If the equation xloga⁡x2=xk−2ak,a≠0 has exactly one solution for r, then the value of k is/are

loga⁡x2loga⁡x=(k−2)loga⁡x−k  (taking log on base a)let loga⁡x=t we getPutting D = 0 (has only one solution), we have    (k−2)2−8k=0 or     k2−12k+4=0 or      k=12±1282 or      k=6±42