Respuesta :

Nirina7
case 1
let   A=log2(0.5)4 = log(0.5)4 x log(0.5)4 =  4log(0.5) x  4log(0.5) =16 log(0.5 +0.5) = 16 log 1 =0 if it is a neperian logarithm

case 2
logn n = 1, if it is  logarithm with n as a basis, 
so A=log2(0.5)4 = 4log2(0.5) = - 4 log2(2) = -4 x1 = -4
 

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