Question

Resolva no campo dos reais, o sistema:

$\Large\begin{cases}log_2x+log_{0,5}(3+y)=1\\x-y^2=3\end{cases}$

Answer 1

$x- y^{2} =3$ ⇒ $x= y^{2} +3$

$\log_2 ( y^{2}+3 )+\log_ \frac{1}{2} (y+3)=1$

$\log_2 ( y^{2}+3 )+ \dfrac{\log_ 2 (y+3)}{\log_ 2 \frac{1}{2}} =1$

$\log_2 ( y^{2}+3 )+ \dfrac{\log_ 2 (y+3)}{\log_ 2 2^{-1} } =1$

$\log_2 ( y^{2}+3 )+ \dfrac{\log_ 2 (y+3)}{-1.1 } =1$

$\log_2 ( y^{2}+3 )-\log_ 2 (y+3) =1$

$\log_2 \frac{y^{2}+3}{y+3} =1$

$\dfrac{y^{2}+3}{y+3}= 2^{1} =2$

$y^{2}+3=2y+6$

$y^{2}-2y-3=0$

$(y+1).(y-3)=0$

$y+1=0$ ⇒ $y=-1$ ⇒ $x= (-1)^{2} +3=1+3=4$
ou
$y-3=0$ ⇒ $y=3$ ⇒ $x= 3^{2} +3=9+3=12$

S = {(x,y) ∈ R | (4, -1), (12, 3)}

Answer 2

Explicação passo-a-passo:

$log_{2}(x) + log_{0,5}(3 + y) = 1 \\ x + {y}^{2} = 3 \\ \\ log_{2}(x) + log_{ {2}^{ - 1} }(3 + y) = 1 \\ x + {y}^{2} = 3 \\ \\ log_{2}(x) - log_{2}(3 + y) = 1 \\ x + {y}^{2} = 3 \\ \\ log_{2}( \frac{x}{3 + y} ) = 1 \\ x + {y}^{2} = 3 \\ \\ \frac{x}{3 + y} = {2}^{1} \\ x + {y}^{2} = 3 \\ \\ \frac{x}{3 + y} = 2\\ x + {y}^{2} = 3$

( x , y ) = ( 2 , 3 )

• Espero ter ajudado.