Example 12. Solve the equation:

a) $\log _{3}(x-2)+\log _{3} x=\log _{3} 8$;

b) $\lg (x-9)+2 \lg \sqrt{2 x-1}=2$;
B) $\frac{1}{2} \log _{5}(x+5)+\log _{5} \sqrt{x-3}=\frac{1}{2} \log _{5}(2 x+1)$.
Solution. a) The equation is equivalent to the system

$$
\begin{aligned}
& \left\{\begin{array} { l } 
{ x - 2 > 0 , } \\
{ x > 0 , } \\
{ \operatorname { log } _ { 3 } ( x ( x - 2 ) ) = \operatorname { log } _ { 3 } 8 }
\end{array} \Leftrightarrow \left\{\begin{array}{l}
x>2, \\
x>0, \\
x(x-2)=8
\end{array} \Leftrightarrow\right.\right. \\
& \Leftrightarrow\left\{\begin{array}{l}
x>2, \\
(x-4)(x+2)=0
\end{array} \Leftrightarrow x=4\right.
\end{aligned}
$$

Thus, the only root of the original equation is the number 4.
b) The equation is equivalent to the system

$$
\begin{aligned}
\left\{\begin{array}{l}
x-9>0 \\
2 x-1>0, \\
\lg (x-9)(2 x-1)=\lg 100
\end{array}\right. & \Leftrightarrow\left\{\begin{array}{l}
x>9 \\
x>1 / 2, \\
(x-9)(2 x-1)=100
\end{array}\right. \\
& \Leftrightarrow\left\{\begin{array}{l}
x>9 \\
2(x-13)(x+7 / 2)=0
\end{array} \Leftrightarrow x=13\right.
\end{aligned}
$$

Therefore, the only root of the original equation is the number 13.

c) The equation is equivalent to the equation

$$
\log _{5}(x+5)+2 \log _{5} \sqrt{x-3}=\log _{5}(2 x+1)
$$

which is equivalent to the system

$$
\begin{aligned}
& \left\{\begin{array} { l } 
{ x - 3 > 0 } \\
{ x + 5 > 0 , } \\
{ \operatorname { log } ( ( x + 5 ) ( x - 3 ) ) = \operatorname { log } _ { 5 } ( 2 x + 1 ) }
\end{array} \Leftrightarrow \left\{\begin{array}{l}
x>3 \\
x>-5 \\
(x+5)(x-3)=2 x+1
\end{array} \Leftrightarrow\right.\right. \\
& \Leftrightarrow\left\{\begin{array} { l } 
{ x > 3 , } \\
{ x ^ { 2 } - 1 6 = 0 }
\end{array} \Leftrightarrow \left\{\begin{array}{l}
x>3 \\
(x-4)(x+4)=0
\end{array} \Leftrightarrow x=4\right.\right.
\end{aligned}
$$

Thus, the only root of the original equation is the number 4.

An equation of the form

$$
\log _{a} \alpha(x)-\log _{a} \beta(x)=\log _{a} f(x)-\log _{a} g(x), \quad a>0, \quad a \neq 1
$$

is equivalent to the equation

$$
\log _{a} \alpha(x)+\log _{a} g(x)=\log _{a} f(x)+\log _{a} \beta(x)
$$

which is equivalent to the system

$$
\left\{\begin{array}{l}
\alpha(x)>0 \\
g(x)>0 \\
f(x)>0 \\
\beta(x)>0 \\
\log _{a}(\alpha(x) g(x))=\log _{a}(f(x) \beta(x))
\end{array}\right.
$$

i.e., the system

$$
\left\{\begin{array}{l}
\alpha(x)>0 \\
g(x)>0 \\
\beta(x)>0 \\
f(x)>0 \\
\alpha(x) g(x)=f(x) \beta(x)
\end{array}\right.
$$