The correct option is B (3,10)
log5(x−3)+12log53<12log5(2x2−6x+7)
log function is defined when
(i) x>3
(ii) 2x2−6x+7>0
As D<0, so 2x2−6x+7>0 ∀ x∈R
Now, 2log5(x−3)+log53<log5(2x2−6x+7)
⇒log53(x−3)2<log5(2x2−6x+7)
⇒3(x−3)2<(2x2−6x+7)
⇒3x2−18x+27<2x2−6x+7
⇒x2−12x+20<0
⇒(x−2)(x−10)<0
⇒x∈(2,10) ...(iii)
From (i),(ii) and (iii), we get
x∈(3,10)