solve-1-4-2-x-1-2-2-x-1-1-

Question Number 187780 by mnjuly1970 last updated on 21/Feb/23

s o l v e

⌊ \frac{1}{4} + 2^{x} ⌋ + ⌊ \frac{1}{2} + 2^{x + 1} ⌋ = 1

Answered by witcher3 last updated on 21/Feb/23

\Leftrightarrow [\frac{1}{4} + 2^{x}] + [2 (\frac{1}{4} + 2^{x})] = 1

y = \frac{1}{4} + 2^{x} > \frac{1}{4}

\Leftrightarrow [y] + [2 y] = 1

\Rightarrow y ⩾ \frac{1}{2} if not [y] + [2 y] ⩽ 0 + 0 = 0

y = \frac{1}{2} + k \Rightarrow [\frac{1}{2} + k] + [1 + 2 k] = 1 \Leftrightarrow [2 k] + [\frac{1}{2} + k] = 0

\Rightarrow [\frac{1}{2} + k] = [2 k] = 0 since k ⩾ 0

\Rightarrow \frac{1}{2} + k \in [0, 1 [& 2 k \in [0, 1 [\Rightarrow [0, \frac{1}{2} [

\Rightarrow y \in [\frac{1}{2}, 1 [

\Rightarrow 2^{x} \in [\frac{1}{4}, \frac{3}{4} [\Rightarrow x \in [- \frac{\ln 4}{2}, - \frac{\ln (\frac{4}{3})}{\ln (2)} [

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