Is there a function $f$ so that $f(x-\exp(y))$ is linear in $y$

Question

I wonder if there is any continuous function $f$ so that $f(x-\exp(y))$ is linear in $y$. I exclude cases where $f$ is a constant or depends on $x$ or $y$. For instance, I do not allow $f(c)=\ln(x-c)$.

Answer 1

There is no such $f$. Indeed, fix $x$ and suppose $f(x - \exp(y)) = a(x) + b(x) y$ for all $y$. Then $f(w) = a(x) + b(x) \log (x - w)$ for all $w < x$. If this also holds for $x'$ then $f(w) = a(x') + b(x') \log(x' - w)$ for all $w < x'$. It follows that $(x - w)/(x' - w)$ is a constant for all $w < \min(x, x')$, which is nonsense.