Solve the following equation:

$$
a \cdot b^{x} \cdot c^{2 x}=\sqrt[3 x]{d} \cdot \sqrt[4 x]{e}
$$

Numerical examples: the values of $a, b, c, d, e$ are respectively: I. 2, 3, 5, 7, 11; $\quad$ II. 5, 3, 2, 1/7, $1 / 11$.
$\alpha$ ) If $b, c, d, e>0$, then the prescribed operations are defined for all real $x$ except $x=0$. Let $a>0$ as well. The logarithms of equal numbers with the same base are also equal, thus

$$
\lg a+x \lg b+2 x \lg c=\frac{\lg d}{3 x}+\frac{\lg e}{4 x}
$$

From this, by rearranging,

$$
\left(\lg \left(b c^{2}\right)\right) \cdot x^{2}+(\lg a) \cdot x-\left(\frac{1}{3} \lg d+\frac{1}{4} \lg e\right)=0
$$

This is indeed a quadratic equation, and its roots are real if

$$
\begin{gathered}
\lg \left(b c^{2}\right) \neq 0, \quad \text { i.e., } \quad b c^{2} \neq 1, \quad \text { and } \\
D=\lg ^{2} a+\left(\lg \left(b c^{2}\right)\right) \cdot\left(\frac{4}{3} \lg d+\lg e\right) \geq 0
\end{gathered}
$$

If $b c^{2}=1$, but $a \neq 1$, then (2) is a linear equation, and its root is real. Finally, if $b c^{2}=a=1$, then the left side of (1) is 1, and if $\sqrt[3]{d} \cdot \sqrt[4]{e}=1$ also holds, then (1) is satisfied by every real number $x \neq 0$; otherwise, there is no solution.

$\beta)$ In the case of the I. value system, (3) and (4) are satisfied, the roots are: $x_{1} \approx-0.624, x_{2} \approx 0.464$.

In the case of the II. system, (4) is not satisfied, there is no real solution.