已知函数f(x)=logx+x-b(a>0,且a不等于1),当2

f(x)=log(a)x+x-bf'(x)=1/[(lna)x]+1lna＜log(0.5)a＜log(0.5)2=-1∴x∈(n,n+1),n∈正整数时,f'(x)＞0即f(x)在x∈(n,n+1),n∈正整数时单调递增f(2)=log(a)2+2-b＜log(2)2+2-3=0f(3)=log(a)3+3-b＞log(3)3+3-4=0f(x0)...