Question Number 187780 by mnjuly1970 last updated on 21/Feb/23
$$ \\ $$$$\:\:{solve} \\ $$$$\:\:\:\:\lfloor\:\:\frac{\mathrm{1}}{\mathrm{4}\:}\:+\:\mathrm{2}^{\:{x}} \:\rfloor\:+\:\lfloor\:\frac{\mathrm{1}}{\mathrm{2}}\:+\:\mathrm{2}^{\:{x}+\mathrm{1}} \:\rfloor=\mathrm{1} \\ $$$$ \\ $$
Answered by witcher3 last updated on 21/Feb/23
![⇔[(1/4)+2^x ]+[2((1/4)+2^x )]=1 y=(1/4)+2^x >(1/4) ⇔[y]+[2y]=1 ⇒y≥(1/2)if not [y]+[2y]≤0+0=0 y=(1/2)+k⇒[(1/2)+k]+[1+2k]=1⇔[2k]+[(1/2)+k]=0 ⇒[(1/2)+k]=[2k]=0 since k≥0 ⇒(1/2)+k∈[0,1[&2k∈[0,1[⇒[0,(1/2)[ ⇒y∈[(1/2),1[ ⇒2^x ∈[(1/4),(3/4)[⇒x∈[−((ln4)/2),−((ln((4/3)))/(ln(2)))[](https://www.tinkutara.com/question/Q187782.png)
$$\Leftrightarrow\left[\frac{\mathrm{1}}{\mathrm{4}}+\mathrm{2}^{\mathrm{x}} \right]+\left[\mathrm{2}\left(\frac{\mathrm{1}}{\mathrm{4}}+\mathrm{2}^{\mathrm{x}} \right)\right]=\mathrm{1} \\ $$$$\mathrm{y}=\frac{\mathrm{1}}{\mathrm{4}}+\mathrm{2}^{\mathrm{x}} >\frac{\mathrm{1}}{\mathrm{4}} \\ $$$$\Leftrightarrow\left[\mathrm{y}\right]+\left[\mathrm{2y}\right]=\mathrm{1} \\ $$$$\Rightarrow\mathrm{y}\geqslant\frac{\mathrm{1}}{\mathrm{2}}\mathrm{if}\:\mathrm{not}\:\left[\mathrm{y}\right]+\left[\mathrm{2y}\right]\leqslant\mathrm{0}+\mathrm{0}=\mathrm{0} \\ $$$$\mathrm{y}=\frac{\mathrm{1}}{\mathrm{2}}+\mathrm{k}\Rightarrow\left[\frac{\mathrm{1}}{\mathrm{2}}+\mathrm{k}\right]+\left[\mathrm{1}+\mathrm{2k}\right]=\mathrm{1}\Leftrightarrow\left[\mathrm{2k}\right]+\left[\frac{\mathrm{1}}{\mathrm{2}}+\mathrm{k}\right]=\mathrm{0} \\ $$$$\Rightarrow\left[\frac{\mathrm{1}}{\mathrm{2}}+\mathrm{k}\right]=\left[\mathrm{2k}\right]=\mathrm{0}\:\mathrm{since}\:\mathrm{k}\geqslant\mathrm{0} \\ $$$$\Rightarrow\frac{\mathrm{1}}{\mathrm{2}}+\mathrm{k}\in\left[\mathrm{0},\mathrm{1}\left[\&\mathrm{2k}\in\left[\mathrm{0},\mathrm{1}\left[\Rightarrow\left[\mathrm{0},\frac{\mathrm{1}}{\mathrm{2}}\left[\right.\right.\right.\right.\right.\right. \\ $$$$\Rightarrow\mathrm{y}\in\left[\frac{\mathrm{1}}{\mathrm{2}},\mathrm{1}\left[\right.\right. \\ $$$$\Rightarrow\mathrm{2}^{\mathrm{x}} \in\left[\frac{\mathrm{1}}{\mathrm{4}},\frac{\mathrm{3}}{\mathrm{4}}\left[\Rightarrow\mathrm{x}\in\left[−\frac{\mathrm{ln4}}{\mathrm{2}},−\frac{\mathrm{ln}\left(\frac{\mathrm{4}}{\mathrm{3}}\right)}{\mathrm{ln}\left(\mathrm{2}\right)}\left[\right.\right.\right.\right. \\ $$