CAT 2024 Slot 3 · Quant · Question 16

Each equation defines one exponent as a logarithm: {3}^{a}=4 means a={\log}_{3}4.

In the same way b={\log}_{4}5, c={\log}_{5}6, d={\log}_{6}7, e={\log}_{7}8 and f={\log}_{8}9.

Rewriting each with the change-of-base rule, a=\frac{\log 4}{\log 3}, b=\frac{\log 5}{\log 4}, and so on.

The product is therefore \frac{\log 4}{\log 3}\times \frac{\log 5}{\log 4}\times \frac{\log 6}{\log 5}\times \frac{\log 7}{\log 6}\times \frac{\log 8}{\log 7}\times \frac{\log 9}{\log 8}.

Every intermediate factor cancels against its neighbour, leaving \frac{\log 9}{\log 3}.

That is {\log}_{3}9, and since 9={3}^{2}, the value is 2.

Shortcut. A chain of logarithms telescopes: {\log}_{3}4\times{\log}_{4}5\times \cdots \times{\log}_{8}9={\log}_{3}9. Only the first base and the last argument survive, so the answer is read off in a single step.