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)=-1/c so z exp(z)=-1/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.
Numerical example
Please convert this into your favorite language. This handles only the equation exp(t) = ct with c>exp(1).
\documentclass{article}
\usepackage{xintexpr}
\newcommand\SolveExpEqualLin[1]{%
\begingroup\long\def\STOP##1\endgroup{}%
\xintdeffloatvar c:= #1;%
\xintifboolexpr{c>exp(1)}
{}
{Bad input $\xintfloateval{c}$ not greater than $\exp(1)$, Aborting!\STOP}%
\xintdeffloatfunc f(t):=(1-t)/(c*exp(-t) - 1);%
\xintdeffloatvar tn:=1/(c-1);%
\xintdeffloatvar eps:=tn * 5e-16;%
\xintloop
\xintdeffloatvar new:=f(tn);%
\xintifboolexpr{abs(new-tn)<eps}
{\iffalse}%
% maybe not update eps?
{\xintdeffloatvar tn, eps :=new, new * 5e-16;\iftrue}
\repeat
\xintdeffloatvar w_1 := tn;%
%
\xintdeffloatfunc g(t):=log(c) + log(t);%
\xintdeffloatvar tn:=log(c);%
\xintdeffloatvar tn:=g(tn);%
\xintdeffloatvar tn:=g(tn);%
\xintdeffloatvar tn:=g(tn);%
\xintdeffloatvar eps:=tn* 5e-16;%
\xintdeffloatfunc k(t):=(g(t)-1)/(1 - 1/t);%
\xintloop
\xintdeffloatvar new:=k(tn);%
\xintifboolexpr{abs(new-tn)<eps}
{\iffalse}%
% maybe not update eps?
{\xintdeffloatvar tn, eps :=new, new * 5e-16;\iftrue}
\repeat
\xintdeffloatvar w_2 := tn;%
%
The solutions to the equation $\exp(t) = \xintfloateval{c}t$ are
$t_1\approx\xintfloateval[-1]{w_1}$ and $t_2\approx\xintfloateval[-1]{w_2}$.
\endgroup
\par
}
\begin{document}
\xintFor* #1 in {\xintSeq{1}{10}}\do{\SolveExpEqualLin{#1}}
\end{document}
With TikZ (general case)
The code is extended to handle all possibilities, I also declared once and for all some functions of two variables for efficiency, and was more careful in termination criterion for Newton's method. I am not nimble with TikZ and can not do easily some obvious improvements the displayed plots are in need of. Incidentally I found a bug of \xintifsgnexpr and had to renounce using it.
\documentclass[tikz,border=1cm]{standalone}
\usepackage{xintexpr}
\xintdeffloatfunc F(t,c) := (1-t)/(c*exp(-t) - 1);
\xintdeffloatfunc G(t,d) := d + log(t);
\xintdeffloatfunc K(t,d) := (G(t,d)-1)/(1 - 1/t);
\newcommand\SolveExpEqualLin[2]{%
\begingroup
\xintdeffloatvar u:= #1;%
\xintdeffloatvar v:= #2;%
% There appears to be a bug in XINT with \xintifsgnexpr
% when used with input starting with a zero. (I was
% skeptical but it seems to be real, I reported to maintainer
% and I am awaiting response).
\xintifboolexpr{u>0}
{\SolveExpEqualLinPos}%
{\xintifboolexpr{u<0}
{\SolveExpEqualLinNeg}
{\SolveExpEqualLinZero}}
\endgroup
}
% negative slope. Always exactly one root.
% Attention to computation of epsilon. If the root is exactly zero
% we are doomed if we let it evolve to be dynamically proportional
% to the computed approximation.
% So we compute eps only at first ieration. The value is then 1/(c-1)
% which is not zero.
\newcommand\SolveExpEqualLinNeg{%
\xintdeffloatvar delta:= v/u;%
\xintdeffloatvar c:= u * exp(delta);% c<0
\xintdeffloatvar tn:= F(0, c);%
% it turned out that it was too small here with 5e-16
% because it could be that iteration fell into
% infinite loop ...6-->...8-->...6-->..6 etc as last
% significant figure. So let's be more cautious.
\xintdeffloatvar eps:=abs(tn) * 1e-15;% attention to sign...
