請解以下算式: (log以2為底的3 - log以¼ 為底的81)*(log以3為底的8 - log以27為底的64) =? 請附上詳答 謝謝?

Sol
｛log2_3－[log(1/4)_81]｝*(log3_8－log27_64)
=｛log3/log2－[log81/log(1/4)]｝*(log8/log3－log64/log27)
=[log3/log2+log81/log4]*[3log2/log3－6log2/(3log3)]
=(log3/log2+2log3/log2)*(3log2/log3－2log2/log3)
=(3log3/log2)*(log2/log3)
=3

[(log 3)/(log 2) -(log 81)/(log 1/4)][(log 8)/(log 3) -(log 64)/(log 27)]
=[(log 9)/(log 4)+(2 log 9)/(log 4)][3(log 8)/(log 27) -2(log 8)/(log 27)]
=[3(log 9)/(log 4)][(log 8)/(log 27)]
=[6(log 3)/(2 log 2)][3(log 2)/(3 log 3)]
=3

-[log (1/4)]
=-[(log 1)-(log 4)]
=-(-log 4)
=log 4