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If log3 2 , log3 (2x – 5), and log3 (2x – 7/2) are in arithmetic progression, determine the value of x.

Simran Bhatia , 10 Years ago
Grade 11
anser 1 Answers
Aditi Chauhan

Last Activity: 10 Years ago

Hello Student,
Please find the answer to your question
Given that log3 2, log3(2x – 5), log3 (2x -7/2) are in A. P.
⇒ 2 log3 (2x – 5) = log3 2 + log3 (2x – 7/2)
⇒ (2x – 5)2 = 2 (2x – 7/2)
⇒ (2x)2 – 10.2x + 25 – 2.2x + 7 = 0
⇒ (2x)2 – 10.2x + 25 – 2.2 x + 7 = 0
⇒ (2x)2 – 12.2x + 32 = 0
Let 2x = y, then we get,
Y2 – 12y + 32 = 0 ⇒ (y – 4) (y – 8) = 0
⇒ y = 4 or 8 ⇒ 2x = 22 or 23 ⇒ x = 2 or 3
But for log3(2x – 5) and log3 (2x – 7/2) to be defined
2x – 5 > 0 and 2x – 7/2 > 0
⇒2x > 5 and 2x > 7/2
⇒ 2x > 5
⇒ x ≠ 2 and therefore x = 3.

Thanks
Aditi Chauhan
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