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 $\frac{1}{c^3}$ ?
 A: 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
$$ x_{\infty} = \log(c)-\frac{\log\log c}{c}+\frac{\frac{\log\log c}{\log c}-\frac{1}{2}\left(\log\log c\right)^2}{c^2}+O\left(\frac{1}{c^3}\right).$$
A: 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.
A: Let's try fixpoint-iteration just for the heck of it.
\begin{align}x&=\ln(c)+\ln\left(1-\frac{\ln(x)}c\right)\approx\ln(c)\\&=\ln(c)+\ln\left(1-\frac1c\ln\left(\ln(c)+\ln\left(1-\frac{\ln(x)}c\right)\right)\right)\approx\ln(c)+\ln\left(1-\frac{\ln(\ln(c))}c\right)\\&=\dots\approx\ln(c)+\ln\left(1-\frac1c\ln\left(\ln(c)+\ln\left(1-\frac{\ln(\ln(c))}c\right)\right)\right)\end{align}
The second line is approximately your approximation. The third line expands into something along the lines of
$$x\approx\ln(c)+\frac{\ln(\ln(c))}c+\frac{\ln^2(c\ln(c))}{2c^2}+o\left(\frac{\ln^2(c\ln(c))}{c^2}\right)$$
(Way too lazy to expand that thing out more, it gets pretty messy and WA doesn't want to do it for me.)

Rewritten, the third line should be:
$$x\approx\ln(c-\ln(\ln(c-\ln(\ln(c)))))$$
And according to WA, this is
$$\small x\approx\ln(c)-\frac{\ln(\ln(c))}c-\frac{\ln^2(\ln(c))}{2c^2}-\frac{\ln^3(\ln(c))}{3c^3}+\frac{3\ln^2(\ln(c))}{2c^3\ln(c)}+\frac{\ln^2(\ln(c))}{c^3\ln^2(c)}+o\left(\frac{\ln^2(\ln(c))}{c^3\ln^2(c)}\right)$$
A: hint Bootstrap step by step the development of $x$ in the equation
$$x = \ln(c) + \ln\left(1 -\frac{\ln(x)}{c}\right)$$
A: The equation $\;\log(x)+e^x=c\;$ is one example, of many, which have iterative solutions. Other answers have given some numerical iterative solutions. For one example among others,
let $\;x_0:=\log(c),\; x_{n+1}:=x_0+\log(1-\log(x_n)/c).\;$ This converges numerically, and  it leads to a series solution in inverse powers of $c.\;$ This is similar to the power series for the Lambert $W$ function. For brevity, set $\;y:=\log(c),\; z:=\log(y).\;$ Let $\;x_0:=y+O(1/c),\;$ and use the iteration previously defined as follows: $\;x_{n+1} := y + \log(1 - (z + \log(x_n/y))/c).\;$
Then we get the sequence values
$\;x_1=y - z/c + O(1/c^2),\; x_2=y - z/c + (2z-z^2y)/(2yc^2) + O(1/c^3),\;$ and $\;x_3 = y - z/c + (2z-z^2y)/(2yc^2) + (-6z +3z^2 +9z^2y -2z^3y^2)/(6y^2c^3) +O(1/c^4).\;$
