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Question Number 58862 by George Mark Samuel last updated on 01/May/19

2^(x+y=) 6^y 3^x =3(2^(y−1) ) solve for x and y

$$\mathrm{2}^{\mathrm{x}+\mathrm{y}=} \mathrm{6}^{\mathrm{y}} \\ $$$$\mathrm{3}^{\mathrm{x}} =\mathrm{3}\left(\mathrm{2}^{\mathrm{y}−\mathrm{1}} \right) \\ $$$$\mathrm{solve}\:\mathrm{for}\:\mathrm{x}\:\mathrm{and}\:\mathrm{y} \\ $$

Answered by MJS last updated on 01/May/19

2^(x+y) =6^y 2^x ×2^y =2^y ×3^y 2^x =3^y xln 2 =yln 3 y=((ln 2)/(ln 3))x 3^x =3(2^(y−1) ) 3^(x−1) =2^(y−1) (x−1)ln 3 =(y−1)ln 2 xln 3−ln 3 =yln 2 −ln2 y=((ln 3)/(ln 2))x−((ln3)/(ln 2))+1 ((ln 2)/(ln 3))x=((ln 3)/(ln 2))x−((ln3)/(ln 2))+1 ((ln 3)/(ln 2))=a (x/a)=ax−a+1 x=a^2 x−a^2 +a x(1−a^2 )=a−a^2 x(1−a)(1+a)=a(1−a) x=(a/(a+1))=((ln 3)/(ln 6)) y=(1/a)x=(1/(a+1))=((ln 2)/(ln 6))

$$\mathrm{2}^{{x}+{y}} =\mathrm{6}^{{y}} \\ $$$$\mathrm{2}^{{x}} ×\mathrm{2}^{{y}} =\mathrm{2}^{{y}} ×\mathrm{3}^{{y}} \\ $$$$\mathrm{2}^{{x}} =\mathrm{3}^{{y}} \\ $$$${x}\mathrm{ln}\:\mathrm{2}\:={y}\mathrm{ln}\:\mathrm{3} \\ $$$${y}=\frac{\mathrm{ln}\:\mathrm{2}}{\mathrm{ln}\:\mathrm{3}}{x} \\ $$$$ \\ $$$$\mathrm{3}^{{x}} =\mathrm{3}\left(\mathrm{2}^{{y}−\mathrm{1}} \right) \\ $$$$\mathrm{3}^{{x}−\mathrm{1}} =\mathrm{2}^{{y}−\mathrm{1}} \\ $$$$\left({x}−\mathrm{1}\right)\mathrm{ln}\:\mathrm{3}\:=\left({y}−\mathrm{1}\right)\mathrm{ln}\:\mathrm{2} \\ $$$${x}\mathrm{ln}\:\mathrm{3}−\mathrm{ln}\:\mathrm{3}\:={y}\mathrm{ln}\:\mathrm{2}\:−\mathrm{ln2} \\ $$$${y}=\frac{\mathrm{ln}\:\mathrm{3}}{\mathrm{ln}\:\mathrm{2}}{x}−\frac{\mathrm{ln3}}{\mathrm{ln}\:\mathrm{2}}+\mathrm{1} \\ $$$$ \\ $$$$\frac{\mathrm{ln}\:\mathrm{2}}{\mathrm{ln}\:\mathrm{3}}{x}=\frac{\mathrm{ln}\:\mathrm{3}}{\mathrm{ln}\:\mathrm{2}}{x}−\frac{\mathrm{ln3}}{\mathrm{ln}\:\mathrm{2}}+\mathrm{1} \\ $$$$\frac{\mathrm{ln}\:\mathrm{3}}{\mathrm{ln}\:\mathrm{2}}={a} \\ $$$$\frac{{x}}{{a}}={ax}−{a}+\mathrm{1} \\ $$$${x}={a}^{\mathrm{2}} {x}−{a}^{\mathrm{2}} +{a} \\ $$$${x}\left(\mathrm{1}−{a}^{\mathrm{2}} \right)={a}−{a}^{\mathrm{2}} \\ $$$${x}\left(\mathrm{1}−{a}\right)\left(\mathrm{1}+{a}\right)={a}\left(\mathrm{1}−{a}\right) \\ $$$${x}=\frac{{a}}{{a}+\mathrm{1}}=\frac{\mathrm{ln}\:\mathrm{3}}{\mathrm{ln}\:\mathrm{6}} \\ $$$${y}=\frac{\mathrm{1}}{{a}}{x}=\frac{\mathrm{1}}{{a}+\mathrm{1}}=\frac{\mathrm{ln}\:\mathrm{2}}{\mathrm{ln}\:\mathrm{6}} \\ $$

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