a) log de base 3 (3.81)
b)log de base 7 (49.343/7)
a) Na base 3:
log (3.81) =
log (3) + log (81) =
1 + 4 = 5
b) Na base 7:
log (49.343/7)=
log(49) + log(343) - log(7)
2 + 3 -1= 4
a) log de base 3: log (3.81) =log (3) + log (81) = 5
b) log de base 7: log (49.343/7)=log(49) + log(343) - log(7)= 4
Comments
a) Na base 3:
log (3.81) =
log (3) + log (81) =
1 + 4 = 5
b) Na base 7:
log (49.343/7)=
log(49) + log(343) - log(7)
2 + 3 -1= 4
a) log de base 3: log (3.81) =log (3) + log (81) = 5
b) log de base 7: log (49.343/7)=log(49) + log(343) - log(7)= 4