## Question 1(i):

**Way 1:**

| Working | Mark | Guidance |
|---------|------|----------|
| $125\sqrt{5} = 5^3 \times 5^{\frac{1}{2}} = 5^{3+\frac{1}{2}}$ | M1 | Writes $125\sqrt{5} = 5^p \times 5^q$ with at least one of $p=3$ or $q=\frac{1}{2}$ and adds their $p$ and $q$ |
| $= 5^{3\frac{1}{2}}$ or $a = 3\frac{1}{2}$ or $3.5$ | A1 | Sight of $a = 3\frac{1}{2}$ or $3.5$ or $5^{3\frac{1}{2}}$ |

**Way 2:**

| Working | Mark | Guidance |
|---------|------|----------|
| $125\sqrt{5} = 5^a \Rightarrow \log_5 125\sqrt{5} = \log_5 5^a \Rightarrow \log_5 125\sqrt{5} = a\log_5 5$ | M1 | Takes logs base 5 of both sides and uses power rule i.e. $\log_5 5^a = a\log_5 5$ or $\log_5 5^a = a$ |
| $= 5^{3\frac{1}{2}}$ or $a = 3\frac{1}{2}$ or $3.5$ | A1 | Sight of $a = 3\frac{1}{2}$ or $3.5$ or $5^{3\frac{1}{2}}$ |

**Way 3:**

| Working | Mark | Guidance |
|---------|------|----------|
| $125\sqrt{5} = 5^a \Rightarrow \log 125\sqrt{5} = \log 5^a \Rightarrow \log 125\sqrt{5} = a\log 5$ | M1 | Takes logs to the same base of both sides and uses the power rule correctly |
| $= 5^{3\frac{1}{2}}$ or $a = 3\frac{1}{2}$ or $3.5$ | A1 | Sight of $a = 3\frac{1}{2}$ or $3.5$ or $5^{3\frac{1}{2}}$ |

**Way 4:**

| Working | Mark | Guidance |
|---------|------|----------|
| $125\sqrt{5} = 5^a \Rightarrow (125\sqrt{5})^2 = (5^a)^2$; $78125 = 5^{2a}$; $2a = \log_5 78125$ or $\log 78125 = 2a\log 5$ | M1 | Squares both sides and takes log base 5 or takes logs in a different base and uses the power rule correctly |
| $= 5^{3\frac{1}{2}}$ or $a = 3\frac{1}{2}$ or $3.5$ | A1 | Sight of $a = 3\frac{1}{2}$ or $3.5$ or $5^{3\frac{1}{2}}$ |

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## Question 1(ii):

**Main Method:**

| Working | Mark | Guidance |
|---------|------|----------|
| $\frac{16(4+\sqrt{8})}{(4-\sqrt{8})(4+\sqrt{8})}$ | M1 | Multiply numerator and denominator by $\pm(4+\sqrt{8})$ or equivalent e.g. $\pm(4+2\sqrt{2})$. May be implied by correct expression in numerator and $16-8$ or full expansion in denominator |
| $= \frac{16(4+2\sqrt{2})}{16-8}$ | A1 | $= \pm\frac{\cdots}{16-8}$ or $= \pm\frac{\cdots}{8}$; must follow M1 |
| $= 8 + 4\sqrt{2}$ | A1 | Fully correct proof with intermediate line showing $16-8$ or $8$, and $\sqrt{8}=2\sqrt{2}$ used (does not need to be explicitly stated). Recovery from invisible brackets allowed |

**Alternative (cancel 2 throughout):**

| Working | Mark | Guidance |
|---------|------|----------|
| $\frac{16}{(4-\sqrt{8})} = \frac{16}{4-2\sqrt{2}} = \frac{8}{2-\sqrt{2}}$; $= \frac{8(2+\sqrt{2})}{(2-\sqrt{2})(2+\sqrt{2})}$ | M1 | Multiply numerator and denominator by $\pm(2+\sqrt{2})$. May be implied by correct expression in numerator and $4-2$ or full expansion in denominator |
| $= \frac{8(2+\sqrt{2})}{4-2}$ | A1 | $= \pm\frac{\cdots}{4-2}$ or $= \pm\frac{\cdots}{2}$; must follow M1 |
| $= 8+4\sqrt{2}$ | A1 | Fully correct proof with intermediate line showing $4-2$ or $2$, and $\sqrt{8}=2\sqrt{2}$ used. Recovery from invisible brackets allowed |

**Alternative (expand from answer):**

| Working | Mark | Guidance |
|---------|------|----------|
| $(8+4\sqrt{2})(4-\sqrt{8}) = \ldots$ | M1 | Attempt to expand to at least 3 terms |
| $= 32 - 8\sqrt{8} + 16\sqrt{2} - 4\sqrt{16}$ | A1 | All terms correct |
| $= 16 \therefore \frac{16}{4-\sqrt{8}} = 8+4\sqrt{2}$ | A1 | Obtains 16 correctly with a conclusion (tick, #, QED etc.) |

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