Task B-4.7. Determine all solutions of the inequality

$$
\sqrt{72 \cdot 3^{x+1}+81} \geq\left|3^{x+1}-9\right|+3^{x} \log _{x}\left(x^{3}\right)
$$
Solution. Since the logarithm is defined only for positive real numbers and the base must be different from one, we have $x>0$ and $x \neq 1$.

(1 point)

Then the given inequality becomes

$$
\sqrt{72 \cdot 3^{x+1}+81} \geq\left|3^{x+1}-9\right|+3^{x} \cdot 3
$$

and we have two cases.

1) If $3^{x+1}-9 \geq 0$, i.e., $x>1(x \neq 1)$, then

$$
\sqrt{72 \cdot 3^{x+1}+81} \geq 3^{x+1}-9+3^{x+1}
$$

which, with the substitution $3^{x+1}=t$, gives

$$
\sqrt{72 \cdot t+81} \geq 2 t-9
$$

Since both sides are positive (as $t>9$), after squaring and simplifying the inequality, we get

$$
4 t^{2}-108 t \leq 0
$$

The solution to this inequality is the interval $[0,27]$.

It follows that $3^{x+1} \leq 3^{3}$, i.e., $x+1 \leq 3$.

Since $x>1$, it follows that $x \in\langle 1,2]$.

2) If $3^{x+1}-9<0$, i.e., $x<1$, then

$$
\begin{gathered}
\sqrt{72 \cdot 3^{x+1}+81} \geq-3^{x+1}+9+3^{x+1} \\
\sqrt{72 \cdot 3^{x+1}+81} \geq 9 \\
72 \cdot 3^{x+1}+81 \geq 81 \\
3^{x+1} \geq 0
\end{gathered}
$$

This inequality, along with all previous conditions, is satisfied for all $0<x<1$. (2 points) The final solution is the interval $\langle 0,2] \backslash\{1\}$.