231. $\log _{4}(x+3)-\log _{4}(x-1)=2-3 \log _{4} 2$.
solution. Representing the number 2 as the logarithm of 16 to the base 4, we rewrite the given equation as

$$
\log _{4}(x+3)-\log _{4}(x-1)=\log _{4} 16-3 \log _{4} 2
$$

From this, we obtain

$$
\log _{4} \frac{x+3}{x-1}=\log _{4} \frac{16}{8}, \text { or } \frac{x+3}{x-1}=2
$$

We solve this equation:

$$
x+3=2(x-1) ; \quad x+3=2 x-2 ; \quad x=5
$$

To verify, substitute the value $x=5$ into the given equation:

$\log _{4}(5+3)-\log _{4}(5-1)=2-3 \log _{4} 2$

$\log _{4} 8-\log _{4} 4=\log _{4} 16-\log _{4} 8 ; \log _{4} \frac{8}{4}=\log \frac{16}{8} ; \log _{4} 2=\log _{4} 2$.

Thus, $x=5$.

Let's consider another type of logarithmic equations.