Consider $a = {\log _2}3 \times {\log _3}4 \times {\log _4}5 \times  \ldots  \times {\log _{31}}32$. Given that $a \in \mathbb{Z}$, find the value of a.

$\frac{{\log 3}}{{\log 2}} \times \frac{{\log 4}}{{\log 3}} \times  \ldots \times \frac{{\log 32}}{{\log 31}}$     M1A1

$= \frac{{\log 32}}{{\log 2}}$     A1

$= \frac{{5\log 2}}{{\log 2}}$     (M1)

$= 5$     A1

hence $a = 5$

Note:     Accept the above if done in a specific base eg ${\log _2}x$.

[5 marks]