## Question 9:

### Part (i):

| Answer/Working | Mark | Guidance |
|---|---|---|
| $4 = \log_3 81$ or $4 = \log_3 3^4$ | B1 | May be implied by e.g. $\log_3\dfrac{x+5}{2x-1} = 4 \Rightarrow \dfrac{x+5}{2x-1} = 3^4$ |
| Combining 2 log terms correctly e.g. $\log_3(x+5) - \log_3 81 = \log_3\dfrac{x+5}{81}$ or $\log_3(x+5)-\log_3(2x-1) = \log_3\dfrac{x+5}{2x-1}$ or $\log_3(2x-1)+\log_3 81 = \log_3 81(2x-1)$ | M1 | Can be awarded following incorrect rearrangement e.g. $\log_3(x+5)-4 = \log_3(2x-1) \Rightarrow \log_3(x+5)+\log_3(2x-1)=4$ |
| $\dfrac{x+5}{81} = 2x-1$ or $\dfrac{x+5}{2x-1} = 3^4$ | A1 | Obtains this equation in any form |
| $x = \dfrac{86}{161}$ | A1 | cao |

**Condone omission of base throughout.**

**Special case:** $\log_3(x+5)-\log_3(2x-1) = 4 \Rightarrow \dfrac{\log_3(x+5)}{\log_3(2x-1)} = 4$ leading to $x = \dfrac{86}{161}$ scores B1(implied) M0 A0 A1.

**Alternative (first 3 marks):**

| Answer/Working | Mark | Guidance |
|---|---|---|
| $\log_3(x+5)-4 = \log_3(2x-1) \Rightarrow 3^{\log_3(x+5)-4} = 3^{\log_3(2x-1)} \Rightarrow 3^{\log_3(x+5)} \times 3^{-4} = 2x-1 \Rightarrow \dfrac{x+5}{81} = 2x-1$ | B1, M1, A1 | B1 for sight of $3^{-4}$; M1 for applying $3^{a\pm b} = 3^a \times 3^{\pm b}$ |

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### Part (ii)(a):

| Answer/Working | Mark | Guidance |
|---|---|---|
| $3^{y+3} = 3^y \times 3^3$ or $2^{1-2y} = 2 \times 2^{-2y}$ or $\dfrac{2}{2^{2y}}$ or $2 \times 4^{-y}$ or $\dfrac{2}{4^y}$ | B1 | One correct index law seen or implied anywhere in working |
| $3^{y+3} \times 2^{1-2y} = 27 \times 3^y \times 2 \times 2^{-2y} = \ldots$ | M1(B1 on EPEN) | Applies both correct index laws to the LHS of the equation |
| $3^y \times 2^{-2y} = \dfrac{108}{27\times2}$ or $\dfrac{3^y}{4^y} = \dfrac{108}{27\times2}$ or $\dfrac{3^y}{2^{2y}} = \dfrac{108}{54}$ | M1 | Isolates terms in $y$ (as powers of 3 and 2(or 4)) on LHS and constants on RHS; must be no incorrect work to combine terms e.g. $3^y \times 3^3 = 27^y$ etc. |
| $(0.75)^y = 2$ * | A1* | cso; reaches given answer with no errors; all steps shown with $2^{2y}$ appearing as $4^y$ at some point |

**Alternative 1 using logs:**

| Answer/Working | Mark | Guidance |
|---|---|---|
| $\log(3^{y+3} \times 2^{1-2y}) = \log 3^{y+3} + \log 2^{1-2y}$ or $\log 3^{y+3} = (y+3)\log 3$ or $\log 2^{1-2y} = (1-2y)\log 2$ | B1 | One correct log law seen or implied; **no bracketing errors allowed for this mark** |
| $\log 3^{y+3} + \log 2^{1-2y} = (y+3)\log 3 + (1-2y)\log 2$ | M1(B1 on EPEN) | Applies correct log laws to LHS; condone missing brackets around $y+3$ and/or $1-2y$ |
| Proceeds to isolate terms in $y$ on LHS and combines constants on RHS e.g. $(y+3)\log 3 + (1-2y)\log 2 = \log 108 \Rightarrow \log\dfrac{3^y}{2^{2y}} = \log\dfrac{108}{3^3\times2}$ | M1 | |
| $(0.75)^y = 2$ * | A1* | cso; with e.g. $2\log 2$ seen as $\log 4$ or $\log 2^2$ at some point |

**Alternative 2 using factors of 108:**

| Answer/Working | Mark | Guidance |
|---|---|---|
| $3^{y+3} \times 2^{1-2y} = 108 = 2^2 \times 3^3 \Rightarrow \dfrac{3^{y+3} \times 2^{1-2y}}{2^2 \times 3^3} = \ldots \Rightarrow 3^y \times 2^{-1-2y} = \ldots$ | B1 | One correct index law seen or implied e.g. $\dfrac{3^{y+3}}{3^3} = 3^y$ or $\dfrac{2^{1-2y}}{2^2} = 2^{-1-2y}$ |
| $3^y \times 2^{-2y} = \ldots$ | M1(B1 on EPEN) | Applies both correct index laws to LHS |
| $3^y \times 2^{-2y} = 2$ | M1 | Isolates terms in $y$ on LHS and constants on RHS |
| $(0.75)^y = 2$ * | A1* | cso; all steps shown with $2^{2y}$ appearing as $4^y$ at some point |

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### Part (ii)(b):

| Answer/Working | Mark | Guidance |
|---|---|---|
| $(0.75)^y = 2 \Rightarrow y = \dfrac{\log 2}{\log 0.75}$ or $(0.75)^y = 2 \Rightarrow y = \log_{0.75} 2$ | M1 | Correct processing to obtain value for $y$; may be implied by awrt $-2.4$ |
| $y = -2.409$ | A1 | Awrt $-2.409$; correct answer implies both marks |

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