If log32, log3(2x - 5), log3(2x - 7/2) are in arithmetic progression, then the value of x is equal to Select one: Ⓒa. 1 O b. 5 c. 7 O d. 3 or 2

I got X=4 or X=2

Hi Taking base as 3 Log3 2 +Log3 (2x-7/2) = 2log3 (2x-5) we get 2(2x-7/2) = (2x-5)^2 4x-7=4x^2+25-20x 4x^2-24x+32=0 x^2-6x+8=0 (x-4)(x-2)=0 but x cannot be 2 as x has to be greater than 7/2 so x=4