Question

Considere a matriz A .

Encontre o conjunto solução da equação
det (A) = 3.

Answer 1

Olá Lucas!

$\\ \begin{vmatrix} \mathsf{\log_2 125} & \mathsf{\log_{16} 36} \\\\ \mathsf{\log_{36} 81} & \mathsf{\log_{125} (x^2 + 5x)} \end{vmatrix} = \mathsf{3} \\\\\\ \mathsf{\log_2 125 \cdot \log_{125} (x^2 + 5x) - \log_{36} 81 \cdot \log_{16} 36 = 3} \\\\\\ \mathsf{\frac{\log 125}{\log 2} \cdot \frac{\log (x^2 + 5x)}{\log 125} - \frac{\log 81}{\log 36} \cdot \frac{\log 36}{\log 16} = 3} \\\\\\ \mathsf{\frac{\log (x^2 + 5x)}{\log 2} - \frac{\log 3^4}{\log 2^4} = 3} \\\\\\ \mathsf{\frac{\log (x^2 + 5x)}{\log 2} - \frac{4 \cdot \log 3}{4 \cdot \log 2} = 3}$

$\\ \mathsf{\log (x^2 + 5x) - \log 3 = 3 \cdot \log 2} \\\\ \mathsf{\log \left ( \frac{x^2 + 5x}{3} \right ) = \log 2^3} \\\\ \mathsf{\log \left ( \frac{x^2 + 5x}{3} \right ) = \log 8}$

 Mas, devemos considerar o logaritmando. Ele não pode assumir qualquer valor, de acordo com a definição.

$\\ \mathsf{x^2 + 5x > 0} \\ \mathsf{x(x + 5) > 0} \\ \boxed{\mathsf{S_1 = \left \{ x \in \mathbb{R} | x < - 5 \vee x > 0 \right \}}}$

 Segue,

$\\ \mathsf{\frac{x^2 + 5x}{3} = 8} \\\\ \mathsf{x^2 + 5x - 24 = 0} \\\\ \mathsf{(x + 8)(x - 3) = 0} \\\\ \boxed{\boxed{\mathsf{S = \left \{ - 8, 3 \right \}}}}$

 Note que: ambos pertencem ao intervalo...

Espero ter ajudado!