Correct Answer - D
The given numbers are in A.P. Therefore,
`2log_(4)(2^(1-x)+1)=log_(2)(5xx2^(x)+1)+1`
or `2log_(2^(2))(2/(2^(x))+1)=log_(2)(5xx2^(x)+1)+log_(2)2`
or `2/2loglog_(2)(2/(2^(x))+1)=log_(2)(10xx2^(x)+2)`
or `2/(2^(x))+1=10xx2^(x)+2`
`rArr2/y+1=10y+2`, where `2^(x)=y`
or `10y^(2)+y-2=0`
or `(5y-2)(2y+1)=0`
`rArry=2//5` or `y=-1//2`
`rArr2^(x)=2//5` or `2^(x)=-1//2`
`rArr x=log_(2)2-log_(2)5` `[because2^(x)` cannot be negative`]`
`rArrx=log_(2)2-log_(2)5`
`rArrx=1-log_(2)5`