Given, log2, log(2n − 1) and log (2n + 3) are in A.P.
∴ log(2n + 3) − log(2n − 1) = log(2n − 1) - log2
⇒ log \(\big(\frac{2^n+3}{2^n−1}\big)\) = log \(\big(\frac{2^n-1}{2}\big)\)
Taking antilog, we get
⇒ \(\frac{2^n+3}{2^n−1}\) = \(\frac{2^n-1}{2}\)
⇒ 2n − 4.2n − 5 = 0
⇒ y2 − 4y − 5 = 0 where y = 2n
⇒ y − 5 and y = −1 (reject)
Hence, 2n = 5
or n log2 = log5 or n = \(\frac{log5}{log2}\)