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Question Number 27258 by hasie09 last updated on 04/Jan/18

$$\mathrm{If}{9}^x-4\times 3^{x+2}+3^5=0,\mathrm{then}\mathrm{the}\mathrm{solution}$$$$\mathrm{set}\mathrm{is}$$

Commented by tawa tawa last updated on 04/Jan/18

$${\left(3^2\right)}^{\mathrm{x}}-\left(4\times 3^{\mathrm{x}}\times 3^2\right)+243=0$$$${\left(3^{\mathrm{x}}\right)}^2-\left(4\times 3^{\mathrm{x}}\times 9\right)+243=0$$$$\mathrm{Let}{3}^{\mathrm{x}}=\mathrm{y}$$$$\therefore {\mathrm{y}}^2-\left(4\times \mathrm{y}\times 9\right)+243=0$$$$\therefore {\mathrm{y}}^2-36\mathrm{y}+243=0$$$$\therefore {\mathrm{y}}^2-27\mathrm{y}-9\mathrm{y}+243=0$$$$\therefore ({\mathrm{y}}^2-27\mathrm{y})-(9\mathrm{y}+243)=0$$$$\therefore \mathrm{y}(\mathrm{y}-27)-9(\mathrm{y}-27)=0$$$$\therefore (\mathrm{y}-27)(\mathrm{y}-9)=0$$$$\therefore \mathrm{y}-27=0\mathrm{or}\mathrm{y}-9=0$$$$\therefore \mathrm{y}=27\mathrm{or}\mathrm{y}=9$$$$\mathrm{Remember}:3^{\mathrm{x}}=\mathrm{y}$$$$\mathrm{when}\mathrm{y}=27$$$$\therefore 3^{\mathrm{x}}=27$$$$\therefore 3^{\mathrm{x}}=3^3$$$$\mathrm{Since}\mathrm{the}\mathrm{base}\mathrm{are}\mathrm{equal}$$$$\therefore \mathrm{x}=3$$$$\mathrm{when}\mathrm{y}=9$$$$\therefore 3^{\mathrm{x}}=9$$$$\therefore 3^{\mathrm{x}}=3^2$$$$\mathrm{Since}\mathrm{the}\mathrm{base}\mathrm{are}\mathrm{equal}$$$$\therefore \mathrm{x}=2$$$$\mathrm{Therefore},\mathrm{x}=3\mathrm{or}\mathrm{x}=2$$

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