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

What are the roots of the function  f(x)=(log(3^x )−2log(3))∙(x^2 −1) with  x∈R?    Mastermind

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

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

±1 ,2

$$\pm\mathrm{1}\:,\mathrm{2} \\ $$

Commented by Mastermind last updated on 02/Aug/22

your full workings sir!

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

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

f(x)=(log(3^x )−2log(3))∙(x^2 −1)=0     log(3^x )−2log(3)=0 ∣ x^2 −1=0     log(3^x )=2log(3) ∣ x^2 =1     log(3^x )=log(3^2 )  ∣  x=±1     3^x =3^2       x=2

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

Commented by Mastermind last updated on 02/Aug/22

Big thanks to you, you′re too much

$$\mathrm{Big}\:\mathrm{thanks}\:\mathrm{to}\:\mathrm{you},\:\mathrm{you}'\mathrm{re}\:\mathrm{too}\:\mathrm{much} \\ $$

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

AnOther way  f(x)=(log(3^x )−2log(3))∙(x^2 −1)=0  (xlog(3)−2log(3))∙(x^2 −1)=0  log(3)(x−2)(x−1)(x+1)=0  (x−2)(x−1)(x+1)=0  x−2=0 ∣ x−1=0 ∣ x+1=0  x=2 ∣ x=1 ∣ x=−1

$${AnOther}\:{way} \\ $$$$\mathrm{f}\left(\mathrm{x}\right)=\left(\mathrm{log}\left(\mathrm{3}^{\mathrm{x}} \right)−\mathrm{2log}\left(\mathrm{3}\right)\right)\centerdot\left(\mathrm{x}^{\mathrm{2}} −\mathrm{1}\right)=\mathrm{0} \\ $$$$\left(\mathrm{xlog}\left(\mathrm{3}\right)−\mathrm{2log}\left(\mathrm{3}\right)\right)\centerdot\left(\mathrm{x}^{\mathrm{2}} −\mathrm{1}\right)=\mathrm{0} \\ $$$$\mathrm{log}\left(\mathrm{3}\right)\left(\mathrm{x}−\mathrm{2}\right)\left(\mathrm{x}−\mathrm{1}\right)\left(\mathrm{x}+\mathrm{1}\right)=\mathrm{0} \\ $$$$\left(\mathrm{x}−\mathrm{2}\right)\left(\mathrm{x}−\mathrm{1}\right)\left(\mathrm{x}+\mathrm{1}\right)=\mathrm{0} \\ $$$$\mathrm{x}−\mathrm{2}=\mathrm{0}\:\mid\:\mathrm{x}−\mathrm{1}=\mathrm{0}\:\mid\:\mathrm{x}+\mathrm{1}=\mathrm{0} \\ $$$$\mathrm{x}=\mathrm{2}\:\mid\:\mathrm{x}=\mathrm{1}\:\mid\:\mathrm{x}=−\mathrm{1} \\ $$

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