Question

Alguem me ajuda
$4^{X} * 8^{Y} = \frac{1}{4} 9^{X}*27 ^{2Y} = 3$

Answer 1

AE manoo,

no sistema de equações exponenciais,

$\begin{cases}4^x\cdot8^y= \dfrac{1}{4}~~(i)\\ 9^x\cdot27^{2y}=3~~(ii) \end{cases}$

podemos usar as propriedades da exponenciação..

$a^m\cdot a^n\Rightarrow a^{m+n}\\\\ \dfrac{1}{a} \Rightarrow \dfrac{1}{a^1}\Rightarrow a^{-1}\\\\ (a^m)^n\Rightarrow a^{m\cdot n}$

_____________


$\Rightarrow\begin{cases}(2^2)^x\cdot(2^3)^y= \dfrac{1}{2^2}~~(i)\\ (3^2)^x\cdot(3^3)^{2y}=3^1~~(ii) \end{cases}\\\\\\ \Rightarrow\begin{cases}2^{2x}\cdot2^{3y}=2^{-2}~~(i)\\ 3^{2x}\cdot3^{6y}=3^1~~(ii)\end{cases}$

$\Rightarrow\begin{cases}2^{2x+3y}=2^{-2}~~(i)\\ 3^{2x+6y}=3^1~~(ii)\end{cases}\\\\\\ \Rightarrow\begin{cases}\not2^{2x+3y}=\not2^{-2}~~(i)\\ \not3^{2x+6y}=\not3^1~~(ii)\end{cases}\\\\\\ \begin{cases}2x+3y=-2~~(i)~~(\times~por~-1~e~soma)\\ 2x+6y=1~~(ii)\end{cases}$

$+\begin{cases}-2x-3y=2~~(i)\\ ~~2x+6y=1~~(ii)\end{cases}\\ ~~~~~~------\\ ~~~~~~~~~~~~~~3y=3\\ ~~~~~~~~~~~~~~y=3/3\\ ~~~~~~~~~~~~~~y=1\\\\\\ 2x+6y=1\\ 2x+6\cdot1=1\\ 2x+6=1\\ 2x=1-6\\ 2x=-5\\\\ x=- \dfrac{5}{2}$

Portanto a solução será..

$\Large\boxed{\text{S}=\left\{\left(- \dfrac{5}{2},~1\right)\right\}}$