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I need to find the asymptote of $r \log_e(\theta)=a$

I understand what is going on. As $\theta$ goes from $0^+$ to $1$, $r$ goes from $0^-$ to $-\infty$. I tried to write the equation as a function of $\theta$ i.e $\theta = e^\frac{a}{r}$ Then I took the two limits of $r$ going to $+\infty$ and $-\infty$

Is this the correct approach to find the asymptotes in polar form? Is there a general approach?

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