Example 1. Solve the equation:

a) $1-\lg 5=\frac{1}{3}\left(\lg \frac{1}{2}+\lg x+\frac{1}{3} \lg 5\right)$;

b) $\log _{1 / 3}(-1 / x)=2 ; \quad$ c) $\lg (2 x-5)^{2}=0$.
Solution. a) Since $1-\lg 5=\lg 10-\lg 5=\lg 2, \lg \frac{1}{2}=$ $=-\lg 2$, the original equation is equivalent to the equation

$$
4 \lg 2-\frac{1}{3} \lg 5=\lg x
$$

i.e., the equation $\lg \frac{16}{\sqrt[3]{5}}=\lg x$. From this, we obtain $x=\frac{16}{\sqrt[3]{5}}$ - the only root of the given equation.

b) The original equation is equivalent to the equation

$$
(-1 / x)=(1 / 3)^{2}
$$

from which $x=-9$. The number (-9) is the only root of the given equation.

c) The original equation is equivalent to the equation

$$
\lg (2 x-5)^{2}=\lg 1
$$

i.e., the equation $(2 x-5)^{2}=1$, which has two roots: $x_{1}=3$, $x_{2}=2$. The numbers 2 and 3 form the set of all solutions to the given equation.

To the simplest logarithmic equations also belongs the equation of the form

$$
\log _{x} A=B, \quad A>0
$$

which

a) when $A \neq 1$ and $B \neq 0$ has a unique root $x=A^{1 / B}$;

b) when $A=1$ and $B=0$ has any positive number, different from one, as a solution;

c) when $A=1$ and $B \neq 0$ has no roots;

d) when $A \neq 1$ and $B=0$ has no roots.