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
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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
1/(log 1/2(1/3))+1/(log 1/5(1/3))=n
n=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))
=2.0969033
n=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))
=2.0969033
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