Question

解方程：log2（4-x）-log4（x-1）=1

Answer 1

函数的定义域为：
{4-x>0
{x-1>0
.......................
1<x<4
原式可化为：
log4(4-x)/log4(2)=1+log4(x-1)=log4[4(x-1)
log4(4-x)/1/2=log4[4(x-1)
log4(4-x)^2=log4[4(x-1)
(4-x)^2=4(x-1)
(x-4)^2=4(x-1)
[(x-1)-3]^2=4(x-1)
(x-1)^2-10(x-1)+9=0
x-1=1或x-1=9(舍去)
所以x=2

Answer 2

4-x>0且x-1>0,所以 1<x<4,由log2（4-x）-log4（x-1）=1得2log4（4-x）-log4（x-1）=1
即log4[(4-x)^2/(x-1)=1 所以(4-x)^2/(x-1)=4即x^2-12x+20=0,解得 x=2(舍) x=10