Example 17. Solve the equation

$$
(2+\sqrt{3})^{x^{2}-2 x+1}+(2-\sqrt{3})^{x^{2}-2 x-1}=\frac{101}{10(2-\sqrt{3})}
$$
Solution. Since $2-\sqrt{3}=\frac{1}{2+\sqrt{3}}$, multiplying both sides of the equation by $2-\sqrt{3}$, we transform it into the form (12):

$$
(2+\sqrt{3})^{x^{2}-2 x}+\frac{1}{(2+\sqrt{3})^{x^{2}-2 x}}=\frac{101}{10}
$$

Let $t=(2+\sqrt{3})^{x^{2}-2 x}$; then equation (15) takes the form

$$
t+1 / t=101 / 10
$$

The roots of this equation are $t_{1}=10, t_{2}=1 / 10$.

Thus, equation (14) is equivalent to the system of exponential equations

$$
(2+\sqrt{3})^{x^{2}-2 x}=10, \quad(2+\sqrt{3})^{x^{2}-2 x}=1 / 10
$$

The first equation of this system is equivalent to the equation

$$
x^{2}-2 x=\log _{2+\sqrt{3}} 10
$$

from which $\quad x_{1}=1+\sqrt{1+\log _{2+\sqrt{3}} 10}, \quad x_{2}=1-\sqrt{1+\log _{2+\sqrt{3}} 10}$. The second equation of the system (16) is equivalent to the equation

$$
x^{2}-2 x+\log _{2+\sqrt{3}} 10=0
$$

which has no roots, since its discriminant $1-\log _{2+\sqrt{3}} 10$ is less than zero.

Thus, the solutions to equation (15) are the numbers $x_{1}$ and $x_{2}$.

Solving some exponential equations reduces to solving algebraic homogeneous equations (see § 3 of Chapter 4).