Question

Calcule (log23)(log34)(log45)...(log3132)

Answer 1

$(log_23)(log_34)(log_45)...(log_{31}32)=(\frac{ln3}{ln2})(\frac{ln4}{ln3})(\frac{ln5}{ln4})...(\frac{ln32}{ln31})$

Cortando todos os numeradores com os denominadores ficamos com:

$\frac{ln32}{ln2}=log_232=\boxed{5}$