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2-x-3-y-6-3-x-4-y-12-




Question Number 99789 by bemath last updated on 23/Jun/20
 { ((2^x .3^y  = 6)),((3^x .4^y  = 12)) :}
$$\begin{cases}{\mathrm{2}^{{x}} .\mathrm{3}^{{y}} \:=\:\mathrm{6}}\\{\mathrm{3}^{{x}} .\mathrm{4}^{{y}} \:=\:\mathrm{12}}\end{cases} \\ $$
Commented by john santu last updated on 23/Jun/20
 { ((3^(y−1)  = 2^(1−x) )),((4^(y−1)  = 3^(1−x)  )) :}  ⇔ ((3/4))^(y−1)  = ((2/3))^(1−x)   ⇔3^(y−1) .3^(1−x)  = 2^(1−x)  . 2^(2y−2)   (3)^(y−x)  = 2^(2y−x−1)    { ((y−x = 0; y=x)),((2y−x−1 = 0 ; 2y= x+1)) :}  ⇒2x = x +1 , x = 1 ∧ y =1
$$\begin{cases}{\mathrm{3}^{{y}−\mathrm{1}} \:=\:\mathrm{2}^{\mathrm{1}−{x}} }\\{\mathrm{4}^{{y}−\mathrm{1}} \:=\:\mathrm{3}^{\mathrm{1}−{x}} \:}\end{cases} \\ $$$$\Leftrightarrow\:\left(\frac{\mathrm{3}}{\mathrm{4}}\right)^{\mathrm{y}−\mathrm{1}} \:=\:\left(\frac{\mathrm{2}}{\mathrm{3}}\right)^{\mathrm{1}−{x}} \\ $$$$\Leftrightarrow\mathrm{3}^{\mathrm{y}−\mathrm{1}} .\mathrm{3}^{\mathrm{1}−{x}} \:=\:\mathrm{2}^{\mathrm{1}−{x}} \:.\:\mathrm{2}^{\mathrm{2}{y}−\mathrm{2}} \\ $$$$\left(\mathrm{3}\right)^{{y}−{x}} \:=\:\mathrm{2}^{\mathrm{2}{y}−{x}−\mathrm{1}} \\ $$$$\begin{cases}{{y}−{x}\:=\:\mathrm{0};\:\mathrm{y}={x}}\\{\mathrm{2}{y}−{x}−\mathrm{1}\:=\:\mathrm{0}\:;\:\mathrm{2y}=\:{x}+\mathrm{1}}\end{cases} \\ $$$$\Rightarrow\mathrm{2x}\:=\:\mathrm{x}\:+\mathrm{1}\:,\:\mathrm{x}\:=\:\mathrm{1}\:\wedge\:\mathrm{y}\:=\mathrm{1}\: \\ $$
Commented by Dwaipayan Shikari last updated on 23/Jun/20
2^x 3^y (1/(3^x 4^y ))=(1/2)  ⇒2^(x−2y+1) 3^(y−x) =1  ⇒2^(x−2y+1) =3^(x−y)   only possible when x−2y+1=0  And x−y=0  so{x=1,y=1
$$\mathrm{2}^{{x}} \mathrm{3}^{{y}} \frac{\mathrm{1}}{\mathrm{3}^{{x}} \mathrm{4}^{{y}} }=\frac{\mathrm{1}}{\mathrm{2}} \\ $$$$\Rightarrow\mathrm{2}^{{x}−\mathrm{2}{y}+\mathrm{1}} \mathrm{3}^{{y}−{x}} =\mathrm{1} \\ $$$$\Rightarrow\mathrm{2}^{{x}−\mathrm{2}{y}+\mathrm{1}} =\mathrm{3}^{{x}−{y}} \\ $$$${only}\:{possible}\:{when}\:{x}−\mathrm{2}{y}+\mathrm{1}=\mathrm{0} \\ $$$${And}\:{x}−{y}=\mathrm{0} \\ $$$${so}\left\{{x}=\mathrm{1},{y}=\mathrm{1}\right. \\ $$

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