If $3^x$ and $5^x$ are both integers, is $x$ an integer? 
Does the following statement hold? $$x\in \mathbb{R}^+ \text{and} \  3^x, 5^x \in \mathbb{Z} \implies x \in \mathbb{Z}$$
In words:
If $x>0$ is a real number, and $3^x$ and $5^x$ are both integers, does
that mean that $x$ is an integer?

This is a slightly modified form of another problem I was working on. A friend of mine claims this is a very hard problem. What do you think?
If one claims it is an open problem, can one show that this problem is equivalent to some other known open problem?
 A: This is probably an open question, as the related problem with $2^x$ and $3^x$ is open. Today, it is known that if $2^x$, $3^x$ and $5^x$ are integers, then $x$ is integer as well--it follows from the six exponentials theorem in transcendental number theory. 
I cannot confirm whether the $3^x$, $5^x$ case follows from the four exponentials conjecture, as I do not know the field; so I would be glad if someone could.
A: I would say yes.
If we assume that $x\not\in\mathbb{Z}$ we can write it as $n+\alpha$ where $n\in\mathbb{Z}$ and $\alpha\in (0,1)$ then $$3^x=3^{n+\alpha}=3^n\cdot 3^{\alpha}$$
We know that $3^n$ is an integer and if $3^x$ is an integer too then $3^\alpha$ must be one too.
Now, as $\alpha\in(0,1)$ we can write it as $\frac{1}{\beta}$ where $\beta>0$ so we get that
$$3^{\alpha}=\sqrt[\beta]{3}$$
which is definitely not an integer which is a contradiction so $x$ must be an integer.
Maybe I'm missing something in the original question but I don't see how the $5^x$ changes anything.
