Question

sabendo que log 2=0,301 e log 3=0,477 e log5=0,699. calcule o valor de:

a)log 10


b)log 64

Answer 1

Resposta:

$log(10) = log(2) + log(5) = \\ = 0.301 + 0.699 = 1 \\ log(64) = 6 \times log(2) = \\ = 6 \times 0.301 =1.806$