Task B-3.1. Solve the equation

$$
\log _{x} 8+\log _{4 x} 4+\log _{2 x} 2=0
$$
## Solution.

For the equation to make sense, it must be $x>0, x \neq 1, x \neq \frac{1}{4}, x \neq \frac{1}{2}$.

We write the initial equation in the form

$$
\frac{3}{\log _{2} x}+\frac{2}{2+\log _{2} x}+\frac{1}{1+\log _{2} x}=0
$$

Introduce the substitution $t=\log _{2} x$.

$$
\frac{3}{t}+\frac{2}{2+t}+\frac{1}{1+t}=0
$$

From this, we get the equation $6 t^{2}+13 t+6=0$

whose solutions are $t_{1}=-\frac{2}{3}$ and $t_{2}=-\frac{3}{2}$.

Further, $\log _{2} x=-\frac{2}{3}$, so $x_{1}=2^{-\frac{2}{3}}, \log _{2} x=-\frac{3}{2}$, so $x_{2}=2^{-\frac{3}{2}}$.

(Or written using roots $x_{1}=\frac{1}{\sqrt[3]{4}}=\frac{\sqrt[3]{2}}{2}, \quad x_{1}=\frac{1}{\sqrt{8}}=\frac{\sqrt{2}}{4}$).