log2, log(2n − 1) and log (2n + 3) are in A.P. Show that n = \(\frac

1.	log2, log(2n − 1) and log (2n + 3) are in A.P. Show that n = \(\frac{log5}{log2}\)
Answer» 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 = \(\frac{log5}{log2}\)

log2, log(2n − 1) and log (2n + 3) are in A.P. Show that n = \(\frac{log5}{log2}\)

Answer»

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 = \(\frac{log5}{log2}\)