解 [log4(3)+log8(3)](log3(2)+log9(2))-log(1/2)(√32的4次方）
=(1/2log2(3)+1/3log2(3))(1/log2(3)+1/log2(9)+log2(32*32))
=5/6log2(3)*(1/log2(3)+1/log2(9)+10)
=5/6log2(3)*(1/log2(3)+1/2log2(3)+10)
=5/6+5/12+50/6log2(3)
=17/12+25/3log2(3)
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