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

It it clear that $\ln(c)$ is a good approximation. 
Experimenting with large numbers , I found out that $$\ln(c)-\frac{\ln(\ln(c))}{c}$$ is an excellent approximation. The difference between this number and the solution seems to be of order $O(\frac{1}{c^2})$. 
A series expansion of the solution for $c\rightarrow\infty$ in terms of $c$ would be vey nice. How can I find the first few terms , lets say upto order $\frac{1}{c^3}$ ?

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
$$ x_{\infty} = \log(c)-\frac{\log\log c}{c}+\frac{\frac{\log\log c}{\log c}-\frac{1}{2}\left(\log\log c\right)^2}{c^2}+O\left(\frac{1}{c^3}\right).$$
