I have the following equations:
Eq. 1) A standard equation for a line
$$y=mx + b,$$
Eq 2) A a parametric equation
$$x=e^{at}-1,$$
$$y=e^{ct}-1,$$
where $a$ and $c$ are known constants.
I want to find the intercepts of Eq .1 with Eq. 2.In the context I am working in I know there are usually two intercepts. I tried substituting $x$ and $y$ to find the intercepts $t$ values, which I could use to get $x$ and $y$ (actually I really just want t anyhow). I get
$$e^{ct}-1=m(e^{at}-1)+b,$$
yet I can't see a way to solve this for t without some kind of numerical method. How do I solve this equation?
Let $z:=e^{at}$ and $\alpha:=\dfrac ca$. The equation is
$$z^\alpha-1=m(z-1)+b.$$
This shows that for most rational $\alpha$, there is no analytical solution.
The slope along the curve is given by
$$\frac{dy}{dx}=\frac cae^{(c-a)t}.$$
If you equate this to $m$, you can draw $t$, which gives the point of tangency of a parallel to the line. Then depending on the side of the line wrt the tangent and wrt the initial point of the curve $(-1,-1)$, you can determine the number of intersections ($0, 1$ or $2$).
In case there are intersections, the tangency point separates them and you can use this information to start numerical iterations.
For convenience, you can translate $x,y$ by $1$ and the equations simplify to
$$x=e^{at},\\y=e^{ct}=x^\alpha$$ and $$y-1=m(x-1)+b\iff y=m'x+b'.$$