Question

alguém me ajuda? por favorrr
Usando apenas log 20= 1,30 log 30 = 1,47 e log 60 = 1,79 calcule:

a) log 5+log 6
b) log 120-6
c) log 180-log 9
d) log 600-log30

Answer 1

log20 = log2 + log10 = log2 + 1 log2 + 1 = 1,30log2 = 1,30 - 1 =
0,30log30 = log3 + log10 = log3 + 1
log3 + 1 = 1,47log3 = 1,47 - 1 = 0,47log60 = log6 + log10 = log6 + 1
log6 + 1 = 1,79log6 = 1,79 - 1 = 0,79
a) log5 + log6 = log[5*6] = log30 = 1,47
b) log120 - 6 = [log3 + log4 + log10] - 6 = [log3 + log2² + log10] - 6 = 
[0,47 + 2log2 + 1] -6 = [0,47 + 2(0,30) + 1] -6 = 0,47 + 0,60 + 1 - 6 =
2,07 - 6 = -3,93
c) log180 - log9 = log[180/9] = log20 = 1,30
d) log600 - log30 = log[600/30] = log20 = 1,30