Question

Log4 5×log5 6×log6 49×log7 32

Answer 1

$log_45*log_56*log_649*log_732=\\\\=\frac{lg5}{lg4}* \frac{lg6}{lg5}* \frac{lg49}{lg6}* \frac{lg32}{lg7}=\\\\= \frac{lg7^2*lg2^5}{lg2^2*lg7}= \frac{2*5*lg7*lg2}{2*lg2*lg7}=5$