$9.237 \log _{x} \log _{2}\left(4^{x}-12\right) \leq 1$.

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$9.237 \log _{x} \log _{2}\left(4^{x}-12\right) \leq 1$.
Solution. $\log _{x} \log _{2}\left(4^{x}-12\right) \leq \log _{x} x$.

1) $\left\{\begin{array}{l}x>1, \\ \log _{2}\left(4^{x}-12\right) \leq x, \\ \log _{2}\left(4^{x}-12\right)>0\end{array} \Leftrightarrow\left\{\begin{array}{l}x>1, \\ 4^{x}-12 \leq 2^{x} \\ 4^{x}-12>1\end{array}, \Leftrightarrow\left\{\begin{array}{l}x>1, \\ \left(2^{x}+3\right)\left(2^{x}-4\right) \leq 0, \\ 4^{x}>13\end{array} \Leftrightarrow\right.\right.\right.$ $\Leftrightarrow\left\{\begin{array}{l}x>1, \\ 2^{x}-4 \leq 0 \Leftrightarrow \log _{4} 1313 .\end{array}\right.$
2) $\left\{\begin{array}{l}00\end{array}\right.$

Answer: $x \in\left(\log _{4} 13 ; 2\right]$.

![](https://cdn.mathpix.com/cropped/2024_05_21_f024bf2ff7725246f3bfg-198.jpg?height=89&width=436&top_left_y=1394&top_left_x=99)

## Solution.

![](https://cdn.mathpix.com/cropped/2024_05_21_f024bf2ff7725246f3bfg-198.jpg?height=168&width=874&top_left_y=1548&top_left_x=100)
$\Leftrightarrow \frac{1}{\sqrt{3}}\frac{x}{2}>\frac{\pi}{6}+\pi n, \\ \frac{5 \pi}{6}+\pi n>\frac{x}{2}>\frac{3 \pi}{4}+\pi n .\end{array}\right.\right.$

Answer: $\quad x \in\left(\frac{\pi}{3}+2 \pi n ; \frac{\pi}{2}+2 \pi n\right) \cup\left(\frac{3 \pi}{2}+2 \pi n ; \frac{5 \pi}{3}+2 \pi n\right)$.

$9.2403^{\frac{2 \cos ^{2} x-6}{2 \cos ^{2} x-1}}>3^{\frac{\cos x}{1-2 \cos ^{2} x}}$.

Solution. $\frac{2 \cos ^{2} x-6}{2 \cos ^{2} x-1}>\frac{\cos x}{1-2 \cos ^{2} x} \Leftrightarrow \frac{2 \cos ^{2} x+\cos x-6}{2 \cos ^{2} x-1}>0 \Leftrightarrow$

$\frac{2(\cos x+2)\left(\cos x-\frac{3}{2}\right)}{2 \cos ^{2} x-1}>0 \Leftrightarrow \frac{1}{2 \cos ^{2} x-1}\frac{3}{2} \Leftrightarrow-\frac{\pi}{3}+2 \pi n1, \\ \log _{3}\left(9^{x}-6\right) \geq x\end{array} \Leftrightarrow\left\{\begin{array}{l}x>1, \\ 9^{x}-6 \geq 3^{x}\end{array} \Leftrightarrow\left\{\begin{array}{l}x>1 \\ \left(3^{x}-3\right)\left(3^{x}+2\right) \geq 0\end{array} \Leftrightarrow x>1\right.\right.\right.$.
2) $\left\{\begin{array}{l}00\end{array} \Leftrightarrow\left\{\begin{array}{l}01\end{array} \Leftrightarrow\left\{\begin{array}{l}0\log _{9} 7\end{array} \Leftrightarrow \log _{9} 7<x<1\right.\right.\right.$.

Answer: $x \in\left(\log _{9} 7 ; 1\right) \cup(1 ;+\infty)$.