The point P lies on the curve with equation y = 4e^{2x-1} - Edexcel - A-Level Maths Pure - Question 2 - 2008 - Paper 5

To find the equation of the tangent line, we need the derivative of the curve. First, we differentiate the curve equation:

$y = 4e^{2x-1}$

Using the chain rule:

$\frac{dy}{dx} = 4 imes e^{2x-1} imes 2 = 8e^{2x-1}$

Next, we evaluate this derivative at the point where $x = \frac{1}{2}( ext{ln}(2) + 1)$:

Since we know that:

$e^{2x-1} = 2$

Then:

$\frac{dy}{dx} = 8 \times 2 = 16$

Thus, the slope of the tangent at point P is 16.

Now we can use the point-slope formula for the tangent line, which passes through point P(\frac{1}{2}( ext{ln}(2) + 1), 8):

$y - 8 = 16(x - \frac{1}{2}( ext{ln}(2) + 1))$

Expanding this equation gives:

$y = 16x - 8 + 8 + 8 ext{ln}(2) = 16x - 8 + 8 ext{ln}(2)$

Thus, we can write the tangent line in the form $y = ax + b$:

$a = 16$ and $b = -8 + 8 ext{ln}(2)$.