The point P lies on the curve with equation y = 4e^(2x+1), and the y-coordinate of P is 8. Find, in terms of ln2, the x-coordinate of P. Find the equation of the tangent to the curve at the point P in the form y = ax + b.

Finding the x-coordinate of P: y = 4e^(2x+1) 8 = 4e^(2x+1) 2 = e^(2x+1) ln2 = 2x+1 (ln2 - 1)/2 = xFinding the tangent equation:differentiate equation of curve y' = 8e^(2x+1)substitute in x-coordinate y' = 8e^(ln2) y' = 16substitute into equation of a straight line y-8 = 16(x-(ln2+1)/2) y-8 = 16x - 8ln2 + 8 y = 16x + 16 - 8ln2