I just read the question Why does $\ln(x) = \epsilon x$ have 2 solutions?, and thought I'd point out a related area of investigation. The equation $\log x=\epsilon x$ has 2 solutions for $\epsilon>0$, and the first one admits a simple asymptotic form, $x=1+\epsilon+3\epsilon^2/2+\cdots$ . The interesting one is the solution near infinity, whose asymptotic form, to leading order, is
$$x=\frac{1}{\epsilon}\left(\frac1\alpha+\beta(1-\alpha+\alpha^2)+\frac12\alpha^2\beta^2\right)+O(\epsilon),$$
where $\alpha^{-1}=\log\epsilon-i\pi$ and $\beta=\log\alpha$. I've simplified this equation as much as possible, whose origin is Mathematica, but how in the world would one go about proving this? Note that the exact solution is $x=-\frac1\epsilon W_{-1}(-\epsilon)$, so this question is related to the asymptotics of the Lambert W function.
Despite your claim that the negative sign makes things easier, I'll write the equation as $$\log(x) = \delta x$$ with $\delta > 0$ and small. You're interested in $x \to +\infty$, so write $x = t/\delta$. The equation becomes $\log(t) - \log(\delta) = t$, or $$\log(t) + \eta = t$$ where $\eta = -\log(\delta) \to +\infty$. Since $\log(t)$ is small compared to $t$ when $t$ is large, the zero-order approximation would be $t = \eta$. If $t = \eta + t_1$ with $t_1 = o(\eta)$, the equation becomes $$ t_1 = \log(\eta + t_1) = \log(\eta (1 + t_1/\eta)) = \log(\eta) + o(1)$$ So the next approximation is $t_1 \approx \log(\eta)$. If $t_1 = \log(\eta) + t_2$, the equation becomes $$\eqalign{ t_2 &= \log(\eta + \log(\eta) + t_2) - \log(\eta) = \log\left(1 + \frac{\log(\eta)}{\eta} + \frac{t_2}{\eta}\right)\cr &= \frac{\log(\eta)}{\eta} + \frac{t_2}{\eta} + \frac{(\log(\eta)+t_2)^2}{2\eta^2} + \ldots}$$ so the next approximation is $t_2 = \log(\eta)/\eta + t_3/\eta$ with $t_3 = o(1)$. I'll stop at this point, but you could continue. Putting it together, $$ x = \frac{-\ln(\delta)}{\delta} + \frac{\ln(-\ln(\delta))}{\delta} - \frac{\ln(-\ln(\delta))}{\delta \ln(\delta)} + \ldots $$