Question

[log4(3)+log8(3)](log3(2)+log9(2))-log(1/2)(√32的4次方）

Answer 1

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)

追问

答案是5/2

追答

啊 那是哪里算错了。。

追问

我也不知道啊，你能在帮忙看看

追答

我再算下 前面看错了。
你题目里面的(√32的4次方）实际表示的根号32的1/4次方吧?
[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(32^1/4))
=5/6log2(3)*(1/log2(3)+1/log2(9))+5/4
=5/6log2(3)*(1/log2(3)+1/2log2(3)+5/4
=5/6log2(3)*(3/2)/log2(3)+5/4
=5/6*3/2+5/4
=5/4+5/4
=5/2