## Question 2:

### Part (a)

| Working | Marks | Guidance |
|---------|-------|----------|
| $2\ln(2x+1)-10=0 \Rightarrow \ln(2x+1)=5 \Rightarrow 2x+1=e^5 \Rightarrow x=\ldots$ | M1 | Proceeds from $2\ln(2x+1)-10=0$ to $\ln(2x+1)=5$ before taking exp to achieve $x$ in terms of $e^5$. Accept alternative using power law: $\ln(2x+1)^2=10\Rightarrow(2x+1)^2=e^{10}\Rightarrow x=g(\sqrt{e^{10}})$ |
| $x=\frac{e^5-1}{2}$ | A1 | cso. Accept $\frac{e^5}{2}-\frac{1}{2}$. Decimal answer 73.7 scores M1A0 unless exact answer also given. $\frac{\sqrt{e^{10}}-1}{2}$ does not score unless simplified. $\frac{\pm e^5-1}{2}$ is M1A0 |

### Part (b)

| Working | Marks | Guidance |
|---------|-------|----------|
| $3^xe^{4x}=e^7 \Rightarrow \ln(3^xe^{4x})=\ln e^7$ | M1 | Takes ln (or log) of both sides and applies addition law. Evidence: $\ln(3^xe^{4x})=\ln3^x+\ln e^{4x}$ or $\ln3^x+4x$ |
| $\ln3^x+\ln e^{4x}=\ln e^7 \Rightarrow x\ln3+4x\ln e=7\ln e$ | M1 | Uses power law of logs (seen at least once in a term with $x$ as index). Score M0M1 possible after incorrect addition/subtraction law |
| $x(\ln3+4)=7 \Rightarrow x=\ldots$ | dM1 | Dependent on both M's. Collects/factorises term in $x$ and proceeds to $x=$. Condone sign slips |
| $x=\frac{7}{(\ln3+4)}$ | A1 | Must use $\ln e=1$. oe |

### Alt 1 Part (b)

| Working | Marks | Guidance |
|---------|-------|----------|
| $3^xe^{4x}=e^7 \Rightarrow 3^x=\frac{e^7}{e^{4x}}=e^{7-4x}$ | | |
| $3^x=e^{7-4x}\Rightarrow x\ln3=(7-4x)\ln e$ | M1,M1 | |
| $x(\ln3+4)=7 \Rightarrow x=\ldots$ | dM1 | |
| $x=\frac{7}{(\ln3+4)}$ | A1 | |

### Alt 2 Part (b) — Using $\log$

| Working | Marks | Guidance |
|---------|-------|----------|
| $3^xe^{4x}=e^7\Rightarrow \log(3^xe^{4x})=\log e^7$ | | |
| $\log3^x+\log e^{4x}=\log e^7\Rightarrow x\log3+4x\log e=7\log e$ | M1,M1 | |
| $x(\log3+4\log e)=7\log e\Rightarrow x=\ldots$ | dM1 | |
| $x=\frac{7\log e}{(\log3+4\log e)}$ | A1 | Must write log as opposed to ln |

### Alt 3 Part (b) — Using $\log_3$

| Working | Marks | Guidance |
|---------|-------|----------|
| $3^x=e^{7-4x}\Rightarrow x=(7-4x)\log_3 e$ | M1,M1 | |
| $x(1+4\log_3 e)=7\log_3 e\Rightarrow x=\ldots$ | dM1 | |
| $x=\frac{7\log_3 e}{(1+4\log_3 e)}$ | A1 | |

### Alt 4 Part (b) — Using $3^x=e^{x\ln3}$

| Working | Marks | Guidance |
|---------|-------|----------|
| $3^xe^{4x}=e^7\Rightarrow e^{x\ln3}e^{4x}=e^7$ | | |
| $\Rightarrow e^{x\ln3+4x}=e^7\Rightarrow x\ln3+4x=7$ | M1,M1 | |
| $x(\ln3+4)=7\Rightarrow x=\ldots \quad x=\frac{7}{(\ln3+4)}$ | dM1 A1 | |