# Is $z=e^{\frac{1}{\log(x)}}$ a solution to a polynomial?

Is $$z=e^{\frac{1}{\log(x)}}$$ a solution to a polynomial? $$x\in\Bbb Q~\cap(0,1).$$

Wolfram is telling me $$z$$ is algebraic but I'm not sure that I believe this.

I believe it would follow from Schanuel's conjecture,

$$\mathbb{Q}(\ln x,\mathrm{e}^{1/\ln x})$$

has transcendence degree $$2$$ whenever $$x$$ is algebraic.

I've tried very hard to solve this problem over the course of several months using the little abstract algebra I know. My first abstract algebra class starts on Monday so I hope to gain better tools to attack this problem.

My main point of attack has been to analyze various classes of algebraic equations that satisfy the condition $$f(0)=0$$ and $$f(1)=1.$$ This is because when you don't know how to think of a single object, sometimes it's better to study a class of similar objects.

Then I've reflected the entire class about $$x=1/2$$ to generate points in $$(0,1)^2,$$ and taken the collection of these points as the elements in the space. I'm specifically looking to see if I can generate points in this way, that satisfy closure w.r.t. the binary operation:$$(\times).$$ For example, if I have points that have algebraic numbers as coordinate pairs $$P_{\in(0,1)^2}(\Bbb A,\Bbb A)$$ then possibly by multiplying two points component wise, closure could be accomplished.

I don't know if this method alone will lead to a definitive answer as to whether $$z$$ is algebraic but it might give a helpful perspective on the road to a solution.

• @N.S. I stated that $x\in\Bbb Q~\cap(0,1).$ – geocalc33 Aug 22 '19 at 22:08
• A polynomial over $\mathbb Q$ you mean? – Zarrax Aug 22 '19 at 22:31
• I'm amazed by the close vote. I think it's pretty clear that the question is whether $z$ is algebraic. – Rob Arthan Aug 22 '19 at 22:32
• Some context – K B Dave Aug 22 '19 at 23:52
• Also FWIW, I think it's unlikely that we can do better than the claim $$\mathrm{e}^{u}\in\mathbb{A}\Rightarrow \{\mathrm{e}^{u^2},\mathrm{e}^{1/u^2},\mathrm{e}^{1/u}\}\not\subset\mathbb{A}\text{,}$$ as follows from the five exponentials theorem. – K B Dave Aug 23 '19 at 3:14