设函数f(x)的定义域为(0,+∞),且满足条件f(4)=1,对于任意x1,x2∈(0,+∞),有f(x1x2)=f(x1)+f(x2),当x1>x2设函数f(x)的定义域为(0，+∞)，且满足条件f(4)=1,对于任意x1,x2∈(0,+∞),有f(x1x2)=f(x1)+f(x2),当x1>x2时，

Question

设函数f(x)的定义域为(0,+∞),且满足条件f(4)=1,对于任意x1,x2∈(0,+∞),有f(x1x2)=f(x1)+f(x2),当x1>x2
设函数f(x)的定义域为(0，+∞)，且满足条件f(4)=1,对于任意x1,x2∈(0,+∞),有f(x1x2)=f(x1)+f(x2),当x1>x2时，有f(x1)>f(x2)。若f(3x+1)+f(2x-6)≤3，求x的取值范围

Answer 1

put x1=4,x2=4
f(16) =f(4) + f(4)
f(16) = 2
put x1=16,x2=4
f(64) = f(16) + f(4)
= 2 + 1 = 3
f(3x+1)+f(2x-6)≤3
f((3x+1)(2x-6))≤f(64)
=> (3x+1)(2x-6)≤ 64
6x^2-16x-6≤ 64
3x^2-8x-36≤0
(4-√124)/ 6≤x≤(4+√124)/ 6