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. ## Mathematical treatise I will focus on the case `u>0`. In order to analyze mathematically it is convenient to make a translation. The equation to solve is `e^x = ux+v`. Let `t= x+v/u`. On finds that the equation becomes `e^t = ct` with `c= u exp(v/u)`. This can be transformed if one so wishes into a Lambert-W function type of equation `-texp(-t)=-c` so `z exp(z)=-c` with `z=-t` I will not use that. Geometrically, as `c>0` we see clearly that there is a `c_0` such that `c<c_0` give no solution, `c=c_0` one, and `c>c_0` gives two. It turns out that `c_o = e`: the line with slope `e` touches the exponential at point `(1,e)`. Thus we can say in advance that for `c>e` we will have a solution `w_1` in `(0,1)` and a solution `w_2>1`. This second solution is the most delicate because it probes when `c` increases the exponential regime. Let's start with `w_1`. It is clear geometrically that it is larger than the abscissa of intersection of our line through the origin of slope `c` and the tangent to the exponential at `t=0` of slope `1`. So `w_1 > 1/(c-1)`. We are in the convex decreasing part of the difference `e^t - ct`, so put `t_0 = 1/(c-1)` and apply Newton iteration, this gives a strictly increasing sequence which should be efficient to find the limit `w_1`. The formula to iterate is `next(t) = (1-t)/(c exp(-t) - 1)`. More challenging is finding `w_2`. We would like a starting point to the right of it. I convert the equation `e^t = c t` into the equation `t = d + log (t)`with `d = log(c) >1`. If we plot the line of equation `y = t-d` we want its intersection the graph of log with abscissa `w_2>1`. The difference `t - d - log(t)` is convex and increasing for `t>1` with zero slope at `t=1` and negative value `1-d` at `t=1`. We look for starting point sufficiently to the right but not too much so that Newton will not start from too far. Now if `d` is large, the solution should be approximately `t=d + log(d)`. In fact we see that iterating `t<- d+log(t)` will converge geometrically, and stay always to the left of the seeked root `w_2`. So perhaps do this a couple of times, then apply Newton which will give us a point on the right of `w_2` hopefully not too far and continue with Newton. So we set `t_0=d`, iterate a bunch of times `t<- d +log(t)`, then switch to `next(t) = ((d-1)+log(t))/(1 - 1/t)`. Posting this for now, a numerical example later.