Consider natural powers of $2$ and their first $k$ digits.
For example find the smallest natural number ( positieve integer ) $n$ such that :
$$2^n = 3...$$
Is solved by $2^5 = 32 $ and thus $n=5$.
The title asks for
$$ 2^n = 9...$$
I think $n=53$ is the smallest solution. But the main point here is how to efficiently solve these questions.
As a sidenote I wonder why considering first digits instead of last ( mod ) is not seen more often in Number theory.
Anyways here is what i considered but im not sure this is an improvement over simple brute force search.
Let $<*>$ denote fractional part. ( example $<e> = 0,71828...$ )
$2^n = 9...$
$ 10^{ log(2) n} = 9 * 10^{s} $
$ 10^{log(2) n} = 10^{s + log(9)} $
$ log(2) n - log(9) = s $
$ < log(2) n - log(9) > < log(10/9) $
$ < log(2) n - log(9) > < 1 - log(9) $
And this happens at $n = 53$.
Probably all this can be simplified alot and be solved more efficiently.
I assume logarithms and digits of logarithms are key ??
Logarithms and continued fractions are the key. If $2^n$ starts with a $9$, we have that $$ 2^n = (9+\theta)\cdot 10^m , \qquad \theta\in[0,1)$$ $$ n\log(2) = m\log(10) + \log(9+\theta) $$ $$ n = m\log_2(10) +\log_2(9+\theta) $$ $$ \frac{n}{m}-\log_2(10) = \frac{\log_2(9+\theta)}{m} $$ $$ \frac{\log_2(9)}{m}\leq \frac{n}{m}-\log_2(10) \leq \frac{\log_2(10)}{m} \tag{A}$$ and we may solve $(A)$ by studying the (best) rational approximations of the constant $$ \log_{2}(10)=3.321928\ldots=[3; 3, 9, 2, 2, 4, 6, 2, 1, 1, 3, 1, 18, 1, 6, 1, 2, 1, 1, 4, 1, 42,\ldots]. $$ For instance, by truncating the continued fraction at the fourth term we get that $0\leq \frac{196}{59}-\log_2(10)\leq \frac{1}{59^2}$, and by interpolating contiguous convergents we may find all the solutions of $(A)$. In our case $53-15\log_2(10)$ belongs to the range $(\log_2(9),\log_2(10))$ and it gives the minimal solution, so $2^{53}$ is the first power of $2$ whose decimal representation begins with the digit $9$.