Question

Dados log2= 0,301 e log3= 0,477 ,calcule:

log12

log24

log3000

log72

log0,6

Answer 1

log(12)=log(3x4)=log(3)+log(4)=log(3)+log(2²)=log(3)+2log(2)=
=0,477+2x0,301=0,477+0602=1,079

log(24)=log(2x12)=log(2)+log(12)=0,301+1,079=1,38

log(3000)=log(3x10³)=log(3)+log(10³)=log(3)+3log(10)=0,477+3x1=3,477

log(72)=log(12x6)=log(12)+log(6)=log(12)+log(3x2)=
=log(12)+log(3)+log(2)=1,079+0,477+0,301=1,857

log(0,6)=log(6/10)=log(6)-log(10)=log(3x2)-1=log(3)+log(2)-1=
=0,477+0,301-1= -0,222

Answer 2

Resposta:

log(12)=log(3x4)=log(3)+log(4)=log(3)+log(2²)=log(3)+2log(2)=

=0,477+2x0,301=0,477+0602=1,079

log(24)=log(2x12)=log(2)+log(12)=0,301+1,079=1,38

log(3000)=log(3x10³)=log(3)+log(10³)=log(3)+3log(10)=0,477+3x1=3,477

log(72)=log(12x6)=log(12)+log(6)=log(12)+log(3x2)=

=log(12)+log(3)+log(2)=1,079+0,477+0,301=1,857

log(0,6)=log(6/10)=log(6)-log(10)=log(3x2)-1=log(3)+log(2)-1=

=0,477+0,301-1= -0,222

Explicação passo-a-passo: