Question

Sabendo-se que log 162= e log 48=, determine o valor de log 6 em função de a e b.

Answer 1

Oi Diogo

log(162) = a
log(48) = b

log(2*81) = log(2) + 4*log(3) = a
log(48) = 4log(2) + log(3) = b 

4log(2) + 16log(3) = 4a
4log(2) + log(3) = b

15log(3) = 4a - b
log(3) = (4a - b)/15 

4log(2) + (4a - b)/15 = b
4log(2) = 15b/15 - 4a/15 + b/15 = 16b/15 - 4a/15
log(2) = 4b/15 - a/15 

log(6) = log(2) + log(3) = 4b/15 - a/15 + 4a/15 - b/15

log(6) = 3b/15 + 3a/15 = b/5 + a/5 = (a + b)/5