If y=&(log_2 3)(log_3 4)… &(log_n[n+1])…(log_3132), then

1977 AMC 12 Problem 18

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18.

\begin{aligned} y={}&(\log_2 3)(\log_3 4)\cdots\\ &(\log_n[n+1])\cdots(\log_{31}32), \end{aligned}

then

4\lt y\lt5

y=5

5\lt y\lt6

y=6

6\lt y\lt7

Rewrite every logarithm using the same base

Adjacent numerators and denominators cancel in the product

By change of base,

\begin{aligned} y&=\frac{\log 3}{\log2}\cdot \frac{\log4}{\log3}\cdots \frac{\log32}{\log31}\\ &=\frac{\log32}{\log2}\\ &=\log_2 32=5. \end{aligned}

Therefore, the correct answer is B.