Let $y=y(x)$ be the solution of the differential equation $d y=e^{\alpha x+y} d x ; \alpha \in N$. If $y\left(\log _e 2\right)=\log _e 2$ and $y(0)=\log _e\left(\frac{1}{2}\right)$, then the value of $\alpha$ is equal to $\_\_\_\_$ .
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