find the values of b for which the equation 2log_1/25(bx+28) = -log₅(12-4x-x²) has only one solution.

2log_1/25(bx+28) = -log₅(12-4x-x²), log_(1/25)(bx+28)²=log₅(12-4x-x²)^-1

(12-4x-x²)^-1=bx+28, (bx+28)(12-4x-x²)=1

= 2 log_1/25 (bx+28) = -log₅(12-4x-x²)

= -2 log₂₅ (bx+28) = -log₅(12-4x-x²)

= - log₅(bx+28) = -log₅(12-4x-x²)

= log₅(bx+28) = log₅(12-4x-x²)

= bx+28 = 12-4x-x²

= x²+ (4+b)x +16

If the quad has only one sol. then D=0

(4+b)²-4(16) =  b² + 8b - 48=0

(b+12)(b-4)=0

b=12,4

method as above ans is -12, 4 for x belongs (-6,2)