题目

解下列方程:（1）log2(4-x)-log4(x-1)=1 （2）2lg(2x-1)=lg(-2x+7)+lg(x+1)

提问时间：2020-08-08

答案

log2(4-x)-log4(x-1)=1
log2(4-x)-log2(x-1)/log2(4)=1
log2(4-x)-log2(x-1)/2=1
2log2(4-x)-log2(x-1)=2
log2(4-x)²-log2(x-1)=2
log2[(4-x)²/(x-1)]=2
则(4-x)²/(x-1)=4
(4-x)²=4(x-1)
16-8x+x²=4x-4
20-12x+x²=0
(x-2)(x-10)=0
x1=2 x2=10
2lg(2x-1)=lg(-2x+7)+lg(x+1)
2lg(2x-1)-lg(-2x+7)-lg(x+1)=0
lg(2x-1)²-lg(-2x+7)-lg(x+1)=0
lg{(2x-1)²/[(-2x+7)(x+1)]}=0
则(2x-1)²/[(-2x+7)(x+1)]=1
(2x-1)²=(-2x+7)(x+1)
4x²-4x+1=-2x²-2x+7x+7
6x²-9x-6=0
2x²-3x-2=0
(2x+1)(x-2)=0
x1=-1/2 x2=2