## Precalculus (10th Edition)

$1$
We know that $\log_a {b}=\frac{\log_c {b}}{\log_c {a}}$ (this is known as the Change-of-Base formula). Hence $\log_2 {3}\cdot \log_3 {4}\cdot...\cdot \log_n {(n+1)}\cdot \log_{n+1} {2}$ Cancel the common factors to obtain $=\frac{\log{3}}{\log{2}}\cdot\frac{\log{4}}{\log{3}}\cdot...\cdot\frac{\log{(n+1)}}{\log{n}}\cdot\frac{\log{2}}{\log({n+1})}=1.$