Question

Sejam x e y números reais não nulos tais que:

$log_{x}( {y}^{\pi} ) + log_{y}( {x}^{e} ) = a$

$\frac{1}{ log_{y}( {x}^{ {\pi}^{ - 1} } ) } - \frac{1}{ log_{x}( {y}^{ {e}^{ - 1} }) } = b$

O valor de $\frac{ {x}^{a + b + 2e} }{ {y}^{a - b + 2\pi} }$ é:

a)$1$

b)$\sqrt{ \frac{\pi}{e} }$

c)$\sqrt{ \frac{a e}{b\pi} }$

d)$a - b$

e)$\frac{ {(a + b)}^{ \frac{e}{\pi} }}{\pi}$

Answer 1

$\begin{cases}\text{log}_{x}y^{\pi}+\text{log}_{y}x^{e}=a\\\dfrac{1}{\text{log}_{y}x^{\pi^{-1}}}-\dfrac{1}{\text{log}_{x}y^{e^{-1}}}\end{cases}$

Vamos simplificar essas equações. Aplicando a propriedade do logaritmo da potência na primeira equação, obtemos:

$\bullet~~\tetx{log}_{x}y^{\pi}=\pi\cdot\text{log}_{x}y$

$\bullet~~\text{log}_{y}x^{e}=e\cdot\text{log}_{y}x$

$\text{log}_{x}y^{\pi}+\text{log}_{y}x^{e}=a~\longrightarrow~=\pi\cdot\text{log}_{x}y+e\cdot\text{log}_{y}x=a$

Agora a segunda equação. Aplicando essa mesma propriedade:

$\bullet~~\tetx{log}_{y}x^{\pi^{-1}}=\pi^{-1}\cdot\text{log}_{y}x\\\bullet~~\text{log}_{x}y^{e^{-1}}=e^{-1}\cdot\text{log}_{x}y\\\\\dfrac{1}{\text{log}_{y}x^{\pi^{-1}}}-\dfrac{1}{\text{log}_{x}y^{e^{-1}}}=b~\longrightarrow~\dfrac{1}{\pi^{-1}\cdot\text{log}_{y}x}-\dfrac{1}{e^{-1}\cdot\text{log}_{x}y}=b$

Mas $\pi^{-1}=\dfrac{1}{\pi}$ e $e^{-1}=\dfrac{1}{e}$. Substituindo:

$\dfrac{1}{\pi^{-1}\cdot\text{log}_{y}x}-\dfrac{1}{e^{-1}\cdot\text{log}_{x}y}=b~\longrightarrow~\dfrac{1}{\dfrac{1}{\pi}\cdot\text{log}_{y}x}-\dfrac{1}{\dfrac{1}{e}\cdot\text{log}_{x}y}=b$

$\dfrac{\pi}{\text{log}_{y}x}-\dfrac{e}{\text{log}_{x}y}=b$

Fazendo mudança de base, obtemos o seguinte:

$\star~~\text{log}_{y}x=\dfrac{\text{log}_{x}x}{\text{log}_{x}y}=\dfrac{1}{\text{log}_{x}y}~~~~~~~~~~~\star~~\text{log}_{x}y=\dfrac{\text{log}_{y}y}{\text{log}_{y}x}=\dfrac{1}{\text{log}_{y}x}$

$\dfrac{\pi}{\text{log}_{y}x}-\dfrac{e}{\text{log}_{x}y}=b~\longrightarrow~\dfrac{\pi}{\dfrac{1}{\text{log}_{x}y}}-\dfrac{e}{\dfrac{1}{\text{log}_{y}x}}=b$ 

$\pi\cdot\text{log}_{x}y-e\cdot\text{log}_{y}x=b$

$\begin{cases}\text{log}_{x}y^{\pi}+\text{log}_{y}x^{e}=a\\\dfrac{1}{\text{log}_{y}x^{\pi^{-1}}}-\dfrac{1}{\text{log}_{x}y^{e^{-1}}}\end{cases}~\longrightarrow~\begin{cases}\pi\cdot\text{log}_{x}y+e\cdot\text{log}_{y}x=a\\\pi\cdot\text{log}_{x}y-e\cdot\text{log}_{y}x=b\end{cases}$

Somando as equações membro a membro:

$a+b=\pi\cdot\text{log}_{x}y+\pi\cdot\text{log}_{x}y+e\cdot\text{log}_{y}x-e\cdot\text{log}_{y}x$

$a+b=2\pi\cdot\text{log}_{x}y$

Subtraindo a segunda equação da primeira:

$a-b=\pi\cdot\text{log}_{x}y-\pi\cdot\text{log}_{x}y+e\cdot\text{log}_{y}x-(-e\cdot\text{log}_{y}x)$

$a-b=e\cdot\text{log}_{y}x+e\cdot\text{log}_{y}x$

$a-b=2e\cdot\text{log}_{y}x$

Vamos determinar $\dfrac{x^{a+b+2e}}{y^{a-b+2\pi}}$:

Primeiro o numerador:

$x^{a+b+2e}=x^{2\pi\cdot\text{log}_{x}y+2e}=(x^{\text{log}_{x}y})^{2\pi}}\cdot x^{2e}=y^{2\pi}\cdot x^{2e}$

Lembre-se que $x^{\text{log}_{x}y}=y$ e $y^{\text{log}_{y}x}=x$

Agora o denominador:

$y^{a-b+2\pi}=y^{2e\cdot\text{log}_{y}x+2\pi}=(y^{\text{log}_{y}x})^{2e}}\cdot y^{2\pi}=x^{2e}\cdot y^{2\pi}$

Logo:

$\dfrac{x^{a+b+2e}}{y^{a-b+2\pi}}=\dfrac{y^{2\pi}\cdot x^{2e}}{x^{2e}\cdot y^{2\pi}}=1$

$\text{Alternativa A}$