log以9为底4的对数+ log以3为底8的对数除以log以1⼀3为底16的对数

2025-02-24 00:06:43
推荐回答(1个)
回答1:

[log9(4)+log3(8)]/log1/3(16)=[lg4/lg9+lg8/lg3]/[lg16/lg1/3]
=[2lg2/2lg3+3lg2/lg3]/[4lg2/(-lg3)]
=4lg2/[- 4lg2]=-1