Question

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Resolver o seguinte sistema:

4^x . (1/8)^y= 16
(1/27)^x . 243^y= 1/243

Answer 1

Resolver o seguinte sistema:

4^x . (1/8)^y= 16
(1/27)^x . 243^y= 1/243



(2^2)^x . (2^-3)^y= 2^4
(3^-3)^x . (3^5)^y= 3^-5

2^2x . 2^3y = 2^2 ==> 2^2x - 3y = 2^4
(3^-3)^x .(3^5)^y= 3^5 ==> 3^-3x + 5y = 3^-5
 

  2x - 3y = 4(3)
-3x + 5y = -5(2)
 
 6x - 9y = 12
-6x +10y = -10
          y = 2


2x = 4+ 3y==> 2x = 4+3.(2)==> 2x =4+6==>6x= 10==> x =5

Answer 2

$\left \{ {{ 4^{x}* ( \frac{1}{8}) ^{y} =16} \atop { (\frac{1}{27})^{x} *243^{y} = \frac{1}{243} }} \right.$

$( 2^{2}) ^{x} * ( 2^{-3}) ^{y} = 2^{4} \\ 2^{2x} * 2^{-3y} = 2^{4} \\ 2x+(-3y)=4 \\ 2x-3y=4$

$(3^{-3})^{x}* (3^{5} )^{y}= 3^{-5} \\ 3^{-3x}* 3^{5y}= 3^{-5} \\ -3x+5y=-5$

$\left \{ {{2x-3y=4} \atop {-3x+5y=-5}} \right.$

$2x-3y=4 \\ 2x=4+3y \\ x=\frac{4+3y}{2}$

$-3x+5y=-5 \\ -3(\frac{4+3y}{2})+5y=-5 \\ -(\frac{12+9y}{2})+5y=-5 \\ -(12+9y)+10y=-10 \\ -12-9y+10y=-10 \\ y=2$

$x= \frac{4+3(2)}{2} \\ x= \frac{4+6}{2} \\ x= \frac{10}{2} \\ x=5$