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Question Number 174495 by Mastermind last updated on 02/Aug/22

$$\mathrm{What}\mathrm{are}\mathrm{the}\mathrm{roots}\mathrm{of}\mathrm{the}\mathrm{function}$$$$\mathrm{f}\left(\mathrm{x}\right)=(\log \left(3^{\mathrm{x}}\right)-2\log \left(3\right))\cdot ({\mathrm{x}}^2-1)\mathrm{with}$$$$\mathrm{x}\in \mathrm{R}?$$$$$$$$\mathrm{Mastermind}$$

Commented by kaivan.ahmadi last updated on 02/Aug/22

$$\pm 1,2$$

Commented by Mastermind last updated on 02/Aug/22

$$\mathrm{your}\mathrm{full}\mathrm{workings}\mathrm{sir}!$$

Answered by Rasheed.Sindhi last updated on 02/Aug/22

$$\mathrm{f}\left(\mathrm{x}\right)=(\log \left(3^{\mathrm{x}}\right)-2\log \left(3\right))\cdot ({\mathrm{x}}^2-1)=0$$$$\log \left(3^{\mathrm{x}}\right)-2\log \left(3\right)=0\mid {\mathrm{x}}^2-1=0$$$$\log \left(3^{\mathrm{x}}\right)=2\log \left(3\right)\mid {\mathrm{x}}^2=1$$$$\log \left(3^{\mathrm{x}}\right)=\log \left(3^2\right)\mid \mathrm{x}=\pm 1$$$$3^{\mathrm{x}}=3^2$$$$\mathrm{x}=2$$$$$$

Commented by Mastermind last updated on 02/Aug/22

$$\mathrm{Big}\mathrm{thanks}\mathrm{to}\mathrm{you},{\mathrm{you}}^{\prime }\mathrm{re}\mathrm{too}\mathrm{much}$$

Answered by Rasheed.Sindhi last updated on 19/Aug/22

$$AnOtherway$$$$\mathrm{f}\left(\mathrm{x}\right)=(\log \left(3^{\mathrm{x}}\right)-2\log \left(3\right))\cdot ({\mathrm{x}}^2-1)=0$$$$(\mathrm{xlog}\left(3\right)-2\log \left(3\right))\cdot ({\mathrm{x}}^2-1)=0$$$$\log \left(3\right)(\mathrm{x}-2)(\mathrm{x}-1)(\mathrm{x}+1)=0$$$$(\mathrm{x}-2)(\mathrm{x}-1)(\mathrm{x}+1)=0$$$$\mathrm{x}-2=0\mid \mathrm{x}-1=0\mid \mathrm{x}+1=0$$$$\mathrm{x}=2\mid \mathrm{x}=1\mid \mathrm{x}=-1$$

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