sendo log 2= a e log 3 = b, calcule, em função de a e b:
a) log54
b) log150
c) log³√12
d) log 40
niltonjr2001:
Na c) é log na base 3 ou log da raiz cúbica de 12?
na raiz cúbica de 12
Soluções para a tarefa
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a)
log 54 = log 27.2
log 54 = log 3³.2
log 54 = log 3³ + log 2
log 54 = 3.log 3 + log 2
log 54 = 3.b + a
log 54 = a + 3b
b)
log 150 = log 300/2
log 150 = log 300 - log 2
log 150 = log 3.100 - log 2
log 150 = log 3 + log 100 - log 2
log 150 = b + 10 - a
log 150 = b - a + 10
c)
log ³√12 = log 12^(1/3)
log ³√12 = (1/3).log 2².3
log ³√12 = (1/3).(2.log 2 + log 3)
log ³√12 = (1/3).(2.a + b)
log ³√12 = (2a + b)/3
d)
log 40 = log 4.10
log 40 = log 2² + log 10
log 40 = 2.log 2 + 1
log 40 = 2a + 1
log 54 = log 27.2
log 54 = log 3³.2
log 54 = log 3³ + log 2
log 54 = 3.log 3 + log 2
log 54 = 3.b + a
log 54 = a + 3b
b)
log 150 = log 300/2
log 150 = log 300 - log 2
log 150 = log 3.100 - log 2
log 150 = log 3 + log 100 - log 2
log 150 = b + 10 - a
log 150 = b - a + 10
c)
log ³√12 = log 12^(1/3)
log ³√12 = (1/3).log 2².3
log ³√12 = (1/3).(2.log 2 + log 3)
log ³√12 = (1/3).(2.a + b)
log ³√12 = (2a + b)/3
d)
log 40 = log 4.10
log 40 = log 2² + log 10
log 40 = 2.log 2 + 1
log 40 = 2a + 1
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