Asked by Evan

log27+log8-log128 and log6-log8

4 years ago

Answered by Evan

I need answer

4 years ago

Answered by Anonymous

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

4 years ago

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