) Dados log 2 = 0,30, log 3 = 0,48 e log de 5 = 0,70, determine:
a) log 32
b) log 30
c) log 60
d) log 1,2
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log(2) = 0.30, log(3) = 0.48, log(5) = 0.70
a) log(32) = log(2^5) = 5*log(2) = 5*0.30 = 1.5
b) log(30) = log(3) + log(2) + log(5) = 0.48 + 0.30 + 0.70 = 1.48
c) log(60) = 2*log(2) + log(3) + log(5) = 0.60 + 0.48 + 0.70 = 1.78
d) log(1.2) = log(12) - log(10) = 2log(2) + log(3) - 1 = 0.60 + 0.48 - 1 = 0.08
log(2) = 0.30, log(3) = 0.48, log(5) = 0.70
a) log(32) = log(2^5) = 5*log(2) = 5*0.30 = 1.5
b) log(30) = log(3) + log(2) + log(5) = 0.48 + 0.30 + 0.70 = 1.48
c) log(60) = 2*log(2) + log(3) + log(5) = 0.60 + 0.48 + 0.70 = 1.78
d) log(1.2) = log(12) - log(10) = 2log(2) + log(3) - 1 = 0.60 + 0.48 - 1 = 0.08
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