Answer
$3$
Work Step by Step
$log_23\cdot log_34\cdot log_45\cdot log_56\cdot log_67\cdot log_78=\frac{ln3}{ln2}\times\frac{ln4}{ln3}\times\frac{ln5}{ln4}\times\frac{ln6}{ln5}\times\frac{ln7}{ln6}\times\frac{ln8}{ln7}=\frac{ln8}{ln2}=3$
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