已知a>2,求证log(a-1)>loga(a+1)求证log(a-1)a>loga(a+1)
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已知a>2,求证log(a-1)>loga(a+1)求证log(a-1)a>loga(a+1)已知a>2,求证log(a-1)>loga(a+1)求证log(a-1)a>loga(a+1)已知a>2,
已知a>2,求证log(a-1)>loga(a+1)求证log(a-1)a>loga(a+1)
已知a>2,求证log(a-1)>loga(a+1)
求证log(a-1)a>loga(a+1)
已知a>2,求证log(a-1)>loga(a+1)求证log(a-1)a>loga(a+1)
因为a>2,
所以lna,ln(a-1)>0
要证
log(a-1)&a>loga&(a+1)
即证
ln(a-1)*ln(a+1)
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