Given that ${\log _{10}}\left( {\frac{1}{{2\sqrt 2 }}\left( {p + 2q} \right)} \right) = \frac{1}{2}\left( {{{\log }_{10}}p + {{\log }_{10}}q} \right),{\text{ }}p > 0,{\text{ }}q > 0$, find $p$ in terms of $q$.

${\log _{10}}\frac{1}{{2\sqrt 2 }}(p + 2q) = \frac{1}{2}({\log _{10}}p + {\log _{10}}q)$

${\log _{10}}\frac{1}{{2\sqrt 2 }}(p + 2q) = \frac{1}{2}{\log _{10}}pq$     (M1)

${\log _{10}}\frac{1}{{2\sqrt 2 }}(p + 2q) = {\log _{10}}{(pq)^{\frac{1}{2}}}$     (M1)

$\frac{1}{{2\sqrt 2 }}(p + 2q) = {(pq)^{\frac{1}{2}}}$     (A1)

${(p + 2q)^2} = 8pq$

${p^2} + 4pq + 4{q^2} = 8pq$

${p^2} - 4pq + 4{q^2} = 0$

${(p - 2q)^2} = 0$     M1

hence $p = 2q$     A1

[5 marks]