(c) Given that, log102,log10(2x−1), log10(2x+3) are in AP. ∴2log10(2x−1)=log102+log10(2x+3) log10(2x−1)2=log102(2x+3) ⇒22x+1−2⋅2x=2⋅2x+6 ⇒(2x)2−4(2x)−5=0 Let 2x=y ⇒y2−4y−5=0 ⇒y=5 or y=−1 (Ignore because 2x cannot be negative) ⇒y=5⇒2x=5 x=log25 Hence, option (c) is correct.