Answer
$3$
Work Step by Step
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 \log_4 {5}\cdot \log_5 {6}\cdot \log_6 {7}\cdot \log_7 {8}\\$
Cancel the common factors to obtain
$=\dfrac{\log{3}}{\log{2}}\cdot\dfrac{\log{4}}{\log{3}}\cdot\dfrac{\log{5}}{\log{4}}\cdot\dfrac{\log{6}}{\log{5}}\cdot\dfrac{\log{7}}{\log{6}}\cdot\dfrac{\log{8}}{\log{7}}\\
=\dfrac{\log{8}}{\log{2}}\\$
Use the Change-of-Base formula to obtain
$\dfrac{\log{8}}{\log{2}}=\log_2 {8}=\log_2 {2^3}$
We know that $\log_a {x^n}=n\cdot \log_a {x}$.
Hence,
$\log_2 {2^3}=3\log_2 {2}=3\cdot1=3$
