IGCSE · Practice questions · Indices II: algebraic and fractional indices

Add, subtract, multiply.

Six original Cambridge-style questions on the laws of indices with algebra, including negative and fractional indices and index equations.

Original questions All questions on this page are original work, written in the Cambridge IGCSE style. They are not from past papers. They test the same concepts and skills the syllabus rewards.

The method

Numbers as normal, powers by the laws.

Multiply: add the powers; multiply the numbers in front.
Divide: subtract the powers; divide the numbers in front.
Bracket to a power: every factor inside gets the power.
Negative and zero: \(x^{-n} = \dfrac{1}{x^{n}}\), \(x^{0} = 1\).
Fractional (Extended): bottom is the root, top is the power.
Equations: same base on both sides, then equate the powers.

Recall

[3 marks]

Simplify (a) \(a^{4} \times a^{6}\) (b) \(b^{9} \div b^{3}\) (c) \((c^{2})^{5}\).

(a) \(a^{10}\) ✓
(b) \(b^{6}\) ✓
(c) \(c^{10}\) ✓

Calculation

[2 marks]

Simplify \(6x^{7}y^{4} \times 5x^{-5}y\).

\(6 \times 5 = 30\); \(x^{7 + (-5)} = x^{2}\); \(y^{4 + 1} = y^{5}\) ✓
\(30x^{2}y^{5}\) ✓

Calculation

[2 marks]

Find the value of \(x\) when (a) \(2^{x} = 32\) (b) \(3^{x} = \dfrac{1}{9}\).

(a) \(32 = 2^{5}\), so \(x = 5\) ✓
(b) \(\dfrac{1}{9} = 3^{-2}\), so \(x = -2\) ✓

Calculation

[3 marks]

(Extended) Without a calculator, find the value of (a) \(64^{\frac{1}{3}}\) (b) \(9^{-\frac{3}{2}}\) (c) \(\left(\dfrac{8}{27}\right)^{\frac{2}{3}}\).

(a) \(\sqrt[3]{64} = 4\) ✓
(b) \(9^{\frac{3}{2}} = 3^{3} = 27\), so \(9^{-\frac{3}{2}} = \dfrac{1}{27}\) ✓
(c) \(\left(\dfrac{2}{3}\right)^{2} = \dfrac{4}{9}\) ✓

Calculation

[3 marks]

(Extended) Simplify (a) \((27x^{6})^{\frac{2}{3}}\) (b) \(4x^{\frac{1}{2}} \times 3x^{-\frac{3}{2}}\).

(a) \(27^{\frac{2}{3}} = 9\) and \(x^{6 \times \frac{2}{3}} = x^{4}\), so \(9x^{4}\) ✓
(b) \(12x^{\frac{1}{2} - \frac{3}{2}} = 12x^{-1} = \dfrac{12}{x}\) ✓

Analysis

[5 marks]

(Extended)

(a) Solve \(9^{x + 1} = 27^{x}\).

(b) Solve \(4^{x} \times 2^{x + 3} = 64\).

(a) \(9^{x + 1} = 3^{2x + 2}\) and \(27^{x} = 3^{3x}\) ✓
\(2x + 2 = 3x\), so \(x = 2\) ✓
(b) \(4^{x} = 2^{2x}\), so the left side is \(2^{2x + x + 3} = 2^{3x + 3}\) ✓
\(64 = 2^{6}\), so \(3x + 3 = 6\) ✓
\(x = 1\) ✓

Mark this once you have attempted all six and checked your working. It records a Practiced badge on the topic and adds a one-time bonus. Revealing the solutions alone does not count.