3 exam-style questions plus unlimited generated ones on this topic, at up to four difficulty levels, with full mark schemes. Notes and worksheets are open to everyone. Practice needs a free account.
Or take a quick quiz: 11 questions on this subtopic in AS Maths (Edexcel), marked as you tap.
The three laws of logarithms let you combine, split and simplify logs, and solve equations that contain them. Expect 'write as a single logarithm', 'express in terms of \(p\) and \(q\)' and 'solve' questions, usually worth 3 to 6 marks and often with the instruction to show all stages of your working.
Key facts
\(\log_a x + \log_a y = \log_a (xy)\)
\(\log_a x - \log_a y = \log_a \left(\dfrac{x}{y}\right)\)
For \(x, y > 0\): \(\log_a x + \log_a y = \log_a (xy)\)
\[\log_a x - \log_a y = \log_a \left(\frac{x}{y}\right)\]
\(k\log_a x = \log_a (x^k)\) for any number \(k\). In particular \(\log_a \dfrac{1}{x} = -\log_a x\) (\(k = -1\)) and \(\log_a \dfrac{1}{\sqrt{x}} = -\dfrac{1}{2}\log_a x\) (\(k = -\dfrac{1}{2}\)).
These come from the laws of indices: multiplying powers of the same base adds the indices, so multiplying numbers adds their logs.
Useful facts: \(\log_a a = 1\), \(\log_a 1 = 0\) and \(\log_a (a^k) = k\)
Writing as a single logarithm
Use the power law first (move any number in front inside as a power), then combine with the addition and subtraction laws.
Plus becomes times, minus becomes divide, a number in front becomes a power.
Check a 'single log' answer on your calculator: pick a value such as \(x = 2\) and evaluate both the original expression and your answer.
Exam technique
In 'show that' questions, use one law per line so every step can be seen.
When you reject a root, write the reason, e.g. '\(x = -2\) is rejected because \(\log_2 (-2)\) is undefined'.
Many log questions say 'show all stages of your working': a calculator's equation solver alone scores nothing.
What each grade needs
What you need to be able to do, from the first marks up to the top grade.
Grade E
Use the addition and subtraction laws\(\log_a x + \log_a y = \log_a xy\) and \(\log_a x - \log_a y = \log_a \dfrac{x}{y}\)
Grade D
Use the power law, including negative powers\(3\log_a x = \log_a x^3\), \(-\log_a x = \log_a \dfrac{1}{x}\) and \(-\dfrac{1}{2}\log_a x = \log_a \dfrac{1}{\sqrt{x}}\)
Grade C
Write an expression as a single logarithmUse the power law first, then combine, writing plain numbers as logs, e.g. \(2 = \log_3 9\)
Grade C
Express logarithms in terms of given logsGiven \(\log_a 2 = p\) and \(\log_a 3 = q\), write \(\log_a 12 = 2p + q\)
Grade B
Solve equations involving logarithmsCombine into one log, then change to index form, e.g. \(\log_2 x(x - 2) = 3 \Rightarrow x(x - 2) = 8\)
Grade A
Solve log equations leading to quadraticsForm and solve a quadratic, then check every root in the original equation, rejecting any that make a log's argument zero or negative.
Sample questions
Written for this site in the style of Edexcel exam questions. They are not taken from real past papers.
Question 1Easy4 marks
A student solves the equation \(\log_2(x + 3) + \log_2 x = 2\) as follows: \(\log_2(x + 3 + x) = 2\), so \(2x + 3 = 4\) and \(x = \frac{1}{2}\).
(a) Explain the error in the student’s working.[1]
(b) Solve the equation correctly, giving a reason for rejecting any value.[3]
Show the answer and mark scheme
(a)Answer: \(\log_2 a + \log_2 b = \log_2(ab)\), not \(\log_2(a + b)\).
B1 for the sum of logs is the log of the product: \(\log_2(x + 3) + \log_2 x = \log_2[x(x + 3)]\)
Worked solution: The law of logarithms is \(\log_2 a + \log_2 b = \log_2(ab)\). The student added \(x + 3\) and \(x\) instead of multiplying them.
A1 for \(x = 1\) only, rejecting \(x = -4\) because \(\log_2 x\) (or \(\log_2(x + 3)\)) is not defined for negative arguments
Worked solution: \(\log_2[x(x + 3)] = 2 \Rightarrow x(x + 3) = 2^2 = 4 \Rightarrow x^2 + 3x - 4 = 0 \Rightarrow (x + 4)(x - 1) = 0\). \(x = -4\) is rejected because \(\log_2(-4)\) is not defined. So \(x = 1\).
Question 2Medium8 marks
(a) Given that \(a\), \(x\) and \(y\) are positive and \(a \ne 1\), prove that \[\log_{a} x + \log_{a} y = \log_{a}(xy).\][3]
(b) Prove similarly that \(k\log_{a} x = \log_{a}(x^{k})\) for any real number \(k\).[2]
(c) Given that \(\log_{a} 2 = s\) and \(\log_{a} 3 = t\), find, in terms of \(s\) and \(t\), the value of \(\log_{a}\left(\dfrac{\sqrt{18}}{a^{2}}\right)\).[3]
Show the answer and mark scheme
(a)Answer: Proof
M1 for letting \(\log_{a} x = p\) and \(\log_{a} y = q\), so \(x = a^{p}\) and \(y = a^{q}\)
M1 for \(xy = a^{p}a^{q} = a^{p + q}\)
A1* for \(\log_{a}(xy) = p + q = \log_{a} x + \log_{a} y\)
Worked solution: Let \(p = \log_{a} x\) and \(q = \log_{a} y\), so that \(x = a^{p}\) and \(y = a^{q}\). Then \(xy = a^{p}a^{q} = a^{p + q}\), so \(\log_{a}(xy) = p + q = \log_{a} x + \log_{a} y\).
(b)Answer: Proof
M1 for \(x^{k} = (a^{p})^{k} = a^{kp}\)
A1* for \(\log_{a}(x^{k}) = kp = k\log_{a} x\)
Worked solution: With \(x = a^{p}\): \(x^{k} = (a^{p})^{k} = a^{kp}\), so \(\log_{a}(x^{k}) = kp = k\log_{a} x\).
Solve the simultaneous equations \[\log_{3} x + \log_{3} y = 5, \qquad x - y = 18.\]
Find the values of \(x\) and \(y\), justifying the rejection of any values.[6]
Show the answer and mark scheme
Answer: \(x = 27, \ y = 9\)
M1 for \(\log_{3} xy = 5\)
A1 for \(xy = 243\)
M1 for substituting \(x = y + 18\) to form a quadratic: \(y^2 + 18y - 243 = 0\)
A1 for \(y = 9\) or \(y = -27\)
B1 for rejecting \(y = -27\) because \(\log_{3} y\) is not defined for \(y \le 0\)
A1 for \(x = 27, \ y = 9\)
Worked solution: \(\log_{3} xy = 5 \Rightarrow xy = 3^{5} = 243\). With \(x = y + 18\): \(y(y + 18) = 243 \Rightarrow (y - 9)(y + 27) = 0\). \(y = -27\) is rejected (the logarithm of a negative number is not defined), so \(y = 9\) and \(x = 27\).