Question

(vii) log, 4 + log, 16 + log, 64 = 12​

Answer 1

Hey MATE!!!

$log(4) + log(16) + log(64) \\ \\ = > log(4) + log( {4}^{2} ) + log( {4}^{3} ) \\ \\ = > log(4) + 2 log(4) + 3 log(4) = 6 log(4) \\ \\ 6 log( {2}^{2} ) = 12 log(2)$

Hence Proved

Hope it helps :))