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Question Number 175333 by mnjuly1970 last updated on 27/Aug/22

$$$$$$solve(x\in \mathbb{R}).$$$$$$$$\lfloor \frac{2^x+1}{3}\rfloor +\lfloor 4^{\frac{x}{2}}\rfloor =4$$$$------$$

Answered by mr W last updated on 27/Aug/22

$$t=\frac{2^x+1}{3}$$$$\lfloor t\rfloor +\lfloor 3t-1\rfloor =4$$$$t=n+f$$$$n+\lfloor 3n+3f-1\rfloor =4$$$$n+3n-1+\lfloor 3f\rfloor =4$$$$5-4n=\lfloor 3f\rfloor ⩾0\Rightarrow 4n⩽5\Rightarrow n⩽1$$$$5-4n=\lfloor 3f\rfloor <3\Rightarrow 4n>2\Rightarrow n⩾1$$$$\Rightarrow n=1$$$$\lfloor 3f\rfloor =5-4n=1$$$$\Rightarrow 1⩽3f<2$$$$\Rightarrow \frac{1}{3}⩽f<\frac{2}{3}$$$$\Rightarrow 1+\frac{1}{3}⩽t<1+\frac{2}{3}$$$$\Rightarrow \frac{4}{3}⩽\frac{2^x+1}{3}<\frac{5}{3}$$$$\Rightarrow 4⩽2^x+1<5$$$$\Rightarrow 3⩽2^x<4$$$$\Rightarrow {\log }_{2}3⩽x<2$$

Commented by mnjuly1970 last updated on 27/Aug/22

$$thankzalotsirW$$

Commented by Tawa11 last updated on 28/Aug/22

$$\mathrm{Great}\mathrm{sir}$$

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