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Say $n$ has $N$ digits and $n^n$ has $N'$ digits, then $$10^{N-1}\leq n \lt 10^N$$ $$\implies n(N-1)\leq n\log n \lt nN$$ $$\implies N' \in [n(N-1),nN)$$

Is it possible to round this down further in terms of elementary functions as an expression relating $N$ and $N'$(or $n$ and $n^n$)?

jamie
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1 Answers1

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We know that the number of digits for a positive integer $x$ is simply $\lfloor \log_{10} x \rfloor +1$. We then have: $$\text{Number of digits in $n^n$} = \lfloor\log_{10} n^n \rfloor+1 = \lfloor n\log_{10} n \rfloor+1$$ However, if you are trying to relate the number of digits in $n$ with the above, your interval is the best you can get, since you have no other information about $n$.

Haran
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  • Thank you, I have edited the question to be more specific. – jamie May 29 '20 at 10:54
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    @jamie As I said above, you cannot get a tight bound for $N'$ in terms of $N$. My answer above gives the precise number of digits that $n^n$ has in terms of $n$. – Haran May 29 '20 at 10:58