How close is the solution of $xe^x+\ln(x)=c$ to $W_0(c)$ for large $c$? Denote $f(c)$ to be the solution of the equation $xe^x+\ln(x)=c$

What is the asymptotic behaviour of $f(c)-W_0(c)$ for large $c$ ?

$W_0(c)$ denotes the lambert-w-function.
It seems to be in the order of $\frac{1}{c}$
 A: We have $\enspace\displaystyle W_0(x) \approx \ln x - \ln\ln x + \frac{\ln\ln x }{\ln x } \enspace$ for large $\,x\,$ .
It follows:
$\displaystyle f(c)-W_0(c) = x - W_0(xe^x + \ln x) $
$\hspace{2cm}\displaystyle \approx \,\, x -\ln (x e^x + \ln x) + \ln\ln (x e^x + \ln x) - \frac{\ln\ln (x e^x + \ln x) }{\ln (x e^x + \ln x) } $
$\hspace{2cm}\displaystyle \approx \,\, \ln(1+\frac{\ln x}{x}) \,\,\approx \,\, \frac{\ln x}{x} $
$\hspace{2cm}\to \,\, 0\enspace$ for $\enspace x\to\infty$ 
Note:
$\displaystyle \lim_{x\to\infty}(1 + x - W_0(xe^x + \ln x))^x = 1$
$\displaystyle \lim_{x\to\infty}(1 + x - W_0(xe^x + \ln x))^{x^2} = \infty$
A: Setting $x=w(1+h)$ with $w=W_0(c)$, $c=we^w$, $\ln(w)=\ln(c)-w$ we get the equation
$$
w(1+h)e^{w+wh}+\ln(w)+\ln(1+h)=we^w
$$
Inserting linear Taylor polynomials one gets
$$
w(1+h)e^w(1+wh)+\ln(w)+h+O(h^2)=we^w\\
h(1+we^w(1+w))+\ln(w)=0\\
h=-\frac{\ln(w)}{1+we^w(1+w)}=-\frac{\ln(w)}{1+c(1+w)}
$$
so that in the dominant terms for large $c$ and thus large $w$
$$
f(c)=w(1+h)+O(h^2)\approx w-\frac{\ln(w)}{c}=W_0(c)\left(1+\frac1c\right)-\frac{\ln(c)}{c}
$$
so that you also get a logarithmic term in the first correction.
