Since log2,log(2n−1) and log(2n+3) are in Arithmetic Progression, we can say that:
log(2n−1)−log2=log(2n+3)−log(2n−1) ⇒log(
2n−1
2
)=log(
2n+3
2n−1
) ⇒
2n−1
2
=
2n+3
2n−1
Let 2n=x ⇒(x−1)2=2(x+3) ⇒x2−2x+1=2x+6 ⇒x2−4x−5=0 ⇒x2−5x+x−5=0 ⇒x(x−5)+(x−5)=0 ⇒(x−5)(x+1)=0 ⇒x−5=0 OR x+1=0 ⇒x=5 OR x=−1. ⇒2n=5OR2n=−1. Since, 2n=−1 is not possible, the only solution is 2n=5⇒log25=n.