1/log1/2为底1/3的对数+1/log1/5为底1/3的对数=n,求n的值属于区间
人气:364 ℃ 时间:2020-06-24 03:58:49
解答
1/(log 1/2(1/3))+1/(log 1/5(1/3))=nn=1/(lg(1/3)/lg(1/2))+1/(lg(1/3)/lg(1/5))=(lg(1/2)/lg(1/3))+(lg(1/5)/lg(1/3))=(lg(1/2)+(lg(1/5))/lg(1/3)=(-lg(2)-(log(5))/(-lg(3))=(lg(2)+(log(5))/(lg(3))=1/(lg(3))=...
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