Equation e tangent line to the graph of the function f(x)=2^(x+1)-log(base2) (x-1) at the point (3,15)?

f(x) = 2ˣ⁺¹ - log₂(x-1)

f'(x) = 2ˣ⁺¹ ln(2) - ln(2)/(x-1)

Slope of the tangent at the point (3,15)
= f'(3)
= 2³⁺¹ ln(2) - ln(2)/(3-1)
= (31/2) ln(2)

The equation of the tangent :
y - 15 = (31/2) ln(2) (x - 3)
2y - 30 = [31 ln(2)](x - 3)
2y - 30 = [31 ln(2)]x - 93 ln(2)
[31 ln(2)]x - 2y + 30 - 93 ln(2) = 0