The variables x and y satisfy the equation a^(2y−1) = b^(x−y), where a and b are constants.

Mark scheme

Marking points from the official mark scheme (May/June 2024). The mark or mark type is shown in brackets.

1. [B1] Use logarithms to obtain a correct expression without powers e.g. (2y-1)ln a = (x - y) ln b. Could use logs to any base e.g. 2y−1= (x - y)log_b. Do not condone missing brackets unless recovered later.
2. [B1] Separate terms and factorise to obtain y(2ln a + ln b) = x ln b + ln a. Or equivalent, e.g. y = (ln b / (ln a²b))x + (ln a / (ln a²b)) or y(2+ log_b) = x log_b + 1.
3. [B1] Clear explanation of linear form. From correct work only. 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 = ... and c = ... correct but never actually mention y = mx + c.
4. [M1] Use a = b³ and log laws to simplify their equation. Denominator reduced to a single log term.
5. [A1] Obtain y = (1/7)x + 3/7 or exact equivalent. Accept y = (x/7) + (3/7) but not y = (x+3)/7.