Question

(1)比较log₂3与log₃4的大小;

(2)求证:log₅6·log₅4＜1;

(3)已知f(x)=log_x(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).

Answer 1

(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)log_n(n+1)＞log_n+1(n+2)?log_n+1(n+2)·log_n+1n＜1,仿(2)的证明思路,此式易证.