a,b,c are terms of ap then

b-a = c-b

 2b = a+c      ..........1

log2 , log2^x-1 , log2^x+3 are in ap so

2log2^x+1 = log2+log2^x+3

loga + logb =logab

now expression is

log(2^x+1)² = log[2(2^x+3)]

 (2^x+1)² = 2(2^x+3)

put 2^x = t

(t+1)² = 2(t+3)

 t = 1,5

2^x = 1       or   2^x = 5   , x = log5/log2

x = 0 cannot be the solution so x = log5/log2 is the required solution