Edexcel A level Maths (9MA0) · Pure mathematics › Exponentials and logarithms
Practise Laws of logarithms. 3 exam-style questions plus unlimited generated ones on this subtopic, at up to four difficulty levels, with full mark schemes and a progress tracker. Free, no account needed.
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 \(\log_{a}(x^{k}) = kp = k\log_{a} x\).