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Inspired by the question on which points of the $\mathbb{C}$ unit circle can be reached by arbitrary products of two example points, I came up wtih the following:

For given $a, b \in \mathbb{R}^+ \backslash \{1\}, a \neq b$, a target value $c \in \mathbb{R}_0^+$ and $\epsilon > 0$, find $k,l \in \mathbb{Z}$ such that $|c - a^k \cdot b^l| < \epsilon$.

This is obviously prone to failure with inappropriate "generator" choices like $a = b^m, m \in \mathbb{Z}$. Now I am wondering under which conditions such an approximation is possible? And how would you compute $k,l$? Searching integer points $(k,l)$ closest to the line $\tilde{k} \cdot \log(a) + \tilde{l} \cdot \log(b) = \log(c)$ would be my first strategy.

An upgrade to the question would be:

Find a sequence $\mathbb{N} \rightarrow \mathbb{Z}^2, n \mapsto (k_n, l_n)$ such that $\lim_{n \rightarrow \infty} a^{k_n} \cdot b^{l_n} = c$.

GDumphart
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1 Answers1

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You can do this if $\log_a(b)$ is irrational.

We want $k_n + l_n\log_a(b)$ to tend to $\log_a(c)$, which is the same as finding $(l_n)$ such that $\lim_{n \rightarrow \infty} \{ l_n\log_a(b)\} = \{ \log_a(c)\}$ (where $\{\cdot\}$ denotes the fractional part).

And we can do this because if $\alpha$ is irrational, the set $\{\{ n\alpha\}:n\in\mathbb N\}$ is dense in $[0,1]$ (see Slade's answer here for a constructive proof).

(If $\log_a(b)$ is rational, you can only do it if $c$ is equal to $a^k\cdot b^l$ for some integers $k,l$.)

TonyK
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