Standard +0.3 This is a straightforward logarithm equation requiring application of log laws (power rule, subtraction rule) to combine terms, then solving the resulting quadratic. The non-standard base 5 adds minimal difficulty since the laws apply identically. The quadratic solving and surd simplification are routine C1/C2 skills, making this slightly easier than average.
Use or state \(\log_5(x+5)^2 - \log_5(2x+2) = \log_5\frac{(x+5)^2}{(2x+2)}\) or \(\log_5(2x+2)+\log_5 5^2 = \log_5 5^2(2x+2)\)
M1
Addition/subtraction law used correctly at least once; coefficient of "2" must have been dealt with
Use or state \(\log_5 25 = 2\)
M1
Connects 2 with 25 OR \(5^2\) correctly
\((x+5)^2 = 25(2x+2)\) or equivalent
A1
Correct equation not involving logs; dependent on all 3 M's
\(x^2 - 40x - 25 = 0\)
A1
Correct 3TQ; dependent on all 3 M's
Solves quadratic; \(x = 20 \pm 5\sqrt{17}\)
M1, A1
M1: solves by formula/calculator/completing the square for surd answer; CSO \(x = 20 \pm 5\sqrt{17}\); withhold final mark if \(x = 20 - 5\sqrt{17}\) rejected
Total
[7]
# Question 6:
| Answer/Working | Mark | Guidance |
|---|---|---|
| Use or state $2\log_5(x+5) = \log_5(x+5)^2$ | M1 | Can be scored without sight of base 5 |
| Use or state $\log_5(x+5)^2 - \log_5(2x+2) = \log_5\frac{(x+5)^2}{(2x+2)}$ or $\log_5(2x+2)+\log_5 5^2 = \log_5 5^2(2x+2)$ | M1 | Addition/subtraction law used correctly at least once; coefficient of "2" must have been dealt with |
| Use or state $\log_5 25 = 2$ | M1 | Connects 2 with 25 OR $5^2$ correctly |
| $(x+5)^2 = 25(2x+2)$ or equivalent | A1 | Correct equation not involving logs; dependent on all 3 M's |
| $x^2 - 40x - 25 = 0$ | A1 | Correct 3TQ; dependent on all 3 M's |
| Solves quadratic; $x = 20 \pm 5\sqrt{17}$ | M1, A1 | M1: solves by formula/calculator/completing the square for surd answer; CSO $x = 20 \pm 5\sqrt{17}$; withhold final mark if $x = 20 - 5\sqrt{17}$ rejected |
| **Total** | **[7]** | |
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