Example 9. Solve the equation:

$$
\lg 2 x=2 \lg (4 x-15)
$$
Solution. The equation is equivalent to the system

$$
\left\{\begin{array} { l } 
{ 4 x - 1 5 > 0 } \\
{ \operatorname { l g } 2 x = \operatorname { l g } ( 4 x - 1 5 ) ^ { 2 } }
\end{array} \Leftrightarrow \left\{\begin{array}{l}
4 x-15>0 \\
2 x=(4 x-15)^{2}
\end{array}\right.\right.
$$

Consider the equation of the last system:

$$
\begin{aligned}
2 x=16 x^{2}-120 x+225 \Leftrightarrow & 16 x^{2}-122 x+225=0 \Leftrightarrow \\
& \Leftrightarrow 16(x-9 / 2)(x-25 / 8)=0 \Leftrightarrow\left[\begin{array}{l}
x=9 / 2 \\
x=25 / 8
\end{array}\right.
\end{aligned}
$$

Taking into account the inequality of the system, we get: the only solution to the original equation is the number $9 / 2$.

An equation of the form

$$
(2 n+1) \log _{a} \alpha(x)=\log _{a} \beta(x), \quad a>0, \quad a \neq 1, \quad n \in \mathbf{N}
$$

is equivalent to the equation

$$
\log _{a} \alpha^{2 n+1}(x)=\log _{a} \beta(x)
$$

which in turn is equivalent to the system

$$
\left\{\begin{array}{l}
\beta(x)>0 \\
\alpha^{2 n+1}(x)=\beta(x)
\end{array}\right.
$$