IGCSE · Practice questions · Laws of logarithms and change of base

Combine, expand, switch.

Six original Cambridge-style questions on the laws of logarithms: product, quotient and power laws, writing constants as logs, combining into a single logarithm, expanding in terms of given logs and changing the base, all without a calculator.

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

Powers up, constants in, then combine.

Three laws: plus means multiply, minus means divide, a coefficient is a power.
Special values: \(\log_a a = 1\), \(\log_a 1 = 0\), \(\log_a a^{k} = k\).
Single log: move coefficients up, write constants as logs in the same base, then combine.
Change of base: \(\log_a b = \dfrac{\log_c b}{\log_c a}\), choosing \(c\) so both logs are easy; \(\log_a b = \dfrac{1}{\log_b a}\).
Never split \(\log (x + y)\), and never treat \((\log x)^{2}\) as \(2\log x\).

Recall

[2 marks]

Without using a calculator, find the value of

(a) \(\log_4 2 + \log_4 32\) (b) \(\log_7 98 - \log_7 2\).

(a) \(\log_4 (2 \times 32) = \log_4 64 = 3\), since \(4^{3} = 64\). ✓
(b) \(\log_7 \dfrac{98}{2} = \log_7 49 = 2\), since \(7^{2} = 49\). ✓

Calculation

[3 marks]

Write \(1 + 3\lg x - \tfrac{1}{2}\lg y\) as a single logarithm to base 10.

Power law: \(3\lg x = \lg x^{3}\) and \(\tfrac{1}{2}\lg y = \lg \sqrt{y}\). ✓
Constant: \(1 = \lg 10\). ✓
Combine: \(\lg 10 + \lg x^{3} - \lg \sqrt{y} = \lg \dfrac{10x^{3}}{\sqrt{y}}\). ✓

Calculation

[4 marks]

Given that \(\log_a 5 = m\) and \(\log_a 2 = n\), express in terms of \(m\) and \(n\)

(a) \(\log_a 40\) (b) \(\log_a \dfrac{25}{4a}\).

(a) \(40 = 2^{3} \times 5\), so \(\log_a 40 = 3\log_a 2 + \log_a 5 = 3n + m\). ✓
(b) \(\log_a \dfrac{25}{4a} = \log_a 5^{2} - \log_a 2^{2} - \log_a a\). ✓
\(= 2m - 2n - 1\), using \(\log_a a = 1\). ✓

Calculation

[4 marks]

Without using a calculator, find the exact value of

(a) \(\log_{27} 9\) (b) \(\log_8 \dfrac{1}{4}\).

(a) Change to base 3: \(\log_{27} 9 = \dfrac{\log_3 9}{\log_3 27} = \dfrac{2}{3}\). ✓
(b) Change to base 2: \(\log_8 \dfrac{1}{4} = \dfrac{\log_2 2^{-2}}{\log_2 2^{3}} = -\dfrac{2}{3}\). ✓
Check: \(27^{2/3} = 3^{2} = 9\) and \(8^{-2/3} = \dfrac{1}{2^{2}} = \dfrac{1}{4}\). ✓
(c) \(\dfrac{2}{3} + \left(-\dfrac{2}{3}\right) = 0\). ✓

Analysis

[4 marks]

Show that \(\dfrac{1}{\log_2 6} + \dfrac{1}{\log_3 6} = 1\).

Swap rule: \(\dfrac{1}{\log_2 6} = \log_6 2\) and \(\dfrac{1}{\log_3 6} = \log_6 3\). ✓
So the left-hand side is \(\log_6 2 + \log_6 3\). ✓
Product law: \(\log_6 (2 \times 3) = \log_6 6\). ✓
\(\log_6 6 = 1\), as required. ✓

Analysis

[6 marks]

(a) Show that \(\log_9 x = \tfrac{1}{2}\log_3 x\) and that \(\log_{27} x = \tfrac{1}{3}\log_3 x\).

(b) Hence show that \(\log_3 x + \log_9 x^{2} + \log_{27} x^{3}\) can be written as \(\log_3 x^{k}\), stating the value of \(k\).

(a) \(\log_9 x = \dfrac{\log_3 x}{\log_3 9} = \dfrac{\log_3 x}{2}\) and \(\log_{27} x = \dfrac{\log_3 x}{\log_3 27} = \dfrac{\log_3 x}{3}\). ✓
(b) Power law then (a): \(\log_9 x^{2} = 2\log_9 x = 2 \times \tfrac{1}{2}\log_3 x = \log_3 x\). ✓
Similarly \(\log_{27} x^{3} = 3\log_{27} x = 3 \times \tfrac{1}{3}\log_3 x = \log_3 x\). ✓
So the expression is \(3\log_3 x = \log_3 x^{3}\), and \(k = 3\). ✓
(c) With \(x = \sqrt{3} = 3^{1/2}\): \(3\log_3 3^{1/2} = 3 \times \tfrac{1}{2} = \dfrac{3}{2}\). ✓

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.