I believe I came up with an answer, however, I'm not too sure if it is valid.
Considering that the derivative of the function is $5-\mathrm{e}^x$, and the function must be equal to zero in order for the tangent line to be horizontal; $\mathrm{e}^x = 5$, so $x$ at $\ln 5$ is a horizontal tangent line.
Would I be correct in my assumptions? Or have I got this all wrong?
The slope of the graph of $y=f(x)$ is given by $f'(x)$. So the slope of the graph of $y=5x-e^x$ is $$f'(x)=5-e^x$$ Where the slope is zero, $f'(x)=0$. So $$5-e^x=0$$ $$e^x=5$$ $$x=\ln 5\approx1.6094$$ So at $x=\ln5\approx1.6094$, the slope of $y=5x-e^x$ is zero.
Here's a graph from Wolfram Alpha: