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

$$$$$$solve$$$$\lfloor \frac{1}{4}+2^x\rfloor +\lfloor \frac{1}{2}+2^{x+1}\rfloor =1$$$$$$

Answered by witcher3 last updated on 21/Feb/23

$$\iff [\frac{1}{4}+2^{\mathrm{x}}]+\left[2(\frac{1}{4}+2^{\mathrm{x}})\right]=1$$$$\mathrm{y}=\frac{1}{4}+2^{\mathrm{x}}>\frac{1}{4}$$$$\iff \left[\mathrm{y}\right]+\left[2\mathrm{y}\right]=1$$$$\Rightarrow \mathrm{y}⩾\frac{1}{2}\mathrm{if}\mathrm{not}\left[\mathrm{y}\right]+\left[2\mathrm{y}\right]⩽0+0=0$$$$\mathrm{y}=\frac{1}{2}+\mathrm{k}\Rightarrow [\frac{1}{2}+\mathrm{k}]+[1+2\mathrm{k}]=1\iff \left[2\mathrm{k}\right]+[\frac{1}{2}+\mathrm{k}]=0$$$$\Rightarrow [\frac{1}{2}+\mathrm{k}]=\left[2\mathrm{k}\right]=0\mathrm{since}\mathrm{k}⩾0$$$$\Rightarrow \frac{1}{2}+\mathrm{k}\in [0,1[\mathrm{\&}2\mathrm{k}\in [0,1[\Rightarrow [0,\frac{1}{2}[$$$$\Rightarrow \mathrm{y}\in [\frac{1}{2},1[$$$$\Rightarrow 2^{\mathrm{x}}\in [\frac{1}{4},\frac{3}{4}[\Rightarrow \mathrm{x}\in [-\frac{\ln 4}{2},-\frac{\ln \left(\frac{4}{3}\right)}{\ln \left(2\right)}[$$

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