I do the calculus exercise. You wonder about the intersection, if any of $e^x$ with $u x + v$. We examine the difference $e^x- ux -v$. It is a strictly convex function.
- if $u<0$, the function is strictly increasing and vanishes at a unique abscissa which will be the one of the your intersection point. Newton method whatever the starting point will converge to the root. Better to start on its right, I have not detailed more.
- if $u=0$ we have horizontal lines, left to reader.
- if $u>0$, this time the difference is still convex but with a minimum. The value is $u - u \log u - v$. If this quantity is negative you have two intersection points, if it vanishes you have a tangent, if it is positive you have no intersection points. To find the intersection points do Newton method with either large positive or large negative starting point.
Hope it helps.
Edit: the starting point is crucial so the above is ill-advised to say "large starting point". Because as long as $e^x$ is large Newton's method will simply do roughly $x\mapsto x-1$ and take a long time to reach the root if you started too far (but I assume your real life examples have some reasonable bounds ,say $-10<x<10$). If $x$ is very negative (and slope $u$ positive), then situation is better, because the studied function whose root is aimed at is quasi linear, so after one iteration we are near $-v/u$ it seems and it should go fast (untested). So it seems to be easier to find the left-intersection point. In the case with negative slope, the sole intersection point can presumably be obtained starting Newton with the $-v/u$ mentioned above. Well, may be not the place for a mathematical treatise, let's leave some work to the AIs.