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We know, log32 , log3(2x - 5) and log3(2x - 7/2) are in A.P.
=> log32 + log3(2x - 7/2) = 2 log3(2x - 5)
=> log3 [2(2x - 7/2)] = log3(2x - 5)2
=> 2(2x - 7/2) = (2x - 5)2
Let 2x = y.
=> 2(y – 7/2) = (y – 5)2y
=> 2y – 7 = y2 – 10y +25
=> y2 – 12y +32 = 0
=> y2 – 4y – 8y +32 = 0
=> (y – 4) (y – 8) = 0
=> y = 4 or y = 8
i.e., 2x = 4 or 2x = 8
=> x = 2 or x = 3
When x = 2, the second term of the AP is not defined.
(2x – 5) will be negative => log is not defined.
Therefore, x=3.
Answered by
1
Answer:
here x is equal to 3
hope its right
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