2B. Solve the equation:

$$
\log _{\frac{1}{8}}(2 x)-4 \log _{\frac{1}{4}} x \cdot \log _{8} x=0
$$
Solution. Given that

$$
\log _{\frac{1}{8}}(2 x)=\log _{\frac{1}{8}} 2+\log _{\frac{1}{8}} x=-\frac{1}{3}+\frac{1}{3} \log _{\frac{1}{2}} x, \log _{\frac{1}{4}} x=\frac{1}{2} \log _{\frac{1}{2}} x \text { and } \log _{8} x=-\frac{1}{3} \log _{\frac{1}{2}} x \text {, }
$$

we rewrite the equation in the form: $2 \log _{\frac{1}{2}}^{2} x+\log _{\frac{1}{2}} x-1=0$. The solutions to the last equation are $\log _{\frac{1}{2}} x_{1}=\frac{1}{2}$ and $\log _{\frac{1}{2}} x_{2}=-1$, from which we obtain $x_{1}=\frac{1}{\sqrt{2}}$ and $x_{2}=2$.