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]