(2)求证:log56·log54＜1;
(3)已知f(x)=logx(x+1),
①比较f(1 024)·f(1 025)·…·f(2 048)与1.1的大小;
②求证:f(n)＞f(n+1)(n∈N
,n≥2).
(2)求证:log56·log54＜1;
(3)已知f(x)=logx(x+1),
①比较f(1 024)·f(1 025)·…·f(2 048)与1.1的大小;
②求证:f(n)＞f(n+1)(n∈N
,n≥2).(1)解析
:log23-log34=
∴log23＞log34.
(2)证明:
=
log524
＜
log525=1.
(3)①解析
:f(1 024)·f(1 025)·…·f(2 048)=
=1.1.②证明:f(n)＞f(n+1)
logn(n+1)＞logn+1(n+2)logn+1(n+2)·logn+1n＜1,仿(2)的证明思路,此式易证.