{ ((2^x .3^y = 6)),((3^x .4^y = 12)) :}


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Question Number 99789 by bemath last updated on 23/Jun/20
${ ((2^x .3^y = 6)),((3^x .4^y = 12)) :}$
${\begin{cases} 2^{x} . 3^{y} = 6 \\ 3^{x} . 4^{y} = 12 \end{cases}$
Commented by john santu last updated on 23/Jun/20
${ ((3^(y−1) = 2^(1−x) )),((4^(y−1) = 3^(1−x) )) :} ⇔ ((3/4))^(y−1) = ((2/3))^(1−x) ⇔3^(y−1) .3^(1−x) = 2^(1−x) . 2^(2y−2) (3)^(y−x) = 2^(2y−x−1) { ((y−x = 0; y=x)),((2y−x−1 = 0 ; 2y= x+1)) :} ⇒2x = x +1 , x = 1 ∧ y =1$
${\begin{cases} 3^{y - 1} = 2^{1 - x} \\ 4^{y - 1} = 3^{1 - x} \end{cases}$ $\Leftrightarrow {(\frac{3}{4})}^{y - 1} = {(\frac{2}{3})}^{1 - x}$ $\Leftrightarrow 3^{y - 1} . 3^{1 - x} = 2^{1 - x} . 2^{2 y - 2}$ ${(3)}^{y - x} = 2^{2 y - x - 1}$ ${\begin{cases} y - x = 0; y = x \\ 2 y - x - 1 = 0; 2 y = x + 1 \end{cases}$ $\Rightarrow 2 x = x + 1, x = 1 \land y = 1$
Commented by Dwaipayan Shikari last updated on 23/Jun/20
$2^x 3^y (1/(3^x 4^y ))=(1/2) ⇒2^(x−2y+1) 3^(y−x) =1 ⇒2^(x−2y+1) =3^(x−y) only possible when x−2y+1=0 And x−y=0 so{x=1,y=1$
$2^{x} 3^{y} \frac{1}{3^{x} 4^{y}} = \frac{1}{2}$ $\Rightarrow 2^{x - 2 y + 1} 3^{y - x} = 1$ $\Rightarrow 2^{x - 2 y + 1} = 3^{x - y}$ $o n l y p o s s i b l e w h e n x - 2 y + 1 = 0$ $A n d x - y = 0$ $s o {x = 1, y = 1$
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