If log32, log3 (2x – 5) and log3 (2x – \(\frac{7}{2}\)) are in A.P

1.	If log32, log3 (2x – 5) and log3 (2x – \(\frac{7}{2}\)) are in A.P., then what is the value of x ?
Answer» Given, log₃2, log₃(2^x – 5) and log₃(2^x – \(\frac{7}{2}\)) are in A.P. ⇒ 2[log₃(2^x-5)] = log₃ 2 + log₃\(\big(2^x-\frac{7}{2}\big)\) ⇒ log₃ (2x – 5)² = log₃ [2 × \(\big(2^x-\frac{7}{2}\big)\)] ⇒ (2^x – 5)² = (2^{x + 1} – 7) ⇒ 2^2x – 10.2^x + 25 = 2.2^x – 7 ⇒ 2^2x – 12.2^x + 32 = 0 ⇒ y² – 12y + 32 = 0 [Let y = 2^x] ⇒ (y – 8) (y – 4) = 0 ⇒ y = 8 or 4 ⇒ 2^x = 8 or 2^x = 4 ⇒ x = 3 or 2

If log32, log3 (2x – 5) and log3 (2x – \(\frac{7}{2}\)) are in A.P., then what is the value of x ?

Answer»

Given, log₃2, log₃(2^x – 5) and log₃(2^x – \(\frac{7}{2}\)) are in A.P.

⇒ 2[log₃(2^x-5)] = log₃ 2 + log₃\(\big(2^x-\frac{7}{2}\big)\)

⇒ log₃ (2x – 5)² = log₃ [2 × \(\big(2^x-\frac{7}{2}\big)\)]

⇒ (2^x – 5)² = (2^{x + 1} – 7) ⇒ 2^2x – 10.2^x + 25 = 2.2^x – 7

⇒ 2^2x – 12.2^x + 32 = 0 ⇒ y² – 12y + 32 = 0 [Let y = 2^x]

⇒ (y – 8) (y – 4) = 0 ⇒ y = 8 or 4 ⇒ 2^x = 8 or 2^x = 4 ⇒ x = 3 or 2