5.1. Find the smallest natural solution of the inequality $2^{x}+2^{x+1}+\ldots+2^{x+2000}>2^{2017}$. Answer. 17.
Solution. The value $x=17$ is clearly a solution. If $x \leqslant 16$, then we have

$$
2^{x}+2^{x+1}+\ldots+2^{x+2000} \leqslant 2^{16}+2^{17}+\ldots+2^{2015}+2^{2016}<2^{2017}
$$

since this inequality reduces sequentially to the following: $2^{16}+2^{17}+\ldots+2^{2015}<$ $2^{2017}-2^{2016}=2^{2016}, 2^{16}+2^{17}+\ldots+2^{2014}<2^{2016}-2^{2015}=2^{2015}, \ldots, 2^{16}+2^{17}<2^{19}-2^{18}=2^{18}$, $2^{16}<2^{18}-2^{17}=2^{17}$, which is true.