\xintloop
\xintdeffloatvar new:=F(tn,c);%
\xintifboolexpr{abs(new-tn)<eps}
{\iffalse}%
{\xintdeffloatvar tn := new;\iftrue}
\repeat
\xdef\roots{\xintfloateval{tn-delta}}%
}
% zero slope
\newcommand\SolveExpEqualLinZero{%
\xintifboolexpr{v<=0}{\xdef\roots{}}{\xdef\roots{\xintfloateval{log(v)}}}%
}
% positive slope
\newcommand\SolveExpEqualLinPos{%
\xintdeffloatvar delta:= v/u;%
\xintdeffloatvar c:= u * exp(delta);%
\xintdeffloatvar d:= log(u) + delta;%
\xintifboolexpr{d-1>1e-15}
{\SolveExpEqualLinPosTwo}
{\xintifboolexpr{d-1<-1e-15}
{\xdef\roots{}}%
{\xdef\roots{\xintfloateval{1 - delta}}}%
}%
}
% positive slope, two roots
\newcommand\SolveExpEqualLinPosTwo{%
\xintdeffloatvar tn:=1/(c-1);%
\xintdeffloatvar eps:=tn * 1e-15;%
\xintloop
\xintdeffloatvar new:=F(tn,c);%
\xintifboolexpr{abs(new-tn)<eps}
{\iffalse}%
{\xintdeffloatvar tn:=new;\iftrue}
\repeat
\xintdeffloatvar w_1 := tn;%
%
\xintdeffloatvar tn:=d;%
\xintdeffloatvar tn:=G(tn, d);%
\xintdeffloatvar tn:=G(tn, d);%
\xintdeffloatvar tn:=G(tn, d);%
\xintdeffloatvar eps:=tn* 1e-15;% careful not too small
\xintloop
\xintdeffloatvar new:=K(tn, d);%
\xintifboolexpr{abs(new-tn)<eps}
{\iffalse}%
{\xintdeffloatvar tn :=new;\iftrue}%
\repeat
\xintdeffloatvar w_2 := tn;%
%
%The solutions to the equation $\exp(t) = \xintfloateval{c}t$ are
%$t_1\approx\xintfloateval[-1]{w_1}$ and $t_2\approx\xintfloateval[-1]{w_2}$.
\xdef\roots{\xintfloateval{w_1-delta}, \xintfloateval{w_2-delta}}%
}
\begin{document}
\begin{tikzpicture}
\def\xmin{-0.5}
\def\xmax{3.6}
\draw[->] (\xmin,0) -- (\xmax,0);
\draw[->] (0,-0.2) -- (0,{exp(\xmax)});
\draw[domain=\xmin:\xmax,samples=200] plot (\x,{exp(\x)});
\xintFor* #1 in {\xintSeq{3}{10}}\do{%
\draw[domain=-0.1:\xmax,samples=2,color=gray,line width=0.1] plot (\x,{#1*\x});
\SolveExpEqualLin{#1}{0}%
\foreach \r in \roots { \fill[red] (\r,{exp(\r)}) circle (1.2pt); }
}
\end{tikzpicture}
\begin{tikzpicture}
\def\xmin{-3}
\def\xmax{1.5}
\def\step{0.25}
\draw[->] (\xmin,0) -- (\xmax,0);
\draw[->] (0,-0.2) -- (0,{exp(\xmax)});
\draw[domain=\xmin:\xmax,samples=200] plot (\x,{exp(\x)});
\edef\tmp{\xinteval{\xmin..[\step]..\xmax}}
\xintFor #1 in {\tmp}\do{%
\draw[domain=\xmin:#1,samples=2,color=blue!50,line width=0.3] plot (\x,{-\x+#1});
\SolveExpEqualLin{-1}{#1}%
\foreach \r in \roots { \fill[blue] (\r,{exp(\r)}) circle (1.2pt); }
}
\end{tikzpicture}
\begin{tikzpicture}
\def\xmin{-3}
\def\xmax{1.5}
\def\step{0.25}
\draw[->] (\xmin,0) -- (\xmax,0);
\draw[->] (0,-0.2) -- (0,{exp(\xmax)});
\draw[domain=\xmin:\xmax,samples=200] plot (\x,{exp(\x)});
\edef\tmp{\xinteval{\xmin..[\step]..\xmax}}
\xintFor #1 in {\tmp}\do{%
\draw[domain=#1:\xmax,samples=2,color=red!50,line width=0.3] plot (\x,{\x-#1});
\SolveExpEqualLin{+1}{-#1}%
\foreach \r in \roots { \fill[red] (\r,{exp(\r)}) circle (1.2pt); }
}
\end{tikzpicture}
\end{document}
