log以9为底4的对数+ log以3为底8的对数除以log以1/3为底16的对数
人气:362 ℃ 时间:2020-05-22 05:04:01
解答
[log9(4)+log3(8)]/log1/3(16)=[lg4/lg9+lg8/lg3]/[lg16/lg1/3]
=[2lg2/2lg3+3lg2/lg3]/[4lg2/(-lg3)]
=4lg2/[- 4lg2]=-1
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