The equation $e^{x-8}+2x-17=0$ has

Clearly$x = 8$ satisfies the given equation. Assume that $f\left(x\right)=e^{x-8}+2x-17=0$ has a real root $\alpha$ 
other than $x = 8.$We may suppose that $a > 8$ (the case for $\alpha <8$ is exactly similar).
 Applying Rolle’s theorem on$[8,\alpha ],$ we $\beta \in (8,\alpha ),$ such that $f^{\mathrm{\prime }}\left(\beta \right)=0.$
But $f^{\mathrm{\prime }}\left(\beta \right)=e^{\beta -8}+2$, so that $e^{\beta -8}=-2$

2 which is not possible. Hence there is no real root other than 8.