## Question 3:

**Part (a):**

| Answer | Mark | Guidance |
|--------|------|----------|
| Use logarithms to obtain a correct expression without powers e.g. $(2y-1)\ln a = (x-y)\ln b$ | B1 | Could use logs to any base e.g. $2y - 1 = (x-y)\log_a b$. Do not condone missing brackets unless recovered later |
| Separate terms and factorise to obtain $y(2\ln a + \ln b) = x\ln b + \ln a$ | B1 | Or equivalent, e.g. $y = x\frac{\ln b}{\ln a^2 b} + \frac{\ln a}{\ln a^2 b}$ or $y(2 + \log_a b) = x\log_a b + 1$ |
| Clear explanation of linear form. From correct work only. | B1 | E.g. equation matches the linear form $y = mx + c$ or $py = qx + r$. Condone if they compare with $y = mx+c$, but do not actually state that it must therefore be a straight line. Stating "this is a linear equation" without comparing to a relevant standard form scores B0. B0 if they have $m = \ldots$ and $c = \ldots$ correct but never actually mention $y = mx + c$ |

**Part (b):**

| Answer | Mark | Guidance |
|--------|------|----------|
| Use $a = b^3$ and log laws to simplify their equation | M1 | $\left[y = x\frac{\ln b}{\ln b^7} + \frac{\ln b^3}{\ln b^7}\right]$ Denominator reduced to a single log term |
| Obtain $y = \frac{1}{7}x + \frac{3}{7}$ | A1 | Accept $y = \frac{x}{7} + \frac{3}{7}$ but not $y = \frac{x+3}{7}$ |

**Alternative Method for Part (b):**

| Answer | Mark | Guidance |
|--------|------|----------|
| Use $a = b^3$ to obtain $b^{3(2y-1)} = b^{x-y}$ or equivalent | (M1) | Or $\log_a b = \frac{1}{3}$ |
| Obtain $y = \frac{1}{7}x + \frac{3}{7}$ | (A1) | Accept $y = \frac{x}{7} + \frac{3}{7}$ but not $y = \frac{x+3}{7}$ |

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