8. (10 points) Find the sum of the real roots of the equation

$$
2 \cdot 3^{3 x}-a \cdot 3^{2 x}-3(a+4) \cdot 3^{x}+18=0
$$
Solution: Let's make the substitution $t=3^{x}$, and since $x \in \mathbb{R}$, then $t>0$. We obtain the following equation

$$
2 t^{3}-a t^{2}-3(a+4) t+18=0
$$

It is obvious that the number $t_{1}=-3$ is a root of the equation and the corresponding $x_{1}$ is not real. We get

$$
(t+3)\left(2 t^{2}-(a+6) t+6\right)=0
$$

The parameter $a$ was chosen such that the remaining two roots $t_{2}$ and $t_{3}$ are real and strictly greater than zero. Then, using Vieta's theorem, we get

$$
3^{x_{2}+x_{3}}=t_{2} \cdot t_{3}=\frac{6}{2}=3
$$

Therefore, $x_{2}+x_{3}=1$.

(10 points) Find the sum of the real roots of the equation

$$
2 \cdot 4^{3 x}-a \cdot 4^{2 x}-4(a+6) \cdot 4^{x}+32=0
$$

Solution: Let's make the substitution $t=4^{x}$, and since $x \in \mathbb{R}$, then $t>0$. We obtain the following equation

$$
2 t^{3}-a t^{2}-4(a+6) t+32=0
$$

It is obvious that the number $t_{1}=-4$ is a root of the equation and the corresponding $x_{1}$ is not real. We get

$$
(t+4)\left(2 t^{2}-(a+8) t+8\right)=0
$$

The parameter $a$ was chosen such that the remaining two roots $t_{2}$ and $t_{3}$ are real and strictly greater than zero. Then, using Vieta's theorem, we get

$$
4^{x_{2}+x_{3}}=t_{2} \cdot t_{3}=\frac{8}{2}=4
$$

Therefore, $x_{2}+x_{3}=1$.

Answers:

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