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I was wondering if there is an implicit form for a logarithmic spiral. For example, if

$$ x=e^{-t}\cos(t)\\y=e^{-t}\sin(t)$$

we can write $x^2+y^2=e^{-2t}$ and $y/x=\tan(t)$ which yields $$x^2+y^2=e^{-2\arctan(y/x)}$$ for $x\neq0$, but of course I would like to find one defined for any $(x,y)\in\mathbb{R}^2$. Any help?

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Rearranging the given equation gives the new equation $$y=x\tan{\bigg(\ln{\bigg(\frac{1}{\sqrt{x^2+y^2}}\bigg)}\bigg)}$$ which is defined for any $(x,y)\in\mathbb{R}^2$.

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    $\begingroup$ No. Moreover, $\ln(1/(\sqrt{x^2+y^2))}=(-1/2)\ln(x^2 + y^2)$. $\endgroup$ – MarianD Mar 6 at 22:50
  • $\begingroup$ @JeanMarie, I'm right with my "No." The given parametric curve is not symmetric with respect to point [0, 0], while this curve is. So the OP's curve is a proper subset of this one. $\endgroup$ – MarianD Mar 6 at 23:31
  • $\begingroup$ @JeanMarie, the error is in one-direction derivation, so from OP's equations folows this equation, but the opposite is not true. Similarly as $x = 2 \implies x^2 = 2^2$, but the second equation has 2 roots. $\endgroup$ – MarianD Mar 6 at 23:41
  • $\begingroup$ @MarianD You are right. The error is in using function $\tan$ for quadrants II and III (generating the symmetrical part) whereas the derivation that has been done is only valid for quadrants I and IV. $\endgroup$ – Jean Marie Mar 6 at 23:42
  • $\begingroup$ @JeanMarie, yes, you pinpoint it. $\endgroup$ – MarianD Mar 6 at 23:49
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The implicit equation you give

$$x^2+y^2=e^{-2\arctan(y/x)}\tag{1}$$

is indeed not defined for $x\neq0$ but has the worst drawback that it generates a double spiral (the same issue met by the "solution" given by Peter Foreman) as shown here :

enter image description here

Fig. 1 : Formulation (1) provides a double spiral instead of a single one.

Nevertheless, there is a solution, suggested by @Blue (who so often has given very interesting remarks) : if we accept to consider "arctan2" as a function "like others", it suffices to transform (1) into

$$x^2+y^2=\exp(-2 \ \text{arctan2}(y,x))\tag{2}$$

(please note the arguments' order for atan2 : $y$, then $x$), or the equivalent form :

$$2 \ \text{arctan2}(y,x)+\log(x^2+y^2)=0\tag{3}$$

Another view of the issue : the spiral can be viewed as the projection of the intersection line of two surfaces as shown on Fig. 2 :

enter image description here

Fig. 2 : The logarithmic spiral as the vertical projection of the intersection line of surfaces with equations $z=$ln$(x^2+y^2)$ (a pseudo-cone) and $z=-2 \arctan \! 2(y,x)$

Remark : "arctan2", where the "2" means "with 2 arguments", finds its origin in numerical analysis (see for example https://en.wikipedia.org/wiki/Atan2 which traces it back to Fortran IV, i.e., around 1960). It is nowadays present in all scientific software, for example atan2 in Matlab, ArcTan[Y,X] with Mathematica... See its particular behavior along the negative $x$-axis "cut" in https://uk.mathworks.com/help/matlab/ref/atan2.html.

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