Find the points $x$ at which $y-\ln(x+y)=0$ defines $y$ implicitly as a function of $x$

Question

Find the points $x$ at which the equation $$y-\ln(x+y)=0$$ defines $y$ implicitly as a function of $x$.

I'm not really sure what is meant by this question. If I define the function $g(x,y) = y - \ln(x+y)$ then it is implicit. Does it mean for me to solve for $x$ like \begin{align*} y-\ln(x+y)=0 &\iff y = \ln(x+y) \\ &\iff e^y = x+y \\ &\iff x = e^y - y \end{align*} So for the function $f^{-1}(y) = x$ described above, we have that $f^{-1} : \mathbb{R} \to [1,\infty)$, so the function $y = f(x)$, the points $x$ at which the equation $y-\ln(x+y)=0$ defines $y$ implicitly as a function of $x$ would be $[1,\infty)$.

I don't know if this right. This is a question out of Numerical Analysis by Cheney.

Answer 1

There is an unique this point: $x=1$

because for all $x>1$ there are two values of $y$ for which $y-\ln(x+y)=0$,

which is impossible if we want to get function.

By the way, for all $x<1$ we have no some value of $y$ for which $y=\ln(x+y)$.

All this follows from your first step: $x=e^y-y$.

Let $x(y)=e^y-y$.

Hence, $x'(y)=e^y-1$, which says that for $y=0$ the function $x(y)$ has a minimal value.

Also, easy to see that $x+y>0$ and we can draw a graph of the equation $x=e^y-y$.