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solve-3-2x-y-12-2-x-y-4-




Question Number 171849 by Mikenice last updated on 21/Jun/22
solve:  3^(2x+y) =12  2^(x−y) =4
$${solve}: \\ $$$$\mathrm{3}^{\mathrm{2}{x}+{y}} =\mathrm{12} \\ $$$$\mathrm{2}^{{x}−{y}} =\mathrm{4} \\ $$
Commented by kaivan.ahmadi last updated on 21/Jun/22
2^(x−y) =4=2^2 ⇒x−y=2⇒x=y+2  on the othere hand  3^(2x+y) =3^(2y+4+y) =12⇒3^(3y+4) =12⇒  (3y+4)log3=log12⇒3y+4=((log12)/(log3))=  ((log3+log4)/(log3))=1+log_3 4⇒y=(−3+log_3 4)/3    ⇒x=2+(−3+log_3 4)/4
$$\mathrm{2}^{{x}−{y}} =\mathrm{4}=\mathrm{2}^{\mathrm{2}} \Rightarrow{x}−{y}=\mathrm{2}\Rightarrow{x}={y}+\mathrm{2} \\ $$$${on}\:{the}\:{othere}\:{hand} \\ $$$$\mathrm{3}^{\mathrm{2}{x}+{y}} =\mathrm{3}^{\mathrm{2}{y}+\mathrm{4}+{y}} =\mathrm{12}\Rightarrow\mathrm{3}^{\mathrm{3}{y}+\mathrm{4}} =\mathrm{12}\Rightarrow \\ $$$$\left(\mathrm{3}{y}+\mathrm{4}\right){log}\mathrm{3}={log}\mathrm{12}\Rightarrow\mathrm{3}{y}+\mathrm{4}=\frac{{log}\mathrm{12}}{{log}\mathrm{3}}= \\ $$$$\frac{{log}\mathrm{3}+{log}\mathrm{4}}{{log}\mathrm{3}}=\mathrm{1}+{log}_{\mathrm{3}} \mathrm{4}\Rightarrow{y}=\left(−\mathrm{3}+{log}_{\mathrm{3}} \mathrm{4}\right)/\mathrm{3} \\ $$$$ \\ $$$$\Rightarrow{x}=\mathrm{2}+\left(−\mathrm{3}+{log}_{\mathrm{3}} \mathrm{4}\right)/\mathrm{4} \\ $$
Answered by cherokeesay last updated on 21/Jun/22
2^(x−y) =4=2^2 ⇔x−y=2⇒x=y+2  3^(2(y+2)+y) =12⇔3^(3y+4) =12⇒y=((2ln2)/(3ln3))−1  and x=((2ln2)/(3ln3))+1
$$\mathrm{2}^{{x}−{y}} =\mathrm{4}=\mathrm{2}^{\mathrm{2}} \Leftrightarrow{x}−{y}=\mathrm{2}\Rightarrow{x}={y}+\mathrm{2} \\ $$$$\mathrm{3}^{\mathrm{2}\left({y}+\mathrm{2}\right)+{y}} =\mathrm{12}\Leftrightarrow\mathrm{3}^{\mathrm{3}{y}+\mathrm{4}} =\mathrm{12}\Rightarrow{y}=\frac{\mathrm{2}{ln}\mathrm{2}}{\mathrm{3}{ln}\mathrm{3}}−\mathrm{1} \\ $$$${and}\:{x}=\frac{\mathrm{2}{ln}\mathrm{2}}{\mathrm{3}{ln}\mathrm{3}}+\mathrm{1} \\ $$

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