已知a>0且a不等于1,x>0,y>0,试比较loga(1/x)*loga(1/y)与与loga根号(y/x)*loga根号(x/y)的大小
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![已知a>0且a不等于1,x>0,y>0,试比较loga(1/x)*loga(1/y)与与loga根号(y/x)*loga根号(x/y)的大小](/uploads/image/z/7105205-29-5.jpg?t=%E5%B7%B2%E7%9F%A5a%3E0%E4%B8%94a%E4%B8%8D%E7%AD%89%E4%BA%8E1%2Cx%3E0%2Cy%3E0%2C%E8%AF%95%E6%AF%94%E8%BE%83loga%281%2Fx%29%2Aloga%281%2Fy%29%E4%B8%8E%E4%B8%8Eloga%E6%A0%B9%E5%8F%B7%28y%2Fx%29%2Aloga%E6%A0%B9%E5%8F%B7%28x%2Fy%29%E7%9A%84%E5%A4%A7%E5%B0%8F)
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已知a>0且a不等于1,x>0,y>0,试比较loga(1/x)*loga(1/y)与与loga根号(y/x)*loga根号(x/y)的大小
已知a>0且a不等于1,x>0,y>0,试比较loga(1/x)*loga(1/y)与与loga根号(y/x)*loga根号(x/y)的大小
已知a>0且a不等于1,x>0,y>0,试比较loga(1/x)*loga(1/y)与与loga根号(y/x)*loga根号(x/y)的大小
loga(1/x)*loga(1/y)
=[-loga(x)][-loga(y)]
=loga(x)*loga(y)
loga √(y/x)*loga √(x/y)
=(1/2)[loga(y)-loga(x)](1/2)[loga(x)-loga(y)]
=-(1/4)[loga(x)-loga(y)]²
loga(1/x)*loga(1/y) - loga √(y/x)*loga √(x/y)
=loga(x)*loga(y) + (1/4)[loga(x)-loga(y)]²
=(1/4)[loga(x)+loga(y)]²≥0
所以loga(1/x)*loga(1/y) ≥ loga √(y/x)*loga √(x/y)
loga(1/x)*loga(1/y)
=[-loga(x)][-loga(y)]
=loga(x)*loga(y)
loga √(y/x)*loga √(x/y)
=(1/2)[loga(y)-loga(x)](1/2)[loga(x)-loga(y)]
=-(1/4)[loga(x)-loga(y)]²