Răspuns:
c) 1
Explicație pas cu pas:
x > 0
[tex]log_{3}(x) \cdot log_{9}(x) \cdot log_{27}(x) \cdot log_{81}(x) = \frac{2}{3} \\ log_{3}(x) \cdot log_{ {3}^{2} }(x) \cdot log_{ {3}^{3} }(x) \cdot log_{ {3}^{4} }(x) = \frac{2}{3} \\ log_{3}(x) \cdot \frac{1}{2} \cdot log_{3}(x) \cdot \frac{1}{3} \cdot log_{3}(x) \cdot \frac{1}{4} \cdot log_{3}(x) = \frac{2}{3} \\ {\Big[log_{3}(x)\Big]}^{4} = {2}^{4} \implies log_{3}(x) = \pm 2[/tex]
[tex]log_{3}(x) = - 2 \implies x_{1} = \frac{1}{9} \\ log_{3}(x) = 2 \implies x_{2} = 9[/tex]
=>
[tex]x_{1}x_{2} = 9 \cdot \frac{1}{9} = \bf 1 \\ [/tex]
