Solve the following equation:

$$
\frac{x}{20}=\left(\frac{5}{2}\right)^{\log _{x} 50}
$$
I. solution. In the exponent, we rewrite $\log _{x} 50$ in terms of base 10 logarithms:

$$
\log _{x} 50=\frac{\lg 50}{\lg x}
$$

We substitute this into the equation and take the base 10 logarithm of both sides:

$$
\lg x-\lg 20=\frac{\lg 50}{\lg x}(\lg 5-\lg 2)
$$

We bring to a common denominator. Using the fact that $\lg 20=1+\lg 2, \lg 50=1+\lg 5$, we obtain the following quadratic equation in $\lg x$:

$$
\lg ^{2} x-(1+\lg 2) \lg x-\lg 5(1+\lg 5)+\lg 2(1+\lg 5)
$$

From this,

$$
\lg x=\frac{\lg 2+1 \pm \sqrt{(1+\lg 2)^{2}+4 \lg 5(1+\lg 5)-4 \lg 2(1+\lg 5)}}{2}
$$

Performing the operations under the square root, we get $(2 \lg 5-\lg 2+1)^{2}$ under the square root. Thus,

$$
\lg x_{1}=\frac{\lg 2+1+2 \lg 5-\lg 2+1}{2}=1+\lg 5=\lg 10+\lg 5=\lg 50, \quad x_{1}=50
$$

and

$$
\lg x_{2}=\frac{\lg 2+1-2 \lg 5+\lg 2-1}{2}=\lg 2-\lg 5=\lg \frac{2}{5}, \quad x_{2}=\frac{2}{5}
$$

Both numbers are solutions to the equation.

II. solution. $x_{1}=50$ is clearly a solution, so in the following, we look for solutions $0<x \neq 1 ; 50$. Using the properties of logarithms:

$$
\begin{aligned}
\frac{x}{20} & =\frac{x}{50} \cdot \frac{5}{2}=\left(\frac{5}{2}\right)^{\log _{x} 50} \\
\frac{x}{50}=\left(\frac{5}{2}\right)^{\log _{x} 50-1} & =\left(\frac{5}{2}\right)^{\log _{x} 50-\log _{x} x}=\left(\frac{5}{2}\right)^{\log _{x} \frac{50}{x}}=\left(\frac{5}{2}\right)^{\frac{1}{\log _{\frac{50}{x} x}^{x}}} \\
\left(\frac{x}{50}\right)^{\log _{\frac{50}{x} x}^{x}} & =\frac{5}{2}, \quad \frac{1}{x}=\frac{5}{2}, \quad x=\frac{2}{5}
\end{aligned}
$$

The solutions to the problem are: $x_{1}=50$ and $x_{2}=\frac{2}{5}$.