AS & A Level · Practice questions · Logarithms and the laws of logarithms

Powers, renamed.

Papers 2 and 3. Six original Cambridge-style questions on logarithms: exact values, the laws, expressions in terms of \(p\) and \(q\), and equations, ending with a simultaneous pair.

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

The method

One log, then index form, then check.

Definition: \(\log_a x = y \iff a^y = x\).
Power law first: move coefficients inside: \(k\log_a x = \log_a x^k\).
Combine: sums become products, differences become quotients.
Undo the log: \(\log_a(\text{stuff}) = c\) means \(\text{stuff} = a^c\).
Check: reject any root that makes the inside of an original log zero or negative.

Recall

[4 marks]

Find the exact value of (a) \(\log_2 64\) (b) \(\log_9 3\) (c) \(\log_5 \dfrac{1}{125}\) (d) \(\log_8 4\).

(a) \(2^6 = 64\), so \(6\). ✓
(b) \(9^{1/2} = 3\), so \(\dfrac{1}{2}\). ✓
(c) \(5^{-3} = \dfrac{1}{125}\), so \(-3\). ✓
(d) \(8^y = 4\) means \(2^{3y} = 2^2\), so \(y = \dfrac{2}{3}\). ✓

Calculation

[3 marks]

Given that \(\log_a x = p\) and \(\log_a y = q\), express in terms of \(p\) and \(q\): (a) \(\log_a(x^2 y)\) (b) \(\log_a \dfrac{\sqrt{x}}{y^3}\) (c) \(\log_a \dfrac{ax}{y}\).

(a) \(2p + q\) ✓
(b) \(\dfrac{1}{2}p - 3q\) ✓
(c) \(1 + p - q\) ✓

Calculation

[2 marks]

Express \(3\log_a 2 + \log_a 5 - \log_a 4\) as a single logarithm in its simplest form.

\(3\log_a 2 = \log_a 8\). ✓
\(\log_a \dfrac{8 \times 5}{4} = \log_a 10\). ✓

Calculation

[3 marks]

Solve the equation \(\log_5(x + 2) + \log_5(x - 2) = 1\).

\(\log_5\left[(x + 2)(x - 2)\right] = 1\), so \(x^2 - 4 = 5\). ✓
\(x^2 = 9\), so \(x = 3\) or \(x = -3\). ✓
\(x = -3\) makes \(x + 2 = -1\) negative: reject. So \(x = 3\). ✓

Analysis

[4 marks]

Solve the equation \(2\log_3 x - \log_3(x - 2) = 2\).

\(\log_3 \dfrac{x^2}{x - 2} = 2\), so \(\dfrac{x^2}{x - 2} = 9\). ✓
\(x^2 = 9x - 18\), so \(x^2 - 9x + 18 = 0\). ✓
\((x - 3)(x - 6) = 0\), so \(x = 3\) or \(x = 6\). ✓
Both exceed 2, so every log is defined: \(x = 3\) or \(x = 6\). ✓

Analysis

[6 marks]

The positive numbers \(x\) and \(y\) satisfy

\(\log_2 x + \log_2 y = 5\) and \(\log_2 x - 2\log_2 y = -1\).

(a) Show that \(xy = 32\) and \(y^2 = 2x\).

(b) Hence find the values of \(x\) and \(y\).

(a) \(\log_2(xy) = 5\), so \(xy = 2^5 = 32\). ✓
\(\log_2 \dfrac{x}{y^2} = -1\), so \(\dfrac{x}{y^2} = 2^{-1} = \dfrac{1}{2}\), i.e. \(y^2 = 2x\). ✓
(b) From \(y^2 = 2x\): \(x = \dfrac{y^2}{2}\). Substitute: \(\dfrac{y^3}{2} = 32\), so \(y^3 = 64\). ✓
\(y = 4\) and \(x = \dfrac{16}{2} = 8\). (Check: \(\log_2 8 + \log_2 4 = 3 + 2 = 5\); \(3 - 2(2) = -1\).) ✓

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.