Edexcel A level Maths (9MA0) · Pure mathematics › Exponentials and logarithms
Practise Logarithms. unlimited generated questions on this subtopic, at up to four difficulty levels, with full mark schemes and a progress tracker. Free, no account needed.
Written for this site in the style of Edexcel exam questions. They are not taken from real past papers.
Question 1Easy5 marks
(a) Write \(4^{2} = 16\) in logarithmic form.[1]
(b) Find the exact value of \(\log_{25} 5\).[2]
(c) Solve the equation \(\log_{x} 16 = 2\).[2]
Show the answer and mark scheme
(a)Answer: \(\log_{4} 16 = 2\)
B1 for \(\log_{4} 16 = 2\)
Worked solution: \(4^{2} = 16 \iff \log_{4} 16 = 2\) (\(\log_a y = x \iff a^x = y\)).
(b)Answer: \(\frac{1}{2}\)
M1 for writing \(25^{x} = 5\) with both sides as powers of a common base
A1 for \(\frac{1}{2}\)
Worked solution: Let \(x = \log_{25} 5\). Then \(25^{x} = 5\), and writing both as powers of the same number gives \(x = \frac{1}{2}\) (check: \(25^{\frac{1}{2}} = 5\)).
(c)Answer: \(x = 4\)
M1 for converting to index form
A1 for \(x = 4\)
Worked solution: \(x^{2} = 16 \Rightarrow x = 4\) (the base of a logarithm must be positive).
Question 2Medium6 marks
The function \(\mathrm{f}\) is defined by \[\mathrm{f}(x) = \ln(2x - 5) + 4, \quad x \gt \frac{5}{2}.\]
(a) State the equation of the asymptote to the graph of \(y = \mathrm{f}(x)\).[1]
(b) Find \(\mathrm{f}^{-1}(x)\) and state its domain.[3]
(c) Solve the equation \(\mathrm{f}(x) = 6\), giving your answer in exact form.[2]
Show the answer and mark scheme
(a)Answer: \(x = \frac{5}{2}\)
B1 for \(x = \frac{5}{2}\)
Worked solution: \(\ln u \to -\infty\) as \(u \to 0^{+}\), so the asymptote is where \(2x - 5 = 0\): \(x = \frac{5}{2}\).
M1 for rearranging \(y = \ln(2x - 5) + 4\) to \(2x - 5 = \mathrm{e}^{y - 4}\)
A1 for \(\mathrm{f}^{-1}(x) = \frac{\mathrm{e}^{x - 4} + 5}{2}\)
B1 for domain \(x \in \mathbb{R}\) (the range of \(\mathrm{f}\))
Worked solution: \(y = \ln(2x - 5) + 4 \Rightarrow 2x - 5 = \mathrm{e}^{y - 4}\), so \(\mathrm{f}^{-1}(x) = \frac{\mathrm{e}^{x - 4} + 5}{2}\). The range of \(\mathrm{f}\) is all real numbers, so the domain of \(\mathrm{f}^{-1}\) is \(x \in \mathbb{R}\).
(c)Answer: \(x = \frac{\mathrm{e}^{2} + 5}{2}\)
M1 for \(2x - 5 = \mathrm{e}^{2}\)
A1 for \(x = \frac{\mathrm{e}^{2} + 5}{2}\)
Worked solution: \(\ln(2x - 5) = 2 \Rightarrow 2x - 5 = \mathrm{e}^{2} \Rightarrow x = \frac{\mathrm{e}^{2} + 5}{2}\).
Question 3Hard9 marks
The diagram shows part of the curve with equation \(y = c + \log_{a}(x + 1)\), where \(a\) and \(c\) are constants and \(a \gt 1\). The line \(x = -1\) (dashed) is an asymptote to the curve. The curve passes through the points \(A(0, -1)\) and \(B(15, 1)\).
[object Object]
(a) Find the value of \(a\) and the value of \(c\).[4]
(b) Find the exact \(x\)-coordinate of the point where the curve crosses the \(x\)-axis.[2]
(c) The curve is the graph of \(y = \mathrm{f}(x)\), \(x \gt -1\). Find \(\mathrm{f}^{-1}(x)\), stating its domain.[3]
Show the answer and mark scheme
(a)Answer: \(a = 4, \ c = -1\)
M1 for substituting both points: \(-1 = c + \log_{a} 1\) and \(1 = c + \log_{a} 16\)
M1 for subtracting: \(\log_{a} 16 - \log_{a} 1 = 2\) or equivalent
A1 for \(a = 4\)
A1 for \(c = -1\)
Worked solution: Subtracting the equations: \(\log_{a}\frac{16}{1} = 2 \Rightarrow a^{2} = 16 \Rightarrow a = 4\). Then \(c = -1 - \log_{4} 1 = -1 - 0 = -1\).
(b)Answer: \(x = 3\)
M1 for \(\log_{4}(x + 1) = 1\)
A1 for \(x = 3\)
Worked solution: \(0 = -1 + \log_{4}(x + 1) \Rightarrow x + 1 = 4 \Rightarrow x = 3\).
Worked solution: \(y = -1 + \log_{4}(x + 1) \Rightarrow x + 1 = 4^{y + 1}\), so \(\mathrm{f}^{-1}(x) = 4^{x + 1} - 1\), with domain the range of \(\mathrm{f}\), i.e. all real \(x\).