202601164

题目

a>0,b>1,aea=a+3,b3lnb=lnb+1,a(b31)

解析

ab

“隐零点问题”

continue



continue

使
a=logbba,a=blogba

continue

aea=a+3ealnea=lnea=3(ea1)lnea=3
b3lnb=lnb+1(b31)lnb=1(b31)3lnb=3
(b31)lnb3=3

continue

f(x)=(x1)lnx
f(ea)=3,f(b3)=3
f(ea)=f(b3)
a>0,b>1ea>1,b3>1

continue

f(x)=3(1,+)
(x1)lnx=3
lnx=3x1(1,+)
(1,+)y=lnxy=3x1

另解

x>1x1>0,lnx>0,y=x1y=lnx
f(x)=(x1)lnx
g(x)=f(x)3=(x1)lnx3
g(x)(1,+)g(e)=(e1)3<0
g(e2)=(e21)23>0
g(x)

综上所述

f(x)=3(1,+)
ea=b3
a(b31)=a(ea1)=aeaa=3