If x = 9 is one of the solutions of ` log_(e)(x^(2)+15a^(2))-log_(e)(a-2)=log_(e)((8ax)/(a-2))`,thenA. ` a= 3/5`B. a = 3C. x = 15D. x = 2

If x = 9 is one of the solutions of ` log_(e)(x^(2)+15a^(2))-log_(e)(a

1.	If x = 9 is one of the solutions of ` log_(e)(x^(2)+15a^(2))-log_(e)(a-2)=log_(e)((8ax)/(a-2))`,thenA. ` a= 3/5`B. a = 3C. x = 15D. x = 2
Answer» Correct Answer - B `log_(e)(x^(2)+15a^(2))-log_(e)(a-2)=log_(e)((8ax)/(a-2))` Here` a gt 2, (8ax)/(a-2) gt 0` `:. x gt 0` Now` (x^(2)+15a^(2))/(a-2) = (8ax)/(a-2)` ` or x^(2) = 8ax+15a^(2) = 0` ` rArr x = 3a, 5a` Now, ` x = 9` ` rArr a = 3, 9/5` But ` a gt 2` ` :. a = 3` For ` a = 3, x = 9, 15`