June 2025 Paper 1 Q6
6 In this question you must show detailed reasoning.
Solve the following equations.
| Scheme | Marks | AO |
|---|---|---|
| DR \(\log_3(x + 32) - \log_3 x - \log_3 2 + \log_3 2 = 2\) | M1 | 3.1a |
| \(\log_3\left(\dfrac{x + 32}{x}\right) = 2\) | M1 | 2.1 |
| \(\dfrac{x + 32}{x} = 9\) or \(3^2\) | M1 | 1.1 |
| \(x = 4\) | A1 | 1.1 |
| [4] |
Notes
M1: Correct use of \(\log(ab) = \log a + \log b\)
M1: Correct use of \(\log\left(\tfrac{a}{b}\right) = \log a - \log b\)
These first two M marks may appear in either order.
M1: Correctly remove logs from their equation
A1: Dependent on all 3 method marks.
Alternative method for M1M1
| Scheme | Marks |
|---|---|
| \(\log_3\left(\dfrac{2(x + 32)}{2x}\right) = 2\) | M2 |
M2: Must see ‘handling’ of 2 in numerator and denominator to award these marks in one step (e.g. writing and cancelling)
Alternative method
| Scheme | Marks |
|---|---|
| \(\log_3\left(\dfrac{x + 32}{2x}\right) + \log_3 2 = 2\) | M1 |
| \(\log_3\left(\dfrac{x + 32}{2x}\right) = 2 - \log_3 2\) \(\dfrac{x + 32}{2x} = 3^{2 - \log_3 2}\ \left[= \dfrac{3^2}{3^{\log_3 2}}\right]\) | M1 |
| \(\dfrac{x + 32}{2x} = \dfrac{9}{2}\ [= 4.5]\) or \(\dfrac{x + 32}{x} = 9\) | M1 |
| \(x = 4\) | A1 |
M1: Correct use of \(\log\left(\tfrac{a}{b}\right) = \log a - \log b\)
M1: Correctly remove logs from all terms involving \(x\)
M1: Correctly evaluate and remove all remaining logs
A1: Dependent on all 3 method marks.
NB candidates who do not show ‘handling’ of the terms in \(\log_3 2\) can score max. 2/4
| Scheme | Marks | AO |
|---|---|---|
| DR \(y^{-\frac{3}{2}} = \tfrac{1}{8}\) oe | M1 | 1.1 |
| \(y = \left(\tfrac{1}{8}\right)^{-\frac{2}{3}}\) or \(y = \sqrt[-\frac{3}{2}]{\tfrac{1}{8}}\) or \(y^{\frac{3}{2}} = 8 \Rightarrow y = 8^{\frac{2}{3}}\) | M1 | 2.1 |
| \(y = 4\) | A1 | 1.1 |
| [3] |
Notes
M1: Correctly removing log.
M1: Manipulate to solvable form - must be correct for their equation. An intermediate step must be seen (between first M1 and solution) otherwise M1M0A0.
A1: www