Example 11. Solve the equation

$$
\lg x=\frac{1}{2} \lg (x+1)
$$
Solution. The equation is equivalent to the equation

$$
2 \lg x=\lg (x+1)
$$

which is equivalent to the system

$$
\begin{aligned}
& \left\{\begin{array} { l } 
{ x > 0 , } \\
{ \operatorname { l g } x ^ { 2 } = \operatorname { l g } ( x + 1 ) }
\end{array} \Leftrightarrow \left\{\begin{array} { l } 
{ x > 0 , } \\
{ x ^ { 2 } = x + 1 }
\end{array} \Leftrightarrow \left\{\begin{array}{l}
x>0, \\
x^{2}-x-1=0
\end{array} \Leftrightarrow\right.\right.\right. \\
& \quad \Leftrightarrow\left\{\begin{array}{l}
x>0, \\
\left(x-\frac{1+\sqrt{5}}{2}\right)\left(x+\frac{\sqrt{5}-1}{2}\right)=0
\end{array} \Leftrightarrow x=\frac{\sqrt{\overline{5}+1}}{2}\right.
\end{aligned}
$$

Thus, the only root of the original equation is the number $\frac{\sqrt{5}+1}{2}$.

## An equation of the form

$$
\log _{a} f(x)+\log _{a} g(x)=\log _{a} m(x), \quad a>0, \quad a \neq 1
$$

is equivalent to the system

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

which in turn is equivalent to the system

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