Prove that there is an integer $k$ such that $2^k$ is starting with $999$

Question

As the title says, I want to prove that there is a natural number $k$ such that $2^k$ is starting with $999$. Can you help me please ?

Answer 1

If $\log_{10}(2^k) = k \log_{10}(2)$ is extremely close to an integer, but less than such integer, we are happy. To get some working values for $k$, it is enough to compute the continued fraction of $$\frac{\log 2}{\log 10}=[0; 3, 3, 9, 2, 2, 4, 6, 2, 1, 1, 3, 1, 18,\ldots ]. $$

By considering the expansion of $[0; 3, 3, 9, 2, 2, 4, 6, 2, 1, 1, 3]$ we get: $$ 254370\cdot\log_{10}(2) = 76572.999997\ldots $$ hence the number $\color{red}{2^{254370}}$ starts with the digits $999$ as wanted.
The same happens with $\color{red}{2^{13301}}$ that is associated with the continued fraction $[0; 3, 3, 9, 2, 2, 4, 6]$.

The least power of two with the wanted property should be $\color{red}{2^{2621}}$ that is associated with the continued fraction $[0, 3, 3, 9, 2, 2, 5]$.