If x = 9 is one of the solutions of loge⁡x2+15a2−loge⁡(a−2)=loge⁡8axa−2 then

loge⁡x2+15a2−loge⁡(a−2)=loge⁡8axa−2 Here a>2,8axa−2>0∴ x>0 Now x2+15a2a−2=8axa−2 or  x2−8ax+15a2=0⇒ x=3a,5anow,x=9⇒     a=3,95 But     a>2∴     a=3For a=3, x=9, 15