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Let $c$ be a natural number such that $c \neq 10^m, m \in \mathbb Z_{\ge0}$. Prove that we can choose some $k \in \mathbb{N}$ such that the $10$th base representation of $ \log_{10} (c^k) \ \text{starts with 7}$.

My idea was dividing value of $\log_{10}(k)$ into multiple intervals, and choosing specific $k$, we can conclude. ( For example, $9 \cdot 10^m \le \log_{10}(c) \le 10^m$,we can let $k=8$, $ |8 \cdot10^m\le \log_{10}(c)<9 \cdot10^m,$ we can let $k=88$ etc.). However, this seems pretty long and inefficient, is there any other way to proof it? Can someone give another idea to aproach? And I think it is possible for other digits than $7$, ( except 0 ofc), and my idea is still the same.

insipidintegrator
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Ioraboi
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    See https://math.stackexchange.com/questions/2230226/show-that-there-are-infinitely-many-powers-of-two-starting-with-the-digit-7. It's all based on the irrationality of $\log_{10}(c)$ for your question. – KCd Oct 23 '22 at 13:05
  • ^ and the equidistribution of $\log_{10}(c^k)=k\log_{10}(c)$, since the set ${k\alpha \mod 1}$ is equidistributed for irrational $\alpha$ – Gareth Ma Oct 23 '22 at 14:09
  • why equidistribution implies such $k$ exists? – Ioraboi Oct 23 '22 at 14:36

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