I tried mapping power functions onto the polar plane (i.e. converting x,y into r and $\theta$). I was successful with power functions representing $y=ax^n$ by $$r=\sqrt[n-1]{\frac {\tan(\theta)}a}\sec(\theta)$$ Fairly reasonable right? So I decided to try the exponential function
$y=ab^{kx}$ $$y=ab^{kx}$$ $$r\sin(\theta)=ab^{kr\cos(\theta)}$$ $$r{\frac {\sin(\theta)}a}=b^{kr\cos(\theta)}$$
Let $k\cos(\theta)$ be $u$ and let ${\frac {\sin(\theta)}a}$ be $q$
$$rq=b^{ru}$$ $$\ln(r)+\ln(q)=ru\ln(b)$$ if we let $U$ represent $u\ln(b)$ we get $$\ln(r)+\ln(q)=rU$$
From $\ln r+\ln q=kr$ Isolating $r$ I get the solution $$r=−\frac {1}UW (−\frac {U}q)$$ Then from Approximation to the Lambert W function $$W(e^{a+x}){\approx}x(1-\frac{\ln x-a}{x+1})$$ with $a=0$ $$W(e^x){\approx}x(1-\frac{\ln x}{x+1})$$ Let $x$ be represented by $\ln e^x$ $$W(e^x){\approx}\ln e^x(1-\frac{\ln\ln e^x}{\ln e^x+1})$$ If we let $-U/q$ be $e^x$ we get $$W(-\frac {U}q){\approx}\ln(-\frac {U}q)(1-\frac{\ln\ln(-\frac {U}q)}{\ln(-\frac {U}q)+1})$$ Putting it all together, we get $$r{\approx}−\frac {1}U\ln(-\frac {U}q)(1-\frac{\ln\ln(-\frac {U}q)}{\ln(-\frac {U}q)+1})$$ Replace U and q $$r{\approx}−\frac {1}{ak\cos(\theta)\ln(b)}\ln(-\frac {{ak\cos(\theta)\ln(b)}}{\sin(\theta)})(1-\frac{\ln\ln(-\frac {{ak\cos(\theta)\ln(b)}}{\sin(\theta)})}{\ln(-\frac {{ak\cos(\theta)\ln(b)}}{\sin(\theta)})+1})$$
$$r{\approx}−\frac {1}{akcos(\theta)ln(b)}(ln{-akcot(\theta)ln(b)})(\frac {ln{-akcot(\theta)ln(b)}}{{-akcot(\theta)ln(b)}+1})$$
$$r{\approx}−\frac {\ln({-ak\cot(\theta)\ln(b)})}{ak\cos(\theta)\ln(b)}\frac {\ln({-ak\cot(\theta)\ln(b)})}{{1-ak\cot(\theta)\ln(b)}}$$ $$r{\approx}\frac {\ln({-ak\cot(\theta)\ln(b)})}{ak\cos(\theta)\ln(b)}\frac {\ln({-ak\cot(\theta)\ln(b)})}{{ak\cot(\theta)\ln(b)-1}}$$
$$r{\approx}\frac {\ln^2({-ak\cot(\theta)\ln(b)})} {ak\cos(\theta)\ln(b)(ak\cot(\theta)\ln(b)-1)}$$
$$r{\approx}\frac {\ln^2({-ak\cot(\theta)\ln(b)})} {a^2k\ln^2(b)\cot(\theta)\cos(\theta)-ak\cos(\theta)\ln(b))}$$
Technically I have an answer, but I really was not expecting something this complicated. Is there any easier way to do this? Or to simplify it? I just feel like this is far too complicated.
