Question

Se log 2 = 0,30 e log 3 = 0,48, calcular: Me ajudem pfvr
a) log raiz quadrada 32, 4
b)log 45
c) log 2,4
d) log 144

Answer 1

Ola Anne

log(2) = 0.30
log(3) = 0.48
log(5) = log(10/2) = log(10) - log(2) = 1 - 0.30 = 0.70

a)

32.4 =  324/10 

log(√(324/10)) = log(√(18²/10) = log(18) - log(√10) 

log(18) = log(2) + 2log(3) = 0.30 + 2*0.48 = 1.26 

log(√10) = log(10)/2 = 1/2 = 0.5 

log(√32.4) = 1.26 - 0.5 = 0.76 

b) 

log(45) = 2log(3) + log(5) = 2*0.48 + 0.70 = 1.66 

c)

2.4 = 24/10 

log(24/10) = 3log(2) + log(3) - log(10) 

3*0.30 + 0.48 - 1 = 1.38 - 1 = 0.38 

d)

log(144) = log(16*9) = 4log(2) + 2log(3) = 1.20 + 0.96 = 2.16