Take three real numbers $x$, $y$, and $\alpha$, of which $\alpha$ is a given irrational number. Can
$$y = \alpha x$$
be solved if the fractional parts of $x$ and $y$ are known (given) but their integer parts are unknown?
Note: this is a restatement of the third question in The power of irrationality: sin(x)+sin(πx)
Edit: If a more general solution is not available, I'm most interested in logarithms of natural numbers, that is, irrational $\alpha$'s of the form ${\rm log}\,2, {\rm log}\,3, \dots$, and their differences.
Edit 2: A similar question, with yet a different perspective, has been asked here: Measure of recurring trajectories on a billiard.
Suppose $\lfloor x\rfloor$ and $\lfloor y\rfloor$ are the integer parts while $\{x\}= x- \lfloor x\rfloor$ and $\{y\}= y- \lfloor y\rfloor$ are the fractional parts.
So you know $\beta=\alpha\{x\} - \{y\}$ and this is equal to $\lfloor y\rfloor-\alpha\lfloor x\rfloor$.
If there were two solutions then $\lfloor y_1\rfloor-\alpha\lfloor x_1\rfloor = \lfloor y_2\rfloor-\alpha\lfloor x_2\rfloor$ making $\alpha= -\frac{\lfloor y_1\rfloor-\lfloor y_2\rfloor}{\lfloor x_1\rfloor-\lfloor x_2\rfloor}$ which would be rational
So there is at most one solution.
If there is one then you can find it by going through the countable integers one at a time to solve $\lfloor y\rfloor=\beta+\alpha\lfloor x\rfloor$. This may take a long time, and if there is no solution (consider the cardinalities) then the process will not halt.