$xy=\ln (x+y)$ implies that $y=\frac{\ln x}{x}+O(\frac{\ln x}{x^3}), x\to+\infty$.

Question

$xy=\ln (x+y)$ implies that $y=\frac{\ln x}{x}+O(\frac{\ln x}{x^3}), x\to+\infty$.

How to prove it? Even the easiest part $\lim_{x\to+\infty}y=0$ I have no idea.

Answer 1

Given that $\, x y = \ln(x+y)\,,$ then it is symmetric in $x$ and $y$. Thus, assume that $\,0<y<x.\,$ This implies that $\,y < \ln(2x)/x\,$ and also $\, y\to +0$ as $x\to+\infty.\,$ Now $$ x y = \ln(x(1+xy/x^2)) = \ln(x) + \ln(1+\ln(x+y)/x^2). \tag1 $$ But, $\,\ln(x+y)\to \ln(x)\,$ as $\,x\to+\infty\,$ since $\, y\to 0$ as $x\to+\infty\,$ and $\,\ln(1+z)\to z\,$ as $\,z\to 0.\,$ Thus, $\,y = \frac{xy}x = \frac{\ln(x)}x + O\Big(\frac{\ln(x)}{x^3}\Big).$

Answer 2

If $$x y=\log(x+y)$$ then $$y=-x-\frac 1x \,W_{-1}\left(-x\,e^{-x^2} \right)$$ where $W(.)$ is Lambert function.

Use the expansion given in the linked page $$W_{-1}(t)= L_1-L_2+\frac{L_2}{L_1}+\cdots$$ where $L_1=\log(-t)$ and $L_2=\log(-L_1)$.