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Romanian Mathematical Olympiad

Algebra Difficulty 5.6 AIME, harder Prove it Romania

a) $\{x \in \mathbb{R} \mid \log_2[x] = [\log_2 x]\} = \bigcup_{m \in \mathbb{N}} [2^m, 2^m + 1)$ .

b) $\{x \in \mathbb{R} \mid 2^{\lfloor x \rfloor} = \lfloor 2^x \rfloor\} = \bigcup_{m \in \mathbb{N}} [m, \log_2 (2^m + 1))$ .

(here, $[a]$ denotes the integer part (floor function) of the real number $a$ ).

Solution

a) The existence of the logarithms requires $x > 1$ .
If $[\log_2 x] = m$ , then $m \in \mathbb{N}$ and $2^m \le x < 2^{m+1}$ . From $\log_2[x] = m$ , follows $[x] = 2^m$ , that is $2^m \le x < 2^m + 1$ .
Conversely, $x \in [2^m, 2^m + 1)$ , $m \in \mathbb{N}$ yields $[\log_2 x] = \log_2[x] = m$ .

b) If $[2^x] = t \in \mathbb{N}$ , then $\log_2 t \le x < \log_2(t+1)$ . From $2^{[x]} = t$ follows $[x] = \log_2 t = m \in \mathbb{N}$ , whence $m \le x < \log_2(2^m + 1)$ .
Conversely, if $x \in [m, \log_2(2^m + 1))$ , $m \in \mathbb{N}$ then $[2^x] = 2^{[x]} = m$ .

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