## Question 9(a)(i):

| Answer | Marks | Guidance |
|--------|-------|----------|
| $u_4 = \log_2 27 + 3\log_2 x$ | M1 | Use $u_4 = a + 3d$. Allow missing/incorrect/inconsistent log bases. Starting with $\log_2 27 + \log_2 x^3$ is M0M0. Starting with $\log_2 27 \times 3\log_2 x$ is M0 (but can get M1 below). Starting with $\log_2 27 + \log_2 x + \log_2 x + \log_2 x$ can get full credit. |
| $= \log_2 27 + \log_2 x^3$ | M1 | Use $b\log a = \log a^b$ on $3\log_2 x$. $u_4$ must still be shown as two terms. Could get M1 if using $a + 4d$. Could get M1 for $\log_2 27 \times 3\log_2 x = \log_2 27 \times \log_2 x^3$ or for $\log_2 27 \cdot 3\log_2 x = \log_2 27 + \log_2 x^3$. Allow missing/incorrect/inconsistent log bases. |
| $= \log_2(27x^3)$ **AG** | A1 | Show $\log_2(27x^3)$ convincingly. Can go straight from $\log_2 27 + \log_2 x^3$ to final answer. CWO, including using base 2 throughout. |
| **[3]** | | **SR – finding consecutive terms (each step must be explicit):** **B1** for $u_2 = \log_2 27 + \log_2 x = \log_2 27x$; **B1** for $u_3 = \log_2 27x + \log_2 x = \log_2 27x^2$; **B1** for $u_4 = \log_2 27x^2 + \log_2 x = \log_2 27x^3$ |

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## Question 9(a)(ii):

| Answer | Marks | Guidance |
|--------|-------|----------|
| $27x^3 = 2^6$ | B1* | State correct equation no longer involving $\log_2 x$. Equation could still involve constant terms such as $\log_2 27$ or $\log_2 3$. Allow truncated or rounded decimals. |
| $x = \frac{4}{3}$ | B1d* | Obtain $\frac{4}{3}$, $1\frac{1}{3}$ or an exact recurring decimal only (not $1.333...$). A0 if cube root still present. Working must be exact, so sight of decimals in method used is B0, even if final answer is exact. Answer only gets full credit. |
| **[2]** | | |

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## Question 9(b)(i):

| Answer | Marks | Guidance |
|--------|-------|----------|
| $\frac{1}{2} < y < 2$ | M1 | Identify at least one of $\frac{1}{2}$ and 2 as end-points. Only one end-point required. Ignore if additional incorrect end-point also given. Ignore any signs used. |
| Obtain $\frac{1}{2} < y < 2$ | A1 | Not two separate inequalities, unless linked by 'and'. A0 for $\frac{1}{2} \leq y \leq 2$. |
| **[2]** | | |

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## Question 9(b)(ii):

| Answer | Marks | Guidance |
|--------|-------|----------|
| $\frac{\log_2 27}{1 - \log_2 y} = 3$ | B1 | State $\frac{\log_2 27}{1 - \log_2 y} = 3$. Allow B1 if no base stated, but B0 if incorrect base. Must be equated to 3 for B1. |
| $\log_2 27 = 3 - 3\log_2 y$; $\log_2 27 = 3 - \log_2 y^3$; $\log_2(27y^3) = 3$ | M1* | Attempt to rearrange equation to $\log_2 f(y) = k$. Must be using $\frac{\log_2 27}{\pm 1 \pm \log_2 y}$ (but allow for no bases). Allow at most 2 manipulation errors (e.g. $+/-$ or $\times\div$ muddles, or slips when expanding brackets) but M0 if other errors (e.g. incorrect use of logs). |
| $27y^3 = 8$ | M1d* | Use $f(y) = 2^k$ as inverse of $\log_2 f(y) = k$. Must have first been arranged to $\log_2 f(y) = k$. No need to go any further than stating $f(y) = 2^k$. |
| Obtain correct exact equation no longer involving $\log_2 y$ | A1* | Equation could still involve constant terms such as $\log_2 27$ or $\log_2 3$. Sight of decimals used is A0, even if answer is exact. |
| $y^3 = \frac{8}{27}$; $y = \frac{2}{3}$ | A1d* | Obtain $\frac{2}{3}$. Allow equiv recurring decimal, but not $0.666...$. A0 if still cube root present. |
| **[5]** | | **SR** answer only is **B3**. Correct $S_\infty = 3$, then answer with no further working is **B3**. |

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