Question

Resolvendo o sistema
║ ㏒₂ x + ㏒₄ y + ㏒₄ z = 1
S= ║ ㏒₃ y + ㏒₉ z + ㏒₉ x = 3
║ ㏒₄ z + ㏒₁₆ x + ㏒₁₆ y = 5
obtém-se que x.y.z vale:
(Usar 55.296=2¹¹.3³ e √6=2,45).

Answer 1

Olá!

$\mathrm{Dado \ o \ sistema \ abaixo:} \\\\ \begin{cases} \mathrm{\log_2 x + \log_4 y + \log_4 z = 1 \qquad \qquad (I)} \\ \mathrm{\log_3 y + \log_9 z + \log_9 x = 3 \qquad \qquad (II)} \\ \mathrm{\log_4 z + \log_{16} x + \log_{16} y = 5 \qquad \quad (III)} \end{cases} \\\\\\ \mathrm{Determinemos \ (x \cdot y \cdot z).}$

 Em (I), temos que:

$\\ \displaystyle \mathrm{\log_2 x + \log_4 y + \log_4 z = 1} \\\\ \mathrm{\log_2 x + \log_{2^2} y + \log_{2^2} z = 1} \\\\ \mathrm{\log_2 x + \frac{1}{2} \cdot \log_2 y + \frac{1}{2} \cdot \log_2 z = 1} \\\\ \mathrm{\log_2 x + \log_2 y^{\frac{1}{2}} + \log_2 z^{\frac{1}{2}} = 1} \\\\ \mathrm{\log_2 x + \log_2 \sqrt{y} + \log_2 \sqrt{z} = 1} \\\\ \mathrm{\log_2 \left ( x \cdot \sqrt{y} \cdot \sqrt{z} \right ) = 1} \\\\ \mathrm{2^1 = x \cdot \sqrt{yz}} \\\\ \mathbf{x\sqrt{yz} = 2}$
 

 De (II),

$\\ \displaystyle \mathrm{\log_3 y + \log_9 z + \log_9 x = 3} \\\\ \mathrm{\log_3 y + \log_{3^2} z + \log_{3^2} x = 3} \\\\ \mathrm{\log_3 y + \frac{1}{2} \cdot \log_3 z + \frac{1}{2} \cdot \log_3 x = 3} \\\\ \mathrm{\log_3 y + \log_3 z^{\frac{1}{2}} + \log_3 x^{\frac{1}{2}} = 3} \\\\ \mathrm{\log_3 y + \log_3 \sqrt{z} + \log_3 \sqrt{x} = 3} \\\\ \mathrm{\log_3 \left ( y \cdot \sqrt{z} \cdot \sqrt{x} \right ) = 3} \\\\ \mathrm{3^3 = y \cdot \sqrt{zx}} \\\\ \mathbf{y\sqrt{xz} = 3^3}$

 Quanto à (III),

$\\ \displaystyle \mathrm{\log_4 z + \log_{16} x + \log_{16} y = 5} \\\\ \mathrm{\log_4 z + \log_{4^2} x + \log_{4^2} y = 5} \\\\ \mathrm{\log_4 z + \frac{1}{2} \cdot \log_4 x + \frac{1}{2} \cdot \log_4 y = 5} \\\\ \mathrm{\log_4 z + \log_4 x^{\frac{1}{2}} + \log_4 y^{\frac{1}{2}} = 5} \\\\ \mathrm{\log_4 z + \log_4 \sqrt{x} + \log_4 \sqrt{y} = 5} \\\\ \mathrm{\log_4 \left ( z \cdot \sqrt{x} \cdot \sqrt{y} \right ) = 5} \\\\ \mathrm{4^5 = z \cdot \sqrt{xy}} \\\\ \mathbf{z\sqrt{xy} = 2^{10}}$

 Por conseguinte, efectuamos a seguinte multiplicação...

$\\ \displaystyle \mathrm{2 \cdot 3^3 \cdot 2^{10} = \left ( x\sqrt{yz} \right ) \cdot \left ( y\sqrt{xz} \right ) \cdot \left ( z\sqrt{xy} \right )} \\\\ \mathrm{2^{11} \cdot 3^3 = xyz \cdot \sqrt{x^2y^2z^2}} \\\\ \mathrm{2^{11} \cdot 3^3 = xyz \cdot xyz} \\\\ \mathrm{(xyz)^2 = 2^{11} \cdot 3^3} \\\\ \mathrm{xyz = \sqrt{\left (2^{10} \cdot 3^2 \right ) \cdot (2 \cdot 3)}} \\\\ \mathrm{xyz = 2^5 \cdot 3^1 \cdot \sqrt{6}} \\\\ \mathrm{xyz \approx 32 \cdot 3 \cdot 2,45} \\\\ \boxed{\boxed{\mathbf{xyz \approx 235,2}}}$