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Question Number 62468 by bshahid010@gmail.com last updated on 21/Jun/19

Commented by mathmax by abdo last updated on 21/Jun/19

$$let{3}^x=t\Rightarrow (k-2)t^2-2t+3>0\mathrm{\forall }t>0\Rightarrow {\mathrm{\Delta }}^{{}^{\prime }}<0\Rightarrow$$$${(-1)}^2-3(k-2)<0\Rightarrow 1-3k+6<0\Rightarrow 7-3k<0\Rightarrow k>\frac{7}{3}$$

Answered by mr W last updated on 21/Jun/19

$$lett=3^x>0$$$$(k-2)9^x-2\cdot 3^x+3$$$$=(k-2)t^2-2t+3$$$$=(k-2)[t^2-\frac{2t}{k-2}+{\left(\frac{1}{k-2}\right)}^2]+3-\frac{1}{k-2}$$$$=(k-2){(t-\frac{1}{k-2})}^2+3-\frac{1}{k-2}$$$$suchthatitisalways>0,wemust$$$$havek-2>0\wedge 3-\frac{1}{k-2}>0$$$$\Rightarrow k>2and$$$$\Rightarrow 3>\frac{1}{k-2}\Rightarrow k-2>\frac{1}{3}\Rightarrow k>2+\frac{1}{3}=\frac{7}{3}$$$$\Rightarrow answerisk>\frac{7}{3}$$

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