Let \(g(x) = {\log _5}\left| {2{{\log }_3}x} \right|\) . Find the product of the zeros of g .

\(g(x) = 0\)

\({\log _5}\left| {2{{\log }_3}x} \right| = 0\)     (M1)

\(\left| {2{{\log }_3}x} \right| = 1\)     A1

\({\log _3}x =  \pm \frac{1}{2}\)     (A1)

\(x = {3^{ \pm \frac{1}{2}}}\)     A1

so the product of the zeros of g is \({3^{\frac{1}{2}}} \times {3^{ - \frac{1}{2}}} = 1\)     A1     N0

[5 marks]

There were many candidates showing difficulties in manipulating logarithms and the absolute value to solve the equation.