Solve the following equation in the set of real numbers: $\log _{a} x=x$, where $a=x^{\log _{4} x}$
Using the definition of logarithm, $x>0$ and $a>0$. Since $\log _{a} 1=0$, it follows that $x \neq 1$, and also $a \neq 1$ is true. (1) can be written, based on the definition of logarithm, as: $a^{x}=x$. From this, considering the condition,

\[
\left(x^{\log _{4} x}\right)^{x}=x
\]

which means

\[
x^{x \cdot \log _{4} x}=x=x^{1}
\]

follows. Here, the base of the power is a positive number different from 1. Due to the monotonicity of exponentiation, the equality of the two powers implies the equality of the exponents. Therefore, $x \cdot \log _{4} x=1$. This equality can be written as $\log _{4} x^{x}=1$, which, according to the definition of logarithm, means that $x^{x}=4$.

Since the power of a positive number less than 1 is also less than 1, if $0<x<1$, then $x^{x}<1$, so $x^{x}=4$ is impossible. For $x$ greater than 1, increasing $x$ increases both the base and the exponent of the function $x^{x}$, which means the function itself increases. This immediately implies that the equation $x^{x}=4$ can have at most one solution. Since $2^{2}=4$, $x=2$ is a solution.

Since we did not check at each step of the solution whether the step was reversible, we now need to check whether $x=2$ indeed satisfies the original equation:

If $x=2$, then $\log _{4} 2=\frac{1}{2}$, so $a=2^{1 / 2}=\sqrt{2}$. We need to check whether $\log _{a} x=x$, or equivalently, $a^{x}=x$ holds. Indeed, $a^{x}=(\sqrt{2})^{2}=2=x$, so we have indeed found a solution.