实数a,b满足0

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实数a,b满足0
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实数a,b满足0
实数a,b满足0

实数a,b满足0
对a^b,b^a取常用对数并做商得
blga/(algb)=b/a*logb(a)>1
因此lg(a^b)>lg(b^a)
a^b>b^a

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