Объяснение:

[tex]log^2x-9 > 0.[/tex]

ОДЗ: х>0       ⇒     x∈(0;+∞).

[tex]log_4^2x-3^2 > 0\\(log_4x+3)*(log_4x-3) > 0\\[/tex]

[tex]\left \{ {{log_4x+3 > 0} \atop {log_4x-3 > 0}} \right.\ \ \ \ \ \left \{ {log_4x > -3} \atop {log_4x > 3}} \right. \ \ \ \ \left \{ {{x > 4^{-3}} \atop {x > 4^3}} \right.\ \ \ \ \left \{ {{x > \frac{1}{64} } \atop {x > 64}} \right. \ \ \ \ \Rightarrow\ \ \ \ \ x\in(64;+\infty).\\[/tex]

[tex]\left \{ {{log_4x+3 < 0} \atop {log_4x-3 < 0}} \right. \ \ \ \ \ \left \{ {{log_4x < -3} \atop {log_4x < 3}} \right. \ \ \ \ \ \left \{ {x < 4^{-3}} \atop {x < 4^3}} \right. \ \ \ \ \left \{ {{x < \frac{1}{64} } \atop {x < 64}} \right. \ \ \ \ \ \Rightarrow\ \ \ \ \ x\in(-\infty;\frac{1}{64} ).[/tex]

Учитывая ОДЗ:

Ответ: х∈(0;1/64)U(64;+∞).