229. $\log _{3}(12 x+4)-\log _{3}(x-7)=\log _{3} 9$.
Solution. Let's write the given equation in the form

$$
\log _{3} \frac{12 x+4}{x-7}=\log _{3} 9
$$

Since the logarithms and their bases are equal, the numbers being logarithmed are also equal:

$$
\frac{12 x+4}{x-7}=9
$$

Assuming $x-7 \neq 0$, we bring the fraction to a common denominator and solve the resulting equation:

$$
12 x+4=9 x-63 ; \quad 3 x=-67 ; \quad x=-22 \frac{1}{3}
$$

Substituting the value $x=-22 \frac{1}{3}$ into the equation, we see that for this value of $x$, the expressions $12 x+4$ and $x-7$ are negative. Since logarithms of negative numbers do not exist, $x=-22 \frac{1}{3}$ is an extraneous root, and the equation has no solutions.