Here, log 2, log(2^x-1) and log(2^x+3) are in A.P.

So, log(2^x-1)-log2=log(2^x+3)-log(2^x-1)

Let 2^x=y

Then above equation is written as

(y-1)²=2(y+3)

∴y²-2y+1=2y+6

∴y²-4y-5=0

∴(y-5)(y+1)=0

∴ y=5 or y=-1

∴ 2^x=5 or 2^x=-1

∴ x=log₂5 or 2^x=-1 is not possible

∴ x=log₂5