# Question 1:

## Part (i) — Way 1
| Answer/Working | Mark | Guidance |
|---|---|---|
| $\sqrt{48} = \sqrt{16}\sqrt{3}$ or $\frac{6}{\sqrt{3}} = 6\frac{\sqrt{3}}{3}$ | M1 | Writes one of the terms of the given expression correctly in terms of $\sqrt{3}$ |
| $\Rightarrow \sqrt{48} - \frac{6}{\sqrt{3}} = 2\sqrt{3}$ | A1 | Correct answer of $2\sqrt{3}$. A correct answer with **no** working implies both marks. |

## Part (i) — Way 2
| Answer/Working | Mark | Guidance |
|---|---|---|
| $\sqrt{48} = 2\sqrt{12}$ or $\frac{6}{\sqrt{3}} = \sqrt{12}$ | M1 | Writes one of the terms correctly in terms of $\sqrt{12}$ |
| $2\sqrt{12} - \sqrt{12} = \sqrt{12} = 2\sqrt{3}$ | A1 | Correct answer of $2\sqrt{3}$. A correct answer with **no** working implies both marks. |

## Part (i) — Way 3
| Answer/Working | Mark | Guidance |
|---|---|---|
| $\sqrt{48} = \frac{12}{\sqrt{3}}$ or $\sqrt{48} = \frac{\sqrt{144}}{\sqrt{3}}$ | M1 | Writes $\sqrt{48}$ correctly as $\frac{12}{\sqrt{3}}$ or $\frac{\sqrt{144}}{\sqrt{3}}$ |
| $\frac{12}{\sqrt{3}} - \frac{6}{\sqrt{3}} = \frac{6}{\sqrt{3}} = 2\sqrt{3}$ | A1 | Correct answer of $2\sqrt{3}$. A correct answer with **no** working implies both marks. |

## Part (i) — Way 4
| Answer/Working | Mark | Guidance |
|---|---|---|
| $\sqrt{48} - \frac{6}{\sqrt{3}} = \frac{\sqrt{3}\sqrt{48} - \ldots}{\sqrt{3}} = \frac{12 - \ldots}{\sqrt{3}}$ | M1 | Writes $\sqrt{48}$ correctly as $\frac{12}{\sqrt{3}}$ or $\frac{\sqrt{144}}{\sqrt{3}}$ |
| $\frac{12-6}{\sqrt{3}} = \frac{6}{\sqrt{3}} = 2\sqrt{3}$ | A1 | Correct answer of $2\sqrt{3}$. A correct answer with **no** working implies both marks. |

## Part (ii) — Way 1
| Answer/Working | Mark | Guidance |
|---|---|---|
| $81 = 3^4$ or $\log_3 81 = 6x - 3$ | B1 | For $81 = 3^4$ or $\log_3 81 = 6x-3$. May be implied by subsequent work. |
| $3^{6x-3} = 3^4$ or $\log_3 81 = 6x-3 \Rightarrow 4 = 6x - 3 \Rightarrow x = \ldots$ | M1 | Solves an equation of the form $6x - 3 = k$ where $k$ is their power of 3 |
| $\Rightarrow x = \frac{4+3}{6} = \frac{7}{6}$ | A1 | $\frac{7}{6}$ or $1\frac{1}{6}$ or $1.1\dot{6}$ |

## Part (ii) — Way 2
| Answer/Working | Mark | Guidance |
|---|---|---|
| $3 = 81^{\frac{1}{4}}$ | B1 | For $3 = 81^{\frac{1}{4}}$. May be implied by subsequent work. |
| $81^{\frac{6x-3}{4}} = 81 \Rightarrow \frac{6x-3}{4} = 1 \Rightarrow x = \ldots$ | M1 | Solves an equation of the form $k(6x-3) = 1$ where $k$ is their power of 81 |
| $\Rightarrow x = \frac{4+3}{6} = \frac{7}{6}$ | A1 | $\frac{7}{6}$ or $1\frac{1}{6}$ or $1.1\dot{6}$ |

## Part (ii) — Way 3
| Answer/Working | Mark | Guidance |
|---|---|---|
| $81 = 9^2$ **and** $3 = 9^{\frac{1}{2}}$ | B1 | For $81 = 9^2$ **and** $3 = 9^{\frac{1}{2}}$. May be implied by subsequent work. |
| $9^{\frac{6x-3}{2}} = 9^2 \Rightarrow \frac{6x-3}{2} = 2 \Rightarrow x = \ldots$ | M1 | Solves an equation of the form $p(6x-3) = q$ where $p$ is their power of 9 for the 3 and $q$ is their power of 9 for the 81 |
| $\Rightarrow x = \frac{4+3}{6} = \frac{7}{6}$ | A1 | $\frac{7}{6}$ or $1\frac{1}{6}$ or $1.1\dot{6}$ |

## Part (ii) — Way 4
| Answer/Working | Mark | Guidance |
|---|---|---|
| $3^{6x-3} = 3^{6x} \times 3^{-3}$ | B1 | For writing $3^{6x-3}$ correctly in terms of $3^{6x}$ |
| $3^{6x} = 81 \times 3^3 = 3^7 \Rightarrow 6x = 7 \Rightarrow x = \ldots$ | M1 | Solves an equation of the form $6x = k$ where $k$ is their $3^3 \times 81$ written as a power of 3 |
| $\Rightarrow x = \frac{7}{6}$ | A1 | $\frac{7}{6}$ or $1\frac{1}{6}$ or $1.1\dot{6}$ |

## Part (ii) — Way 5
| Answer/Working | Mark | Guidance |
|---|---|---|
| $\log 3^{6x-3} = \log 81$ | B1 | Takes logs of both sides |
| $6x - 3 = \frac{\log 81}{\log 3}$, $6x - 3 = 4 \Rightarrow x = \ldots$ | M1 | Solves an equation of the form $6x - 3 = k$ where $k$ is their $\frac{\log 81}{\log 3}$ |
| $\Rightarrow x = \frac{7}{6}$ | A1 | $\frac{7}{6}$ or $1\frac{1}{6}$ or $1.1\dot{6}$ |

> **Note:** The question does not specify the form of the final answer and so if answers are left un-simplified as e.g. $\frac{\log_3 81 + 3}{6}$, $\frac{\log_3 2187}{6}$ then allow full marks if correct.

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