Question

Usando Apenas log2=0,30,log3=0,47 e log5=0,69,calcule:
a) log4+log5
b) log5+log6
c) log150−log5
d) log120−log6

Answer 1

Resposta:

a) 1,29

b) 1,46

c) 1,46

d) 1,29

Explicação passo-a-passo:

a) log4+log5=

log(2.2)+log5=

log2+log2+log5=

0,30+0,30+0,69=1,29

b) log5+log6=

log5+log(2.3)=

log5+log2+log3=

0,69+0,30+0,47=1,46

c) log150−log5=

log(2.3.5.5)−log5=

log2+log3+log5+log5-log5=

log2+log3+log5=

0,30+0,47+0,69=1,46

d) log120−log6=

log(2.2.2.3.5)-log(2.3)=

log2+log2+log2+log3+log5-log2-log3=

log2+log2+log5=

0,30+0,30+0,69=1,29