Question

considerando que 10^0,301=2 e 100,477=, veja como podemos calcular log6

Answer 1

$-\ \textgreater \ \ 10^{0,301}=2 \ \implies \ log2=0,301\\ -\ \textgreater \ \ 10^{0,477}=3 \ \implies \ log3=0,477\\ -\ \textgreater \ \ 6=2.3\\\\ log6= log(2.3)= log2+log3=0,301+0,477=0,778$