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Question Number 130337 by EDWIN88 last updated on 24/Jan/21

$${({\mathrm{x}}^2-4\mathrm{x}+3)}^{{\mathrm{x}}^2-6\mathrm{x}+4}⩽1$$

Answered by mathmax by abdo last updated on 24/Jan/21

$$\mathrm{e}\Rightarrow {\mathrm{e}}^{({\mathrm{x}}^2-6\mathrm{x}+4)\ln ({\mathrm{x}}^2-4\mathrm{x}+3)}⩽1$$$${\mathrm{x}}^2-4\mathrm{x}+3=0\to {\mathrm{\Delta }}^{{}^{\prime }}=4-3=1\Rightarrow {\mathrm{x}}_1=\frac{2+1}{1}=3\mathrm{and}{\mathrm{x}}_2=\frac{2-1}{1}=1$$$$\mathrm{the}\mathrm{equation}\mathrm{is}\mathrm{dfined}\mathrm{on}]-\mathrm{\infty },1[\cup ]3,+\mathrm{\infty }[$$$$\mathrm{e}\Rightarrow ({\mathrm{x}}^2-6\mathrm{x}+4)\ln ({\mathrm{x}}^2-4\mathrm{x}+3)⩽0$$$${\mathrm{x}}^2-6\mathrm{x}+4=0\to {\mathrm{\Delta }}^{{}^{\prime }}=9-4=5\Rightarrow {\mathrm{x}}_1=3+\sqrt{5}\mathrm{and}{\mathrm{x}}_2=3-\sqrt{5}$$$$\ln ({\mathrm{x}}^2-4\mathrm{x}+3)>0\Rightarrow {\mathrm{x}}^2-4\mathrm{x}+3>1\Rightarrow {\mathrm{x}}^2-4\mathrm{x}+2>0\Rightarrow$$$${\mathrm{x}}^2-4\mathrm{x}+4-2>0\Rightarrow {(\mathrm{x}-2)}^2-2>0\Rightarrow (\mathrm{x}-2-\sqrt{2})(\mathrm{x}-2+\sqrt{2})>0\Rightarrow$$$$\mathrm{x}\in ]-\mathrm{\infty },2-\sqrt{2}[\cup ]2+\sqrt{2},+\mathrm{\infty }[$$$$\ln ({\mathrm{x}}^2-4\mathrm{x}+3)<0\Rightarrow {\mathrm{x}}^2-4\mathrm{x}+3<1\Rightarrow {\mathrm{x}}^2-4\mathrm{x}+2<0\Rightarrow$$$$]2-\sqrt{2},2+\sqrt{2}[\mathrm{the}\mathrm{problem}\mathrm{lead}\mathrm{to}\mathrm{find}\mathrm{sgn}\mathrm{of}({\mathrm{x}}^2-6\mathrm{x}+4)\ln ({\mathrm{x}}^2-4\mathrm{x}+3)$$$$...\mathrm{be}\mathrm{continued}...$$

Answered by MJS_new last updated on 24/Jan/21

$$p\in \mathbb{R}\wedge q\in \mathbb{R}\wedge r\in \mathbb{Z}$$$$p^q=1$$$$\left(1\right)p=1$$$$\left(2\right)q=0\wedge p\ne 0$$$$\left(3\right)p=-1\wedge q=2r$$$$\left(1\right)x^2-4x+3=1\Rightarrow x=2\pm \sqrt{2}$$$$\left(2\right)x^2-6x+4=0\Rightarrow x=3\pm \sqrt{5}$$$$x^2-4x+3\ne 0\mathrm{in}\mathrm{both}\mathrm{cases}$$$$\left(3\right)x^2-4x+3=-1\Rightarrow x=2$$$$x^2-6x+4=-4\mathrm{in}\mathrm{this}\mathrm{case}$$$$$$$$\mathrm{both}p\mathrm{and}q\mathrm{are}\mathrm{parabolas}\Rightarrow$$$$\mathrm{solution}\mathrm{for}{p}^q⩽1\mathrm{is}$$$$2-\sqrt{2}⩽x⩽3-\sqrt{5}\vee x=2\vee 2+\sqrt{2}⩽x⩽3+\sqrt{5}$$

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