Solve the following equation in the set of real numbers:

$$
\log _{2 x} 4 x+\log _{4 x} 16 x=4
$$
From the definition of logarithm, it follows that $x>0, 2 x \neq 1, 4 x \neq 1$ must hold. Using the known identities, we can rewrite our equation in terms of base-2 logarithms:

$$
\frac{2+\log _{2} x}{1+\log _{2} x}+\frac{4+\log _{2} x}{2+\log _{2} x}=4
$$

By introducing the new variable $\log _{2} x=y$, we arrive at the following quadratic equation:

$$
\begin{gathered}
(2+y)(2+y)+(4+y)(1+y)=4(1+y)(2+y) \\
2 y^{2}+3 y=y(2 y+3)=0
\end{gathered}
$$

From this,

$$
\begin{aligned}
y_{1}=\log _{2} x_{1}=0, & \text { i.e., } x_{1}=1 \\
y_{2}=\log _{2} x_{2}=-\frac{3}{2}, & \text { i.e., } x_{2}=2^{-\frac{3}{2}}=\frac{1}{(\sqrt{2})^{3}}
\end{aligned}
$$

By substitution, we can verify that both roots satisfy the equation.