Question

Log 1,2 (log 64 32)+9^log 3 5

Answer 1

$log_{1,2} (log_{64} 32)+9^{log_3 5} = log_{1,2} ( frac{5}{6} log_{2} 2)+3^{log_3 5^2} =$
$= log_{1,2} 1,2^{-1}+5^2 = -1+ 25 = 24$

Answer 2

Будет = (- 1 + 25) = 24