Question

Log12 1+ log⁴ 64 + log8 8 + log100

Answer 1

$log12 \: 1 = x = >{12}^{x} = 1 = > {12}^{x} = {12}^{0} = > x = 0 \\ log4 \: 64 = > log4 \: {4}^{3} = > 3 \\ log8 \: 8 = 1 \\ log100 = log {10}^{2} = 2 \\ 0 + 3 + 1 + 2 = 6$