Question

Solve the question:​

Answer 1

Answer:

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.

Answer 2

Answer:

here x is equal to 3

hope its right