\). The power belongs only to the letter it touches.

Stage 1 · Learn

Check your understanding

Four quick checks. Each correct answer earns XP and records this skill on your profile.

Quick check+10 XP

Simplify \(x^{5} \times x^{3}\).

Quick check+10 XP

Simplify \((x^{4})^{3}\).

Quick check+10 XP

What is the value of \(5^{0}\)?

Quick check+10 XP

Which is equal to \(x^{-2}\)?

Section 03

Fractional powers (Extended).

Since \(x^{\frac{1}{2}} \times x^{\frac{1}{2}} = x^{1}\), the power \(\frac{1}{2}\) must mean square root. In general the bottom of the fraction is the root and the top is the power.

Worked example Fractional power of a term (Extended)
Simplify \((8x^{6})^{\frac{2}{3}}\).
  1. Number: \(8^{\frac{2}{3}} = (\sqrt[3]{8})^{2} = 2^{2} = 4\).
  2. Letter: \((x^{6})^{\frac{2}{3}} = x^{6 \times \frac{2}{3}} = x^{4}\).
Answer: \(4x^{4}\)
Worked example Multiplying fractional powers (Extended)
Simplify \(4x^{\frac{1}{2}} \times 3x^{-\frac{3}{2}}\).
  1. Numbers: \(4 \times 3 = 12\).
  2. Powers: \(\frac{1}{2} + (-\frac{3}{2}) = -1\), so \(x^{-1} = \dfrac{1}{x}\).
Answer: \(12x^{-1} = \dfrac{12}{x}\)
Your turn
(Extended) Simplify \(\left(\dfrac{16x^{8}}{81}\right)^{\frac{3}{4}}\).
Examiner trap

\(8^{\frac{2}{3}}\) is not \(8 \times \frac{2}{3}\). The fraction is an instruction: cube root, then square.

Section 04

The unknown in the power.

Write both sides as powers of the same base. If \(a^{p} = a^{q}\) then \(p = q\), so you can solve an ordinary equation in the powers.

Worked example Same base
Solve \(2^{x} = 32\).
  1. \(32 = 2^{5}\), so \(2^{x} = 2^{5}\).
  2. Compare powers.
Answer: \(x = 5\)
Worked example A fractional answer (Extended)
Solve \(32^{x} = 2\).
  1. \(32 = 2^{5}\), so \((2^{5})^{x} = 2^{5x}\).
  2. \(2^{5x} = 2^{1}\), so \(5x = 1\).
Answer: \(x = \dfrac{1}{5}\)
Worked example Different bases (Extended)
Solve \(5^{x + 1} = 25^{x}\).
  1. \(25 = 5^{2}\), so \(25^{x} = 5^{2x}\).
  2. \(5^{x + 1} = 5^{2x}\), so \(x + 1 = 2x\).
  3. Subtract \(x\) from both sides.
Answer: \(x = 1\)
Your turn
(Extended) Solve \(4^{x} = 8\).
Examiner trap

\(\dfrac{1}{27}\) is \(3^{-3}\), not \(3^{3}\). A fraction 'one over' needs a negative power.

Stage 2 · Exam

Exam-style questions

Unlocks once the four checks above are done. Worth more XP, written to IGCSE Paper 2 and Paper 4 standard.

Finish the four checks above to unlock the exam questions
Exam style+20 XP

Simplify \((3x^{2}y)^{3}\).

Exam style+20 XP

Simplify \(12a^{5} \div 3a^{-2}\).

Exam style+20 XP

(Extended) Without a calculator, find the value of \(8^{-\frac{2}{3}}\).

Exam style+20 XP

Solve \(3^{2x - 1} = \dfrac{1}{27}\).

Skill unlocked

Index laws, applied.

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Go deeper · practice
Six original Cambridge-style questions
The three laws with letters, negative and zero powers, fractional powers (Extended) and equations with the unknown in the power.