A peculiar limit? Consider $A^N = X$, where $A$ and $N$ are positive integers and $A > 1$. The ratio between $N$ and the number of digits of $X$ apparently have a limit(!). See these:


*

*$2^3=8$  (ratio $3 : 1$)

*$2^{13}=8192$  (ratio $3.25 : 1$)

*$2^{23}=8388608$  (ratio $3.2857 : 1$ )

*$2^{33}= \ldots$  and so on. 


I used $2$ as a base because no other bases such as $3,4,5,\ldots$ give minimum number of digits for $X$, in other words, $2$ is the only base we need to calculate this limit, And if my calculation is correct, the first $5$ digits of this limit is $3.3219$, Is my calculation correct ?

Let's assume we're working in our usual base ten number system. From Antonio's comment we have that the number of digits in the number $A^n$ is $\lfloor log_{10} A^n \rfloor + 1 = \lfloor nlog_{10} A \rfloor + 1$.
Now $\lfloor nlog_{10} A \rfloor + 1 = nlog_{10} A + \Delta$ for some $\Delta \in \{0, 1\}$. So the ratio is
$$ \frac{n}{nlog_{10} A + \Delta}$$
Clearly as $n$ goes off to infinity, the $\Delta$ becomes negligible and we're just left with:
$$ \frac{1}{log_{10} A} = log_{A} 10$$
as your limit. You can plug in $A = 2$ for your specific case and you should get the right answer.
