Example 15. Solve the equation

$$
27^{x}+12^{x}=2 \cdot 8^{x}
$$
Solution. This equation is similar in appearance to equation (6): the exponent of the bases is the same, but the bases 27, 12, and 8 of the three consecutive terms of a geometric progression do not form one.

Consecutive (but four) terms of a geometric progression are the numbers $27, 18, 12$, and 8. Therefore, we can assume that the term containing $18^{x}$ enters the given equation with a zero coefficient.

We divide all terms of equation (11) by $8^{x}$ and obtain

$$
(3 / 2)^{3 x}+(3 / 2)^{x}=2
$$

Let $t=(3 / 2)^{x}$; then we have the equation $t^{3}+t-2=0$. Since $t^{3}+t-2=(t-1)\left(t^{2}+t+2\right)$, this equation has a unique root $t_{1}=1$. Thus, the original equation is equivalent to the equation

$$
(3 / 2)^{x}=1
$$

the only root of which is $x=0$.

An equation of the form

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

where $\alpha, \beta, c$ are real numbers, and the bases $a$ and $b$ are positive numbers that are reciprocals of each other (i.e., $a b=1$), can be solved as follows.

Introduce the variable $t=a^{f(x)}$ and, using the equality $a b=1$, transition from equation (12) to the equation

$$
\alpha t^{2}+c t+\beta=0
$$

Then equation (12) will be equivalent to the combination of two exponential equations:

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

where $t_{1}, t_{2}$ are the roots of equation (12'). If equation (12') has no solutions, then equation (12) also has no solutions.