3. The point P lies on the curve with equation y=ln(x/3) The x-coordinate of P is 3. Find an equation of the normal to the curve at the point P in the form y = ax + b, where a and b are constants.

P- (3,0) y=ln(x/3)     u=x/3    y=ln(u) ​​​​​​            du = 1/3  dy = 1/u = 3            dx       du dy= du x dy dx dx  du   = 1/3 x 3 = 1 gradient at normal = -1 equation at normal = y = m(x) + c                  0 = -3 + c                  3 = c Answer: equation at normal = y = -x + 3