Find integer values of $m$ and $n$ for which

$$m - n{\log _3}2 = 10{\log _9}6$$

METHOD 1

$m - n{\log _3}2 = 10{\log _9}6$

$m - n{\log _3}2 = 5{\log _3}6$    M1

$m = {\log _3}\left( {{6^5}{2^n}} \right)$    (M1)

${3^m}{2^{ - n}} = {6^5} = {3^5} \times {2^5}$    (M1)

$m = 5,{\text{ }}n =  - 5$    A1

Note:     First M1 is for any correct change of base, second M1 for writing as a single logarithm, third M1 is for writing 6 as $2 \times 3$.

METHOD 2

$m - n{\log _3}2 = 10{\log _9}6$

$m - n{\log _3}2 = 5{\log _3}6$    M1

$m - n{\log _3}2 = 5{\log _3}3 + 5{\log _3}2$    (M1)

$m - n{\log _3}2 = 5 + 5{\log _3}2$    (M1)

$m = 5,{\text{ }}n =  - 5$    A1

Note:     First M1 is for any correct change of base, second M1 for writing 6 as $2 \times 3$ and third M1 is for forming an expression without ${\log _3}3$.

[4 marks]

The first stage on this question was to change base, so each logarithm was written in the same base. Some candidates chose to move to base 10 or base e, rather than the more obvious base 3, but a few still successfully reached the correct answer having done this. A large majority though did not seem to know how to change the base of a logarithm.

Simplifying the expression further was a struggle for many candidates.