IGCSE · Practice questions · Indices: the laws of indices

Same base, add the powers.

Six original Cambridge-style questions on indices: zero, negative and (Extended) fractional indices and the laws of indices.

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

Same base first, then the laws.

Multiply: add indices. Divide: subtract. Power of a power: multiply.
Zero: \(a^{0} = 1\). Negative: \(a^{-n} = \dfrac{1}{a^{n}}\); a fraction flips.
Fractional (Extended): bottom = root, top = power: root first.
Equations: write both sides as powers of the same base, then compare indices.
Check: a negative index never makes the answer negative.

Recall

[3 marks]

Find the value of (a) \(6^{0}\) (b) \(4^{-2}\) (c) \(10^{-3}\).

(a) 1 ✓
(b) \(\dfrac{1}{4^{2}} = \dfrac{1}{16}\) ✓
(c) \(\dfrac{1}{1000} = 0.001\) ✓

Calculation

[3 marks]

Simplify, leaving each answer in index form: (a) \(7^{5} \times 7^{-2}\) (b) \((5^{-2})^{3}\) (c) \(9^{4} \div 9^{6}\).

(a) \(7^{5 + (-2)} = 7^{3}\) ✓
(b) \(5^{-2 \times 3} = 5^{-6}\) ✓
(c) \(9^{4-6} = 9^{-2}\) ✓

Calculation

[3 marks]

Find the value of (a) \(2^{-3} \times 2^{4}\) (b) \((2^{3})^{2}\) (c) \(2^{3} \div 2^{4}\).

(a) \(2^{1} = 2\) ✓
(b) \(2^{6} = 64\) ✓
(c) \(2^{-1} = \dfrac{1}{2}\) ✓

Calculation

[3 marks]

(Extended) Find the value of (a) \(81^{\frac{1}{2}}\) (b) \(125^{\frac{2}{3}}\) (c) \(16^{-\frac{1}{2}}\).

(a) \(\sqrt{81} = 9\) ✓
(b) \(\sqrt[3]{125} = 5\), \(5^{2} = 25\) ✓
(c) \(\sqrt{16} = 4\), so \(\dfrac{1}{4}\) ✓

Analysis

[3 marks]

Find \(x\) when (a) \(3^{x} = \dfrac{1}{81}\) (b) \(5^{x} = 1\) (c) (Extended) \(4^{x} = 32\).

(a) \(\dfrac{1}{81} = 3^{-4}\), so \(x = -4\) ✓
(b) \(5^{0} = 1\), so \(x = 0\) ✓
(c) \(2^{2x} = 2^{5}\), so \(2x = 5\) and \(x = \dfrac{5}{2}\) ✓

Analysis

[4 marks]

A digital image is \(2^{11}\) pixels wide and \(2^{10}\) pixels high. Each pixel uses \(2^{5}\) bits of memory. There are \(2^{3}\) bits in a byte, and \(2^{20}\) bytes in a megabyte (MB).

(a) Find the number of bits in the image, as a power of 2. (b) Find the size of the image in megabytes.

(a) Pixels: \(2^{11} \times 2^{10} = 2^{21}\). Bits: \(2^{21} \times 2^{5} = 2^{26}\) ✓
(b) Bytes: \(2^{26} \div 2^{3} = 2^{23}\) ✓
Megabytes: \(2^{23} \div 2^{20} = 2^{3} = 8\) MB ✓

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.