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Question Number 143627 by bobhans last updated on 16/Jun/21

Answered by Olaf_Thorendsen last updated on 16/Jun/21

$$\mathrm{X}=\frac{(3^x-4^x)(3^x-4.4^x)}{{\log }_2\left(6(x-\frac{1}{2})(x-\frac{4}{3})\right)}$$$$etc...$$

Answered by Canebulok last updated on 18/Jun/21

$$$$$$\mathit{S}\mathit{o}\mathit{l}\mathit{u}\mathit{t}\mathit{i}\mathit{o}\mathit{n}:$$$$\Rightarrow \frac{4\left(4^{2x}\right)-5\left(4^x\right)\left(3^x\right)+\left(3^{2x}\right)}{{log}_2(6x^2-11x+4)}\le 0$$$$Wehavetwopossiblecasesinthisequation,$$$$Case1:$$$$\Rightarrow 4\left(4^{2x}\right)-5\left(4^x\right)\left(3^x\right)+\left(3^{2x}\right)\le 0$$$$Let:$$$$k=4^{x}andy=3^x$$$$$$$$\Rightarrow 4k^2-5ky+y^2\le 0$$$$\Rightarrow y^2-5ky+4k^2\le 0$$$$\Rightarrow (y-4k)(y-k)\le 0$$$$Thus;$$$$\Rightarrow y-4k\le 0$$$$\Rightarrow y\le 4k$$$$\Rightarrow 3^x\le 4^{x+1}$$$$\Rightarrow xLn\left(3\right)\le (k+1)Ln\left(4\right)$$$$\Rightarrow \frac{Ln\left(3\right)}{Ln\left(4\right)}\le \frac{x+1}{x}$$$$\Rightarrow \frac{Ln\left(3\right)}{Ln\left(4\right)}\le 1+\frac{1}{x}$$$$\Rightarrow \frac{Ln\left(3\right)}{Ln\left(4\right)}-1\le \frac{1}{x}$$$$\Rightarrow \frac{Ln\left(3\right)-Ln\left(4\right)}{Ln\left(4\right)}\le \frac{1}{x}$$$$\Rightarrow x\le \frac{Ln\left(4\right)}{Ln\left(3\right)-Ln\left(4\right)}$$$$$$$$Case2:Asymptotes$$$$\Rightarrow \frac{1}{{log}_2(6x^2-11x+4)}\le 0$$$$\Rightarrow {log}_2(6x^2-11x+4)<0$$$$\Rightarrow 6x^2-11x+4>2^0$$$$\Rightarrow 6x^2-11x+4>1$$$$\Rightarrow 6x^2-11x+3>0$$$$\Rightarrow x^2-\frac{11}{6}x+\frac{1}{2}>0$$$$Byusingthequadraticformula:$$$$\Rightarrow x=\frac{\left(\frac{11}{6}\right)\pm \sqrt{{(-\frac{11}{6})}^2-4\left(\frac{1}{2}\right)}}{2}$$$$\Rightarrow x_1=\frac{3}{2}$$$$\Rightarrow x_2=\frac{1}{3}$$$$$$$$Thus;$$$$\Rightarrow (x-\frac{3}{2})(x-\frac{1}{3})>0$$$$\Rightarrow x>\frac{3}{2}$$$$\Rightarrow x>\frac{1}{3}$$$$$$$$\sim Kevin$$

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