## Question 10:

### Part (i)(a):

| Answer/Working | Mark | Guidance |
|---|---|---|
| $2a$ | B1 | |

### Part (i)(b):

| Answer/Working | Mark | Guidance |
|---|---|---|
| $\log_2\left(\frac{\sqrt{3}}{16}\right) = \log_2\sqrt{3} - \log_2 16 = \frac{1}{2}a - 4$ | M1A1 | M1: uses subtraction rule to write $\log_2\sqrt{3} - \log_2 16$. A1: $\frac{1}{2}a-4$. Note: if they reach $\frac{1}{2}a-4$ correctly then say $=a-8$, score M1A0 |

### Part (ii):

| Answer/Working | Mark | Guidance |
|---|---|---|
| $3^x \times 2^{x+4} = 6 \Rightarrow \log_2 3^x + \log_2 2^{x+4} = \log_2 6$ | M1 | Takes logs (same base) of both sides and applies addition rule on lhs |
| Collecting terms in $x$ on one side, factorising and dividing e.g. $x(\log_2 3 + \log_2 2) = \log_2 6 - 4 \Rightarrow x = ...$ | dM1 | Correct method to make $x$ subject; more than one term in $x$ required. Dependent on first M |
| $\log_2 a^b = b\log_2 a$ | B1 | Correct power law seen for logs. Can score independently |
| $x = \frac{a-3}{a+1}$ | A1 | Or e.g. $\frac{3-a}{-a-1}$ |

# Alternative Methods for Solving $3^x \times 2^{x+4} = 6$

## Alternative 1: Not requiring logs

| Working | Mark | Guidance |
|---------|------|----------|
| $a = \log_2 3 \Rightarrow 3 = 2^a$ | **B1** | State or use $a = \log_2 3 \Rightarrow 3 = 2^a$ |
| Writes $2^{ax} \times 2^{x+4} = 3 \times 2 = 2^a \times 2$ | **M1** | Attempts to write all terms as powers of 2 |
| $2^{ax+x+4} = 2^{a+1} \Rightarrow ax+x+4 = a+1 \Rightarrow x(a+1) = a-3$ | **dM1** | Combines and equates powers and makes $x$ the subject |
| $x = \dfrac{a-3}{a+1}$ or e.g. $\dfrac{3-a}{-a-1}$ | **A1** | Correct expression |

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## Alternative 2: Using change of base

| Working | Mark | Guidance |
|---------|------|----------|
| $3^x \times 2^{x+4} = 6 \Rightarrow 2^{x+3} = 3^{1-x}$, divides by 2 and $3^x$, takes logs of **both** sides | **M1** | |
| $\log 2^{x+3} = \log 3^{1-x}$, e.g. $(x+3)\log 2 = (1-x)\log 3$ | **B1** | |
| $(x+3)\dfrac{\log_2 2}{\log_2 10} = (1-x)\dfrac{\log_2 3}{\log_2 10} \Rightarrow x+3 = a(1-x) \Rightarrow x(a+1) = a-3$ | **dM1** | Changes to base 2 correctly and makes $x$ the subject |
| $x = \dfrac{a-3}{a+1}$ or e.g. $\dfrac{3-a}{-a-1}$ | **A1** | Correct expression |

---

## Alternative 3: Using logs base 6

| Working | Mark | Guidance |
|---------|------|----------|
| $3^x \times 2^x = \dfrac{3}{8} \Rightarrow 6^x = \dfrac{3}{8}$, divides by $2^4$, writes $3^x \times 2^x$ as $6^x$, takes $\log_6$ of both sides | **M1** | May be implied by e.g. $6^x = \dfrac{3}{8} \Rightarrow x = \log_6\dfrac{3}{8}$ |
| $\log_6 6^x = x\log_6 6$ | **B1** | May be implied |
| $x = \log_6\dfrac{3}{8} = \dfrac{\log_2\frac{3}{8}}{\log_2 6}$ | **dM1** | Changes to log base 2 correctly and makes $x$ the subject |
| $\dfrac{\log_2\frac{3}{8}}{\log_2 6} = \dfrac{\log_2 3 - \log_2 8}{\log_2 3 + \log_2 2} = \dfrac{a-3}{a+1}$ | **A1** | Correct expression |