Example 12. Solve the equation

$$
8^{2 / x}-2^{(3 x+3) / x}+12=0
$$
The solution is as follows. Since

$$
2^{(3 x+3) / x}=2^{3+3 / x}=8 \cdot 2^{3 / x}=8 \cdot\left(2^{3}\right)^{1 / x}=8 \cdot 8^{1 / x},
$$

the equation (5) is equivalent to the equation

$$
8^{2 / x}-8 \cdot 8^{1 / x}+12=0
$$

Let $8^{1 / x}=t$; then the equation (5) is equivalent to the system of equations

$$
8^{1 / x}=6, \quad 8^{1 / x}=2,
$$

where the numbers 6 and 2 are the roots of the quadratic equation

$$
t^{2}-8 t+12=0
$$

From this system, we find two roots of the equation (5): $x_{1}=3 \log _{6} 2, x_{2}=3$.

Solving exponential equations in which there are three powers with different bases that are consecutive terms of a geometric progression, and these bases are raised to the same power depending on $x$, reduces to solving quadratic equations. Such equations have the form

$$
\alpha a^{f(x)}+\beta b^{f(x)}+\gamma c^{f(x)}=0
$$

where $\alpha \neq 0, \beta, \gamma$ are real numbers, $f(x)$ is some function, and the bases $a, b$ and $c$ satisfy the condition $b^{2}=a c$.

Equations of this type are reduced to solving the system of exponential equations

$$
(a / b)^{f(x)}=t_{1}, \quad(a / b)^{f(x)}=t_{2}
$$

where $t_{1}, t_{2}$ are the roots of the quadratic equation $\alpha t^{2}+\beta t+\gamma=0$.