199. a) $\log _{4} x=-3$;
b) $\log _{x} \frac{1}{8}=\frac{3}{2}$.
Solution. a) By the definition of logarithm, we write $4^{-3}=x$, from which $x=\frac{1}{64}$.

6) According to the definition of logarithm, we obtain the equation $x^{3 / 2}=\frac{1}{8}$. Since $\frac{1}{8}=2^{-3}$ and $x^{3 / 2}=\sqrt{x^{3}}$, the equation becomes $\sqrt{x^{3}}=2^{-3}$. Squaring both sides:

$$
\left(\sqrt{x^{3}}\right)^{2}=\left(2^{-3}\right)^{2} ; \quad x^{3}=2^{-6} ; \quad x=2^{-2} ; \quad x=\frac{1}{4}
$$