Find the value of $\log_2{3} \cdot \log_3{4} \cdot \log_4{5} \cdot \log_5{6} \cdot \log_6{7} \cdot \log_7{8}$.
First, we should use the property of logarithms that says that $ \log_a{b}=\frac{\log_c{b}}{\log_c{a}} $. That way we can change each logarithm to base 2, which will make the multiplication easier. Right. So the expression becomes $\frac{\log_2{3}}{\log_2{2}} \cdot \frac{\log_2{4}}{\log_2{3}} \cdot \frac{\log_2{5}}{\log_2{4}} \cdot \frac{\log_2{6}}{\log_2{5}} \cdot \frac{\log_2{7}}{\log_2{6}} \cdot \frac{\log_2{8}}{\log_2{7}}$. So when we multiply, everything cancels out except $\log_2{8}$, so the answer is $\log_2{8}=3$.

