Big Powers of 2

We want to find $N \in \mathbb{N}$ such that $\begin{array}{rcccl} 10^N & \le & 2^k & < & 1.00001 \times 10^N \\ N & \le & k \log_{10}2 & < & N + \log_{10}1.00001 \\ N & \le & (10^{100} + n)\log_{10}2 & < & N + \log_{10}1.00001 \\ 0 & \le & \big\{ \{10^{100}\log_{10}2\} + n\log_{10}2 \big\} & < &\log_{10}1.00001 \end{array}$ where $\{x\}$ is the fractional part of $x$ . Since (thanks to Mathematica) $\{ 10^{100} \log_{10}2\} \,=\, 0.4068447719...$ , we want to find the smallest $n$ such that $0 \; \le \; \big\{ 0.4068447719 + n \log_{10}2 \big\} \; < \; 4.34292 \times 10^{-6} \; .$ A computer evaluation of $F(n) \,=\, \big\{ 0.4068447719 + n \log_{10}2\big\}$ for $1 \le n \le 15000$ shows that $F(13974) = 4.18 \times 10^{-6}$ , while the next smallest value of $F(n)$ is $F(673) = 3.19 \times 10^{-5}$ . Thus the answer is $n = 13974$ .