7.099. $\lg (3-x)-\frac{1}{3} \lg \left(27-x^{3}\right)=0$.
## Solution.

Domain of definition: $3-x>0, x<3$.

## Rewrite the equation as

$$
3 \lg (3-x)=\lg \left(27-x^{3}\right), \lg (3-x)^{3}=\lg \left(27-x^{3}\right)
$$

Then $(3-x)^{3}=27-x^{3} \Rightarrow x^{2}-9 x=0$, from which $x_{1}=0, x_{2}=9 ; x_{2}=9$ does not satisfy the domain of definition.

Answer: 0.