# Do any ordered pairs of algebraic numbers satisfy this function?

Q: Does there exist an algebraic number $$x,$$ s.t. $$f(x)$$ is also an algebraic number?

$$f(x)=\exp\bigg(\frac{1}{\ln(x)}\bigg)$$ for $$x\ne0,1.$$

I would like to prove that the set of points $$(x,f(x))$$ is the empty set, for algebraic $$x.$$

Are there any promising approaches to solve this?

• If $x\ne1$ then there is a positive answer $x=0 \implies f(0)=1$ – Ultradark Nov 8 at 15:50
• When did the logarithm become defined on $\{0\}$? In fact, this $f$ is maximally defined on $(0,\infty)$ if we are thinking of it as a real function. If we are thinking of it as a complex function (valid since the setting is algebraics), then there is freedom to choose a branch cut, but not the branch point: $0$. – Eric Towers Nov 8 at 15:52
• wolfram alpha says that $f(0)=1$ – Ultradark Nov 8 at 15:53
• Then apparently, Wolfram Alpha has misled you by silently taking $\lim_{x \rightarrow 0^+} f(x)$ instead of correctly observing $\ln 0$ is undefined, so $f$ is undefined at $x = 0$. – Eric Towers Nov 8 at 15:56
• Always be wary of trusting a computer. Programmers make errors too. – Paul Sinclair Nov 9 at 3:58