Log 384/5 + log 81/32 + 3log 5/3 + log 1/9

Answer 1

factor 384, 81,32,9
you should be able to factor into 3,5,and 2 factors. Example:
384=3*128=3*4*32=3*4*4*4*2=3* 2^7
so the log of 384=log3+7log2 so the first term Log 384/5 becomes
7log2+log3-log5
do that same thing with the other terms.

Answer 2

Log 384/5 + log 81/32 + 3log 5/3 + log 1/9
by the rules of logs we get
= log 384 - log 5 + log 81 - log 32 + 3log 5 - 3log 3 + log 1 - log 9
= log 3 + log 128 - log 5 + 4log 3 - 5log 2 + 3log 5 - 3log 3 + 0 - 2 log3
= 0log3 + 7log2 - log5 - 5log2 + 3log5
= 2log2 + 2log5
= log4 + log25 = log(4*25)
= log 100
= 2

or, easier way

Log 384/5 + log 81/32 + 3log 5/3 + log 1/9
= log(384/5*81/32*125/27*1/9)
= log 100
= 2