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.