Is $2^k = 2013...$ for some $k$? [duplicate]

I'm wondering if some power of $2$ can be written in base $10$ as $2013$ followed by other digits.

Formally, does there exist $k,q,r \in \mathbb N$ such that $$2^k=2013 \cdot 10^q+r \,\,\,; \,\,\,r<10^q $$

I'm not sure if it's true or not. I would go for a 'no', but I can't prove it.

Thanks for your help.

Since $\dfrac{\log 2}{\log 10}$ is irrational, there are such $k$. We want

$$2.013 \cdot 10^m \leqslant 2^k < 2.014\cdot 10^m,$$

and taking the logarithm to the base $10$ that becomes

$$m + \frac{\log 2.013}{\log 10} \leqslant k\cdot\frac{\log 2}{\log 10} < m + \frac{\log 2.014}{\log 10}$$

or

$$\frac{\log 2.013}{\log 10} \leqslant \left\lbrace k\frac{\log 2}{\log 10}\right\rbrace < \frac{\log 2.014}{\log 10},$$

where $\{x\} = x - \lfloor x\rfloor$ is the fractional part. It is a well-known fact that for irrational $\alpha$ the sequence of $\{ k\alpha\}$ is dense in $[0,1]$.

Prelude> [n | n <- [1 .. 100000], take 4 (show (2^n)) == "2013"]
[1363,3499,5635,14664,16800,18936,30101,32237,34373,43402,45538
,47674,56703,58839,60975,63111,72140,74276,76412,85441,87577,89713]