If log3 2, log3 (2x – 5), and log3 (2x – 7/2) are in A.P., the va

1.	If log3 2, log3 (2x – 5), and log3 (2x – 7/2) are in A.P., the value of x is(a) 0 (b) \(\frac{1}{3}\)(c) 2 (d) 3
Answer» Answer : (d) 3 log₃2, log₃ (2^x – 5) and log₃ (2^x - \(\frac{7}{2}\)) are in A.P ⇒ 2 log₃ (2^x – 5) = log₃ 2 + log₃ (2^x – 7/2) (∵ a, b, c in A.P ⇒ 2b = a + c) ⇒ log₃ (2^x – 5)² = log₃[2(2^x – 7/2)] (∵ a log b = log b^a and log a + log b = log ab) ⇒ (2^x – 5)² = 2^x+1 – 7 Let 2^x= y. Then, (y – 5)² = 2y – 7 ⇒ y² – 10y + 25 = 2y – 7 ⇒ y² – 12y + 32 = 0 ⇒ (y – 8) (y – 4) = 0 ⇒ y = 8 or 4 ⇒ 2^x = 8 or 2^x = 4 2^x = 8 ⇒ x = 3 (∵ 2^x = 4 shall make the term log₃(2^x – 5) negative which is not possible)

If log3 2, log3 (2x – 5), and log3 (2x – 7/2) are in A.P., the value of x is(a) 0 (b) \(\frac{1}{3}\)(c) 2 (d) 3

Answer»

Answer : (d) 3

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

⇒ 2 log₃ (2^x – 5) = log₃ 2 + log₃ (2^x – 7/2) (∵ a, b, c in A.P ⇒ 2b = a + c)

⇒ log₃ (2^x – 5)² = log₃[2(2^x – 7/2)] (∵ a log b = log b^a and log a + log b = log ab)

⇒ (2^x – 5)² = 2^x+1 – 7

Let 2^x= y. Then,

(y – 5)² = 2y – 7

⇒ y² – 10y + 25 = 2y – 7

⇒ y² – 12y + 32 = 0

⇒ (y – 8) (y – 4) = 0

⇒ y = 8 or 4

⇒ 2^x = 8 or 2^x = 4

2^x = 8

⇒ x = 3 (∵ 2^x = 4 shall make the term log₃(2^x – 5) negative which is not possible)