log8(9)/log2(3) 都忘记了说详细过程log底可以怎么算
2个回答
换底公式
原式=(lg9/lg8)/(lg3/lg2)
=(2lg3/3lg2)/(lg3/lg2)
=(2/3)(lg3/lg2)/(lg3/lg2)
=2/3
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