The curve \(y=f(x)\) passes through the point \(\left(\dfrac12,\dfrac72\right)\) and is such that

\(f'(x)=e^{2x-1}.\)

(i) Find the equation of the curve.

(ii) Find the value of \(x\) for which \(f''(x)=4\), giving your answer in the form \(a+b\ln\sqrt2\), where \(a\) and \(b\) are constants.