The equation logx+1⁡(x−0.5)=logx−0.5⁡(x+1) has

log2⁡(x−0.5)log2⁡(x+1)=log2⁡(x+1)log2⁡(x−0.5) or     log2⁡(x+1)2=log2⁡(x−0.5)2∴    log2⁡(x+1)=±log2⁡(x−0.5) If     log2⁡(x+1)=log2⁡(x−0.5). Then,     x+1=x−0.5, hence no solution  If     log2⁡(x+1)=log⁡(x−0.5)−1. Then,     x+1=1x−(1/2)=22x−1 or     (x+1)(2x−1)=2 or     2x2+x−3=0 or     2x2+3x−2x−3=0 or     (x−1)(2x+3)=0⇒    x=1