设f（x)=logx (x-1) =(ln(x-1)) / (lnx)=(lg(x-1)) / (lgx)
求导f'(x)=((lnx/(x-1)-ln(x-1)/x)/(lnx)^2
由于lnx>ln(x-1) x>(x-1)所以((lnx/(x-1)-ln(x-1)/x)>0
则导数f'(x)=((lnx/(x-1)-ln(x-1)/x)/(lnx)^2>0
所以f（x)=logx (x-1) =(ln(x-1)) / (lnx)单调增
f(6)