Question

se log 2 =0,301 e log 3=0,477 , entao log2 6 e:

Answer 1

log2 6=
log2 (2×3)=
(log2 2)+(log2 3)=
(log 2 / log 2)+(log 2 / log 3)=
(0,301/0,301)+(0,301/0,477)=
1 + (0,301/0,477)=
1 + 1,585=
2,585