Standard +0.3 This is a straightforward logarithm equation requiring application of log laws (difference rule, power rule) to combine terms, then converting to exponential form and solving a quadratic. It's slightly above average difficulty due to the non-standard base and the quadratic step, but follows a standard procedure taught in C2 with no novel insight required.
Correct use of subtraction rule (or addition rule) for logs. N.B. \(2\log_3 x - \log_3(x-2) = 2\log_3 \frac{x}{x-2}\) is M0
\(\frac{x^2}{x-2} = 9\)
A1 o.e.
Correct equation without logs (allow any correct equivalent including \(3^2\) instead of 9)
Solves \(x^2 - 9x + 18 = 0\) to give \(x =\) ...
M1
Attempting to solve \(x^2 - 9x + 18 = 0\)
\(x = 3\), \(x = 6\)
A1
Both correct answers
Alternative Method: \(\log_3 x^2 = 2 + \log_3(x-2)\) is B1, so \(x^2 = 3^{2+\log_3(x-2)}\) needs to be followed by \((x^2) = 9(x-2)\) for M1 A1. M1 is for complete method i.e. correct use of powers after logs are used correctly.
# Question 2:
| Answer/Working | Marks | Guidance |
|---|---|---|
| $2\log x = \log x^2$ | B1 | Correct use of power rule (may be implied) |
| $\log_3 x^2 - \log_3(x-2) = \log_3 \frac{x^2}{x-2}$ | M1 | Correct use of subtraction rule (or addition rule) for logs. N.B. $2\log_3 x - \log_3(x-2) = 2\log_3 \frac{x}{x-2}$ is M0 |
| $\frac{x^2}{x-2} = 9$ | A1 o.e. | Correct equation without logs (allow any correct equivalent including $3^2$ instead of 9) |
| Solves $x^2 - 9x + 18 = 0$ to give $x =$ ... | M1 | Attempting to solve $x^2 - 9x + 18 = 0$ |
| $x = 3$, $x = 6$ | A1 | Both correct answers |
**Alternative Method:** $\log_3 x^2 = 2 + \log_3(x-2)$ is B1, so $x^2 = 3^{2+\log_3(x-2)}$ needs to be followed by $(x^2) = 9(x-2)$ for M1 A1. M1 is for complete method i.e. correct use of powers after logs are used correctly.
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