Solve the equation \(2 - {\log _3}(x + 7) = {\log _{\tfrac{1}{3}}}2x\) .

\({\log _3}\left( {\frac{9}{{x + 7}}} \right) = {\log _3}\frac{1}{{2x}}\)     M1M1A1

Note: Award M1 for changing to single base, M1 for incorporating the 2 into a log and A1 for a correct equation with maximum one log expression each side.

\(x + 7 = 18x\)     M1

\(x = \frac{7}{{17}}\)     A1

[5 marks] 

Some good solutions to this question and few candidates failed to earn marks on the question. Many were able to change the base of the logs, and many were able to deal with the 2, but of those who managed both, poor algebraic skills were often evident. Many students attempted to change the base into base 10, resulting in some complicated algebra, few of which managed to complete successfully.