TITLE: Approximation of elements in arithmetic progressions by logarithms of integers
QUESTION [9 upvotes]: For fixed $a,b,c \in \mathbb{R}$ with $ac \neq 0$, it seems to me that one can find an increasing sequence of integers $\{\alpha_n\}$ such that the quantity $c \log \alpha_n$ becomes arbitrarily close to elements of the set
$$
A = \{ ak+b \,\colon k \in \mathbb{Z} \}.
$$
First, is this true?  If so, my question is: how good is the approximation?
For example, given any $\epsilon > 0$, is it possible to find infinitely-many integers $n$ such that, for some constant $C$,
$$
\textrm{dist}(c \log n,A) := \inf_{k \in \mathbb{Z}} |c \log n - ak-b| \leq C n^{-\epsilon}?
$$
How about
$$
\textrm{dist}(c \log n,A) \leq C e^{-n}?
$$
I am also interested in results which might say something like "There are at most finitely-many $n$ satisfying
$$
\textrm{dist}(c \log n,A) \leq C e^{-n^2}
$$
for any positive constant $C$" to illustrate the "best-possible" nature of a less restrictive bound.

Motivation
I am trying to determine the behavior of a quantity like
$$
\left|\left(1-e^{i c \log n}\right)g(n)\right|^{1/n},
$$
where $g(n)$ is well-behaved.  Until now I have simply been excluding all $n$ for which $1-e^{i c \log n}$ lies in some small fixed neighborhood of the origin, but doing this I lose infinitely-many $n$.
If, for example, it turns out that, for some positive constant $C$, there are only finitely-many $n$ satisfying
$$
\left| 1-e^{i(\theta + c \log n)} \right| \leq C n^{-1-\epsilon}
\tag{1}
$$
for any $\epsilon > 0$, then all but finitely-many $n$ satisfy
$$
C n^{-1} \leq \left| 1-e^{i(\theta + c \log n)} \right| \leq 2.
$$
In that case we could exclude at most finitely-many $n$ to obtain the desirable property
$$
\left| 1-e^{i(\theta + c \log n)} \right|^{1/n} \to 1
$$
as $n \to \infty$.
Now, if we let
$$
B = \{2\pi k - \theta \,\colon k \in \mathbb{Z}\},
$$
then equation $(1)$ is equivalent to the existence of a positive constant $C_1$ such that only finitely-many $n$ satisfy
$$
\textrm{dist}(c \log n,B) \leq C_1 n^{-1-\epsilon}
$$
for any $\epsilon > 0$.

REPLY [2 votes]: PART I
We can solve this problem by constructing such a sequence. First we simplify our terms. So, we divide both the terms by the quantity $c$ and let the new terms be $\log\alpha_{n}$ and $a'k+b'$. 
So, now the problem is to find a sequence $\{\alpha_{n}\}$ such that
\begin{equation}
\log(\alpha_{n+1}) - (a'k_{n+1}+b') < \log(\alpha_{n}) - (a'k_{n}+b')
\end{equation}
I say the following is such a sequence
\begin{equation}
\alpha_{n} = \lceil {e^{a'n+b'}} \rceil
\end{equation}
PROOF
\begin{equation}
\alpha_{n} - {e^{a'n+b'}}  \leq 1
\end{equation}
Hence, 
\begin{align}
&\log(\alpha_{n}) - (a'n+b')  \\
&= \log\left(\frac{\alpha_{n}}{e^{a'n+b'}}\right) \\
&\leq \log\left(\frac{e^{a'n+b'} + 1}{e^{a'n+b'}}\right) \\
&= \log\left(1 + \frac{1}{e^{a'n+b'}}\right) 
\end{align}
Now,
\begin{align}
\log\left(1 + \frac{1}{e^{a'n+b'}}\right) < \log\left(1 + \frac{1}{e^{a'(n+1)+b'}}\right) 
\end{align}
And hence, the proof for the first part.