A 10 seconds question!

$\log_4{9}\times \log_{81}{5} \times \log_7{16}\times \log_{25}{49} \\ \quad = \dfrac {\log{9}}{\log{4}}\times \dfrac {\log{5}}{\log{81}}\times \dfrac {\log{16}}{\log{7}}\times \dfrac {\log{49}}{\log{25}} \\ \quad = \dfrac {\log{9}}{\log{4}}\times \dfrac {\log{5}}{2\log{9}}\times \dfrac {2\log{4}}{\log{7}}\times \dfrac {2\log{7}}{2\log{5}} \\ \quad = \boxed{1}$

Another solution, diferent and not direct as the solution of Chew-Seong Cheong:

I did basically everything Chew-Seong Cheong did... only with ln's. Remember folks: $log_{4} 9 = \frac{\log{9}}{\log{4}} = \frac{\ln{9}}{\ln{4}}$ And the same rules with powers moving to the front apply.

writing in change of base form every thing gets cancelled