Is there an efficient way of testing if the resulting value of an exponential gives an integer without actually expanding the equation.
For example:
$ {\log(12) - \log(4)}=1.09861\ldots $
and is a transcendental number.
But if we take the exponential, we get an integer:
$ e^{\log(12) - \log(4)}=e^{1.09861\ldots}=3 $
This can be easily shown by proving that the equation is equivalent to $12/3$.
However is there a way to prove this while staying in the logarithm nation... We are assuming that the two numbers $12$ and $4$ are very large...
Can I find say numerical stability bounds to this simple equation:
$ 2.71828^{2.48491-1.38629}=3\ldots $
such that if I use $n$ digits, I will get an integer close the the $n$-th digit.
This may look trivial, but let's assume we do not have say $\log(12)$, but only $2.48491 \ldots$ up to some precision of k-digits. Same thing for $\log(4)$, etc.
If you write your log numbers as $x$ and $y$. Then you can find the upper and lower bound (depending on sign) with $\exp(x\pm\epsilon_x-y\mp\epsilon_y)$ where $\epsilon$ are the errors (something like $0.5/10^{k+1}$). If an integer is not included in the range then you can be certain that it is not an integer.