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Question Number 18818 by khamizan833@yahoo.com last updated on 30/Jul/17

Answered by daffa22 last updated on 30/Jul/17

$${(x^2-6x+9)}^{x^2-4}={(x^2-6x+9)}^0$$$$x^2-4=0$$$$(x-2)(x+2)=0$$$$x=2\vee x=-2$$$$forx\in R,x=\{2,-2\}$$

Commented by khamizan833@yahoo.com last updated on 30/Jul/17

$$\mathrm{Good}\mathrm{job}\mathrm{brothee}.$$$$\mathrm{thanks}\mathrm{for}\mathrm{your}\mathrm{solution}.$$

Commented by daffa22 last updated on 30/Jul/17

$$sorryimadeamistake,$$$$i{don}^{\prime }tknowwhatx\in Rmeans,$$$$thanksforyourcomment$$

Commented by FilupS last updated on 30/Jul/17

$$x\in R\mathrm{means}x\in \mathbb{R}$$$$\mathrm{It}\mathrm{means}x\mathrm{is}\mathrm{in}\mathrm{the}real\mathrm{set}\mathrm{of}\mathrm{numbers}.$$$$\mathrm{i}.\mathrm{e}.\mathrm{Im}\left(x\right)=0$$

Answered by mrW1 last updated on 30/Jul/17

$$\left(1\right)$$$${\mathrm{x}}^2-4=0$$$$\Rightarrow \mathrm{x}=\pm 2$$$$\left(2\right)$$$${\mathrm{x}}^2-6\mathrm{x}+9=1$$$$\Rightarrow {(\mathrm{x}-3)}^2=1$$$$\Rightarrow \mathrm{x}-3=\pm 1$$$$\Rightarrow \mathrm{x}=3\pm 1$$$$\Rightarrow \mathrm{x}=4,2$$$$\left(3\right)$$$${\mathrm{x}}^2-6\mathrm{x}+9=-1\mathrm{and}{\mathrm{x}}^2-4=\mathrm{even}$$$$\Rightarrow \mathrm{no}\mathrm{real}\mathrm{solution}$$$$$$$$\mathrm{all}\mathrm{solutions}:$$$$\mathrm{x}=-2,2,4$$

Commented by khamizan833@yahoo.com last updated on 30/Jul/17

$$\mathrm{agree}\mathrm{with}\mathrm{your}\mathrm{solution}.$$$$\mathrm{thank}\mathrm{mr}.$$

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