Asymptotic behaviour of the solution of $\ \ln(x)+e^x=c\ $ for large $c\ $?

Hint. Use methods of perturbation theory. Let $x_0 = \ln c$ and $x = x_0 + \delta$. Then $$ c = e^x + \ln x = e^{x} + \ln x_0 + \ln \left(1 + \frac{\delta}{x_0} \right). $$

Last term is of order $O(\delta / x_0)$. So, $$x_1 = \ln (c - \ln x_0) = \ln (c - \ln \ln c) = \ln c - \frac{\ln \ln c}{c} + o(\ln \ln c / c).$$

In the same manner you can obtain $x_2$, $x_3$, and so on.

For large values of $x$ the function $\log(x)+e^x$ is convex, and Newton's method with starting point $x_0=\log(c)$ converges pretty fast to the solution of $\log(x)+e^x=c$. The iteration

$$ x_{n+1} = x_n-\frac{\log(x_n)+e^{x_n}-c}{\frac{1}{x_n}+e^{x_n}} $$ produces $$ x_1 = \log(c)-\frac{\log\log c}{c+\frac{1}{\log c}} $$ and $$ x_{\infty} = \log(c)-\frac{\log\log c}{c}+\frac{\frac{\log\log c}{\log c}-\frac{1}{2}\left(\log\log c\right)^2}{c^2}+O\left(\frac{1}{c^3}\right).$$