If log5⁡2,log5⁡2x−5 and log5⁡2x−7/2 are in A.P., then x is equal to

As log5⁡2,log5⁡2x−5,  log5⁡2x−7/2 are in A.P., we get     2log5⁡2x−5=log5⁡2+log5⁡2x−7/2⇒   2x−52=22x−7/2⇒    2x2−122x+32=0 ⇒ 2x−42x−8=0⇒     2x=22,23                     ⇒ x=2,3.Clearly, x≠2.        Therefore x=3