Standard +0.3 This is a standard C2 logarithm question requiring systematic application of log laws (power rule, subtraction rule) and algebraic manipulation to reach a given polynomial form, then solving. The 'show that' structure provides a clear target, making it slightly easier than average. The domain restrictions (x > 6.5) add minor complexity but the quadratic factorizes neatly to (x-8)². Slightly above routine recall but below average A-level difficulty.
Must be seen in part (b). Attempt to solve the given quadratic (usual rules). Having additional solutions such as \(x=-8\) loses the A mark. \(x=8\) with no working scores both marks
## Question 7:
### Part (a):
| Answer | Marks | Guidance |
|--------|-------|----------|
| $2\log_3(x-5) = \log_3(x-5)^2$ | B1 | Marks may be awarded if equivalent work seen in part (b) |
| $\log_3(x-5)^2 - \log_3(2x-13) = \log_3\frac{(x-5)^2}{2x-13}$ | M1 | $\log_3(x-5)^2 - \log_3(2x-13) = \frac{\log_3(x-5)^2}{\log_3(2x-13)}$ is M0 |
| $\log_3 3 = 1$ seen or used correctly | B1 | |
| $\log_3\left(\frac{P}{Q}\right)=1 \Rightarrow P=3Q \Rightarrow \frac{(x-5)^2}{2x-13}=3 \Rightarrow (x-5)^2=3(2x-13)$ | M1 | 'Cancelling logs' loses this M mark |
| $x^2 - 16x + 64 = 0$ $(*)$ | A1 cso | |
### Part (b):
| Answer | Marks | Guidance |
|--------|-------|----------|
| $(x-8)(x-8)=0 \Rightarrow x=8$ | M1 A1 | Must be seen in part (b). Attempt to solve the given quadratic (usual rules). Having additional solutions such as $x=-8$ loses the A mark. $x=8$ with no working scores both marks |