Question

sendo log2 = 0,3 e log3 = 0,48 determine log 2 + log 4 + log 8 / log 6 + log 9

Answer 1

Sabemos que:

log a.b = log a + log b
log a/b = log a - log b
log aⁿ = n.log a

log 2 = 0,3
log 3 = 0,48

Logo:

(log 2 + log 4 + log 8)/(log 6 + log 9)

(log 2 + log 2² + log 2³)/(log 2.3 + log 3²)

(log 2 + 2.log 2 + 3.log 2)/(log 2 + log 3 + 2.log 3)

(6.log 2)/(log 2 + 3.log 3)

(6.0,3)/(0,3 + 3.0,48)

1,8/(0,3 + 1,44)

1,8/1,74 ≈ 1,0345