Answer
$\dfrac{1-\ln 2}{2}$
Work Step by Step
We have: $y=x-e^{2x-1}$
We differentiate both sides with respect to $x$.
$y^{\prime}=\dfrac{d}{dx} [x-e^{2x-1}] \\=1-2e^{2x-1}$
The tangent line to the graph will be horizontal when $ \dfrac{dy}{dx} =0$
Therefore, $1-2e^{2x-1}=0 \\ \ln (e^{2x-1})=\ln (\dfrac{1}{2}) \\ 2x-1=-\ln (2)\\ x=\dfrac{1-\ln 2}{2}$
