$7.27 \lg (3-x)-\frac{1}{3} \lg \left(27-x^{3}\right)=0$.
7.27 We have $\lg (3-x)^{3}=\lg \left(27-x^{3}\right)$, which is equivalent to the system

$$
\left\{\begin{array} { l } 
{ 3 - x > 0 , } \\
{ ( 3 - x ) ^ { 3 } = 2 7 - x ^ { 3 } , }
\end{array} \text { or } \left\{\begin{array}{l}
x<3, \\
(3-x)^{3}=(3-x)\left(9+3 x+x^{2}\right)
\end{array}\right.\right.
$$

Further, we get

$$
(3-x)\left(9-6 x+x^{2}-9-3 x-x^{2}\right)=0, \text { or }(3-x)(-9 x)=0 \text {, }
$$

from which $x_{1}=0, x_{2}=3$ (does not satisfy the inequality $x<3$). Therefore, $x=0$.

Answer: $x=0$.