So I have two logarithmic spirals in parametric form $$ x(t) = ae^{bt}\cos t \\ y(t)=ae^{bt}\sin t $$ and $$ x'(t) = \alpha e^{\beta t}\cos t \\ y'(t)=\alpha e^{\beta t}\sin t $$ With $\beta$ and $b$ having opposite signs so the spirals grow in opposite directions. Setting the $x$'s and $y$'s equal and solving for $t$ I get $$ t = \frac{1}{b-\beta}\ln\frac{\alpha}{a} $$ Evaluating $(x(t),y(t))$ for that value does produce a single point of intersection for the two spirals, but I can't seem to find the general form that would give all the points of intersection.
I've fiddled around with it in polar form, both as $r(\theta)$ and as $\theta(r)$, and solved for the intersection points, but I couldn't figure out the general form in those cases either. (I tried integer coefficients, and multiples of $\pi$ and $2\pi$ to no avail.)
Eventually I realized I would prefer the parameterized approach because it'll be easiest to work in terms of $t$.
But I still feel lost in how to find the other points of intersection.
In this reply I will give a solution for the intersection of two logarithmic spirals.
Consider two spirals in the complex plane with different angular regions, e.g.,
$$ z=e^{(b+i)u}\\ w=e^{(\beta+i)v} $$
and we seek all the points where $z=w$. We can expand these and separate the real and imaginary parts to obtain
$$ e^{bu}\cos u=e^{\beta v}\cos v\quad \quad \quad (1)\\ e^{bu}\sin u=e^{\beta v}\sin v\quad \quad \quad (2) $$
If we multiply (1) by $\sin u$ and (2) by $\cos u$ and subtract, we obtain
$$0=e^{\beta v}[\sin u\cos v-\cos u\sin v]=e^{\beta v}\sin(u-v)\quad \quad \quad (3)$$
Similarly, if we multiply (1) by $\cos u$ and (2) by $\sin u$ and add, we obtain
$$e^{bu}=e^{\beta v}[\cos u\cos v-\sin u\sin v]=e^{\beta v}\cos(u-v)\quad \quad \quad (4)$$
Equation (3) tells us that $(u-v)=n\pi$. Equation (4) tells us that $n$ must be even since $\cos n\pi$ alternates in sign, but (4) must be positive. So we conclude that
$$v=u+2n\pi$$
(assuming that $v$ the larger of the two).
Finally, going back to the beginning, we must also have $|z|=|w|$, and therefore
$$e^{bu}=e^{\beta v}=e^{\beta(u+2n\pi)}$$
so that
$$ u=\frac{2n\pi\beta}{b-\beta}\\ v=u+2n\pi $$
The figure below shows a sample calculation (cleverly chosen to give simple results). Here
$$ b=\frac{2\ln\varphi}{\pi},\quad \text{the golden spiral}\\ \beta=\frac{\ln\varphi}{2\pi}\\ $$
$$ u=\frac{2n\pi}{3}\quad \{120,240,360,480,600,720^\circ...\}\\ v=\frac{8n\pi}{3}\quad \{480,960,1440,1920,2400,2880^\circ...\} $$