Equation log4⁡(3−x)+log0.25⁡(3+x)=log4⁡(1−x)+log0.252x+1 has

log4⁡(3−x)+log0.25⁡(3+x)=log4⁡(1−x)+log0.25⁡(2x+1)⇒  log4⁡(3−x)−log4⁡(3+x)=log4⁡(1−x)−log4⁡(2x+1)⇒ log4⁡(3−x)+log4⁡(2x+1)=log4⁡(1−x)+log4⁡(3+x)⇒ (3−x)(2x+1)=(1−x)(3+x)⇒ 3+5x−2x2=3−2x−x2⇒ x2−7x=0⇒ x=0,7Only x = 0 is the solution and x=7 is to be rejected.