Question

a>2,log以(a-1)为底a的对数与log以a为底（a+1）对数比较大小

Answer 1

a>2,所以a-1>1

令 f(x) = log(x-1) x = lnx/ln(x-1) (x>1)

所以 f'(x) = [ln(x-1) /x - lnx/(x-1)]/[ln(x-1)]^2 =

= [(x-1)ln(x-1) - xlnx ]/x(x-1)[ln(x-1)]^2 < 0

（因为 (x-1)ln(x-1)< xlnx )

所以f(x）是减函数

所以 log以(a-1)为底a的对数 > log以a为底（a+1）对数