Question

Resolva em C, o sistema

$\Large\begin{cases}2\cdot log_{ \tfrac{1}{2} }(2y+x)=1,5\\log_{0,25}(5x-y)=-1\end{cases}$

Answer 1

Manipulando a primeira equação:

$2\cdot log_{(\frac{1}{2})}(2y+x)=1,5\\\\2\cdot log_{(2^{-1})}(x+2y)=\frac{3}{2}\\\\\frac{2}{-1}\cdot log_{2}(x+2y)=\frac{3}{2}\\\\-2\cdot log_{2}(x+2y)=\frac{3}{2}\\\\log_{2}(x+2y)=-\frac{3}{4}\\\\2^{-\frac{3}{4}}=x+2y\\\\x+2y=(\frac{1}{2})^{\frac{3}{4}}\\\\x+2y=\sqrt[4]{\frac{1}{2^{3}}}\\\\x+2y=\sqrt[4]{\frac{1}{8}}\\\\x+2y=\frac{1}{\sqrt[4]{8}}$

Racionalizando (multiplicar o numerador e denominador por raiz quarta de 8³):

$x+2y=\frac{1\cdot\sqrt[4]{8^{3}}}{\sqrt[4]{8}\cdot\sqrt[4]{8^{3}}}\\\\x+2y=\frac{\sqrt[4]{512}}{8}\\\\x+2y=\frac{\sqrt[4]{2^{8}\cdot2}}{8}\\\\x+2y=\frac{2^{2}\sqrt[4]{2}}{8}\\\\\boxed{\boxed{x+2y=\frac{\sqrt[4]{2}}{2}}}$

Manipulando a segunda equação:

$log_{(0,25)}(5x-y)=-1\\\\log_{(\frac{1}{4})}(5x-y)=-1\\\\log_{(2^{-2})}(5x-y)=-1\\\\\frac{1}{-2}\cdot log_{2}(5x-y)=-1\\\\log_{2}(5x-y)=-(-2)\\\\log_{2}(5x-y)=2\\\\2^{2}=5x-y\\\\\boxed{\boxed{5x-y=4}}$
_________________

$\begin{cases}x+2y=\dfrac{\sqrt[4]{2}}{2}\\\\5x-y=4\end{cases}$

Multiplicando a segunda equação por 2:

$\begin{cases}x+2y=\dfrac{\sqrt[4]{2}}{2}\\\\10x-2y=8\end{cases}$

Somando membro à membro:

$x+10x=\frac{\sqrt[4]{2}}{2}+8\\\\11x=\frac{\sqrt[4]{2}+2\cdot8}{2}\\\\\boxed{\boxed{x=\frac{\sqrt[4]{2}+16}{22}}}$



$5x-y=4\\\\y=5x-4\\\\y=\frac{5(\sqrt[4]{2}+16)}{22}-4\\\\y=\frac{5\sqrt[4]{2}+80}{22}-4\\\\y=\frac{5\sqrt[4]{2}+80-22\cdot4}{22}\\\\y=\frac{5\sqrt[4]{2}+80-88}{22}\\\\\boxed{\boxed{y=\frac{5\sqrt[4]{2}-8}{22}}}$

Answer 2

Explicação passo-a-passo:

• Primeira equação

$\sf 2\cdot log_{\frac{1}{2}}~(2y+x)=1,5$

$\sf 2\cdot log_{2^{-1}}~(x+2y)=1,5$

Lembre-se que $\sf log_{b^{\beta}}~a=\dfrac{1}{\beta}\cdot log_{b}~a$

Assim:

$\sf \dfrac{2}{-1}\cdot log_{2}~(x+2y)=1,5$

$\sf -2\cdot log_{2}~(x+2y)=1,5$

$\sf log_{2}~(x+2y)=\dfrac{1,5}{-2}$

$\sf log_{2}~(x+2y)=-0,75$

$\sf log_{2}~(x+2y)=\dfrac{-3}{4}$

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

$\sf x+2y=\sqrt[4]{2^{-3}}$

$\sf x+2y=\sqrt[4]{\dfrac{1}{8}}$

$\sf x+2y=\dfrac{1}{\sqrt[4]{8}}$

Racionalizando:

$\sf x+2y=\dfrac{1}{\sqrt[4]{8}}\cdot\dfrac{\sqrt[4]{8^3}}{\sqrt[4]{8^3}}$

$\sf x+2y=\dfrac{\sqrt[4]{512}}{8}$

$\sf x+2y=\dfrac{\sqrt[4]{(2^2)^4\cdot2}}{8}$

$\sf x+2y=\dfrac{4\sqrt[4]{2}}{8}$

$\sf x+2y=\dfrac{\sqrt[4]{2}}{2}$

• Segunda equação

$\sf log_{0,25}~(5x-y)=-1$

$\sf log_{2^{-2}}~(5x-y)=-1$

$\sf \dfrac{1}{-2}\cdot log_{2}~(5x-y)=-1$

$\sf log_{2}~(5x-y)=(-1)\cdot(-2)$

$\sf log_{2}~(5x-y)=2$

$\sf 5x-y=2^2$

$\sf 5x-y=4$

Temos:

$\sf \begin{cases} \sf x+2y=\dfrac{\sqrt[4]{2}}{2} \\ \\ \sf 5x-y=4 \end{cases}$

Da segunda equação:

$\sf 5x-y=4~\Rightarrow~y=5x-4$

Substituindo na primeira equação:

$\sf x+2\cdot(5x-4)=\dfrac{\sqrt[4]{2}}{2}$

$\sf x+10x-8=\dfrac{\sqrt[4]{2}}{2}$

$\sf 11x=8+\dfrac{\sqrt[4]{2}}{2}$

$\sf 11x=\dfrac{16+\sqrt[4]{2}}{2}$

$\sf x=\dfrac{\frac{16+\sqrt[4]{2}}{2}}{11}$

$\sf x=\dfrac{16+\sqrt[4]{2}}{2}\cdot\dfrac{1}{11}$

$\sf \red{x=\dfrac{16+\sqrt[4]{2}}{22}}$

Assim:

$\sf y=5x-4$

$\sf y=5\cdot\left(\dfrac{16+\sqrt[4]{2}}{22}\right)-4$

$\sf y=\dfrac{80+5\sqrt[4]{2}}{22}-4$

$\sf y=\dfrac{80-88+5\sqrt[4]{2}}{22}$

$\sf \red{y=\dfrac{-8+5\sqrt[4]{2}}{22}}$