If (x+1)log10⁡(x+1)=100(x+1) then

(x+1)log10⁡(x+1)=100(x+1)⇒ log10⁡(x+1)log10⁡(x+1)=log10⁡(100(x+1)) log10⁡(x+1)log10⁡(x+1)=2+log10⁡(x+1) Let log10⁡(x+1)=y⇒ y2−y−2=0⇒y=2 or −1⇒log10⁡(x+1)=2,−1⇒x+1=100,1/10⇒x=99 or −9/10