## Question 6:

### Part (a):

| Answer/Working | Mark | Guidance |
|---|---|---|
| $\ln(4-2x)(9-3x) = \ln(x+1)^2$ | M1, M1 | First M1: addition law on lhs, accept sign slips, subtraction law if term taken to rhs. Second M1: power rule, writing $2\ln(x+1)$ as $\ln(x+1)^2$, condone invisible brackets |
| $36-30x+6x^2 = x^2+2x+1$ and $5x^2-32x+35=0$ | A1 | Undoes logs to obtain 3TQ. Equals zero may be implied by subsequent solution |
| Solve $5x^2-32x+35=0$ to give $x=\frac{7}{5}$ (ignore solution $x=5$) | M1A1 | Solves quadratic by any allowable method (cannot be version of $(4-2x)(9-3x)=0$). Deduces $x=1.4$ or equivalent. Accept both $x=1.4$ and $x=5$. Extra solutions in range $-1<x<2$ scores A0 |

### Part (b):

| Answer/Working | Mark | Guidance |
|---|---|---|
| Take $\log_e$'s to give $\ln 2^x + \ln e^{3x+1} = \ln 10$ | M1 | — |
| $x\ln 2 + (3x+1)\ln e = \ln 10$ | M1 | — |
| $x(\ln 2 + 3\ln e) = \ln 10 - \ln e$ | dM1 | — |
| Uses $\ln e = 1$ | M1 | Note: 4th M mark may occur on line 2 |
| $x = \frac{-1+\ln 10}{3+\ln 2}$ | A1 | — |

# Question 6(b):

| Answer/Working | Mark | Guidance |
|---|---|---|
| Takes logs of both sides AND splits LHS using addition law | M1 | If one term taken to other side, award for logs of both sides AND subtraction law |
| Takes both powers down using power rule | M1 | Not wholly dependent on first M1, but logs of both sides must have been taken |
| Collects $x$ terms on one side and factorises, leading to $x = \ldots$ | dM1 | Dependent on both previous M's. Condone slips in signs but process must be correct |
| Uses $\ln e = 1$ | M1 | Must be part of equation, not just a statement |
| $x = \dfrac{-1 + \ln 10}{3 + \ln 2} = \left(\dfrac{\ln 10 - 1}{3 + \ln 2}\right) = \left(\dfrac{\log_e 10 - 1}{3 + \log_e 2}\right)$ | A1 | DO NOT ISW HERE |

**Note 1:** Taking $\log_{10}$ of both sides scores M1M1dM1M0A0 (3 out of 5). Answer $= x = \dfrac{-\log e + \log 10}{3\log e + \log 2}$

**Note 2:** Writing $x = \dfrac{-1 + \log 10}{3 + \log 2}$ without reference to natural logs: award M4 but hold last A1 (4 out of 5)

---

# Question 6(b) Alt 1:

| Answer/Working | Mark | Guidance |
|---|---|---|
| Writes lhs in $e$'s: $2^x e^{3x+1} = 10 \Rightarrow e^{x\ln 2} e^{3x+1} = 10$ | 1st M1 | Seeing $2^x = e^{x\ln 2}$ in equation is sufficient |
| $\Rightarrow e^{x\ln 2 + 3x + 1} = 10$, $\quad x\ln 2 + 3x + 1 = \ln 10$ | 2nd M1, 4th M1 | Uses addition law on lhs to produce single exponential |
| $x(\ln 2 + 3) = \ln 10 - 1 \Rightarrow x = \ldots$ | dM1 | Takes ln of both sides, attempts to solve linear equation in $x$ |
| $x = \dfrac{-1 + \ln 10}{3 + \ln 2}$ | A1 | **(5)** |

---