Question

prove the above question.

Answer 1

$log( \frac{7}{6} ) + log( \frac{8}{7} ) + log( \frac{9}{8} ) + \\ log( \frac{10}{9} ) + log( \frac{12}{10} ) \\ log( \frac{7}{6} \times \frac{8}{7} \times \frac{9}{8} \times \frac{10}{9} \times \frac{12}{10} ) \\ log( \frac{12}{6} ) = log(2)$