log27+log8-log128 and log6-log8
I need answer
log27+log8-log128
log (27*8/128) = log (3^3 *2^3 / 2^3*2^4) = log (3^3 / 2^4) = 3 log 3 - 4 log 2
and log6-log8 = log (3/4) = log 3 - 2 log 2
log (27*8/128) = log (3^3 *2^3 / 2^3*2^4) = log (3^3 / 2^4) = 3 log 3 - 4 log 2
and log6-log8 = log (3/4) = log 3 - 2 log 2