Question

解方程log4(3-x)+log0.25(3+x)=log4(1-x)+log0.25(2x+1)

Answer 1

log4(3-x)+log0.25(3+x)=log4(1-x)+log0.25(2x+1)
log4(3-x)-log4(1-x)=log0.25(2x+1)-log0.25(3+x)
log4[(3-x)/(1-x)]=log0.25[(2x+1)/(3+x)]
log4[(3-x)/(1-x)]=log1/4[(2x+1)/(3+x)]
log4[(3-x)/(1-x)]=-log4[(2x+1)/(3+x)]
log4[(3-x)/(1-x)]+log4[(2x+1)/(3+x)]=0
log4[(3-x)(2x+1)/(3+x)(1-x)]=0
(3-x)(2x+1)/(3+x)(1-x)=1
(3-x)(2x+1)=(3+x)(1-x)
6x+3-2x²-x=3-3x+x-x²
-x²+7x=0
-x(x-7)=0
x=0或x=7
x=7时，3-x=-4＜0，原方程无意义，所以x=7舍去
所以x=0

Answer 2

log4(3-x)-log4 (3+x)=log4 (1-x)-log4 (2x+1)
log4(3-x)/(3+x)=log4 (1-x)/(2x+1)
(3-x)/(3+x)=(1-x)(2x+1)
(3-x)(2x+1)=(3+x)(1-x)
5x-2x^2+3=3-2x-x^2
x^2-7x=0
x(x-7x)=0
x=0或x=7

Answer 3

log4(3-x)-log4(3+x)=log4(1-x)-log4(2x+1)
log4(3-x)/(3+x)=log4(1-x)/(2x+1)
(3-x)/(3+x)=(1-x)/(2x+1)
(3-x)(2x+1)=(1-x)(3+x)
-2x2+5x+3=-x2-2x+3
x2-7x=0
x1=0,x2=7