The correct option is A log211
x+1−2log2(2x+3)+2log4(10−2−x)=0
⇒x+1=log2(2x+3)2−log2(10−2−x)
⇒x+1=log2(2x+3)2(10−2−x)
⇒2⋅2x=(2x+3)2(10−2−x)
Assuming 2x=t
⇒2t=(t+3)2(10−t−1)
⇒20t−2=(t+3)2
⇒t2−14t+11=0 ⋯(1)
If t1=2x1, t2=2x2 are roots of the equation (1), then
t1t2=2x1⋅2x2=11
∴x1+x2=log211