## Question 9:

**(a)** Solve $2^{3x} = 5$

| $3x\log 2 = \log 5$ | M1 | Taking logs |
|---|---|---|
| $x = \frac{\log 5}{3\log 2} = 0.774$ | A1 | cao, 3 s.f. |

**(b)** Given $\log_a k - \log_a 2 = \frac{2}{3}$, express $a$ in terms of $k$

| $\log_a \frac{k}{2} = \frac{2}{3}$ | M1 | Combining logs |
|---|---|---|
| $a^{\frac{2}{3}} = \frac{k}{2}$ | M1 | Converting to index form |
| $a = \left(\frac{k}{2}\right)^{\frac{3}{2}}$ | A1 A1 | |

**(c)(i)** Expand $(1+2x)^3$

| $(1+2x)^3 = 1 + 3(2x) + 3(2x)^2 + (2x)^3$ | M1 | |
|---|---|---|
| $= 1 + 6x + 12x^2 + 8x^3$ | A1 A1 | |

**(c)(ii)** Using $\log_2[(1+2n)^3 - 8n] = \log_2(1+2n) + \log_2[4(1+n^2)]$

| $(1+2n)^3 - 8n = (1+2n) \cdot 4(1+n^2)$ | M1 | Removing logs |
|---|---|---|
| $1 + 6n + 12n^2 + 8n^3 - 8n = 4(1+2n)(1+n^2)$ | M1 | Using expansion from (c)(i) |
| $1 - 2n + 12n^2 + 8n^3 = 4(1 + n^2 + 2n + 2n^3)$ | | |
| $8n^3 + 12n^2 - 2n + 1 = 8n^3 + 4n^2 + 8n + 4$ | M1 | Expanding and simplifying |
| $8n^2 - 10n - 3 = 0$ | A1 | Correct quadratic |
| $(4n+1)(2n-3) = 0$ | | |
| $n = -\frac{1}{4}$ or $n = \frac{3}{2}$ | A1 | Both values |

The images you've shared show only **blank answer spaces** (pages 22, 23, and 24) from what appears to be an AQA MPC2 June 2015 exam paper. These pages contain:

- Lined answer spaces for **Question 9**
- A blank page stating "There are no questions printed on this page"

**There is no mark scheme content visible in these images.** The pages shown are from the student question paper (answer booklet), not the mark scheme.

To find the mark scheme for AQA MPC2 June 2015, you would need to access it directly from the **AQA website** or **PMT (Physics & Maths Tutor)** at physicsandmathstutor.com, where past paper mark schemes are freely available.