## Question 6:

### Way 1:
| Answer/Working | Marks | Guidance |
|---|---|---|
| $\log_x(7y+1) - \log_x 2y = \log_x\!\left(\frac{7y+1}{2y}\right)$ | B1 | Combines logs correctly |
| $1 = \log_x x$ | B1 | Correct statement (may be implied) |
| $\frac{7y+1}{2y} = x$ | M1 | Remove logs to obtain this equation or equivalent |
| $2yx = 7y + 1 \Rightarrow y(2x-7) = 1$ | dM1 | Isolate $y$ correctly to give $y$ as a function of $x$. Allow sign errors only. Dependent on previous method mark |
| $y = \frac{1}{2x-7}$ or $\frac{-1}{7-2x}$ | A1 | Cao |

### Way 2:
| Answer/Working | Marks | Guidance |
|---|---|---|
| $\log_x(7y+1) = 1 + \log_x 2y$ | — | — |
| $\log_x(7y+1) = \log_x x + \log_x 2y$ | B1 | $1 = \log_x x$ (may be implied) |
| $\log_x x + \log_x 2y = \log_x 2xy$ | B1 | Combines logs correctly |
| $7y + 1 = 2xy$ | M1 | Remove logs to obtain this equation or equivalent |
| $2yx = 7y+1 \Rightarrow y(2x-7) = 1$ | dM1 | Isolate $y$ correctly. Allow sign errors only. Dependent on previous method mark |
| $y = \frac{1}{2x-7}$ or $\frac{-1}{7-2x}$ | A1 | Cao |

### Way 3:
| Answer/Working | Marks | Guidance |
|---|---|---|
| $\log_x\!\left(\frac{7y+1}{2y}\right) = \frac{\log_{10}\!\left(\frac{7y+1}{2y}\right)}{\log_{10} x}$ | B1 | Correct change of base |
| $\log_{10}\!\left(\frac{7y+1}{2y}\right) = \log_{10} x$ | — | — |
| $\frac{7y+1}{2y} = x$ | M1 | Remove logs to obtain this equation or equivalent |
| Then as above | — | — |