$7.30 \lg (\lg x)+\lg \left(\lg x^{3}-2\right)=0$.

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$7.30 \lg (\lg x)+\lg \left(\lg x^{3}-2\right)=0$.
7.30 For the existence of logarithms, it is necessary that the inequalities $x>0$, $\lg x>0$, $3 \lg x-2>0$ are satisfied simultaneously, i.e., $x>0, x>1, \lg x>\frac{2}{3}$. Hence, $\lg x>\frac{2}{3}$. Now we transition to the equivalent system of equations given by

$$
\left\{\begin{array}{l}
\lg x>\frac{2}{3} \\
\lg x(3 \lg x-2)=1
\end{array}\right.
$$

After substituting the variable $y=\lg x$, we obtain the equation $3 y^{2}-2 y-1=0$, whose roots are $y_{1}=1, y_{2}=-\frac{1}{3}$. Of these values, only the first satisfies the inequality $\lg x>\frac{2}{3}$. Therefore, $\lg x=1$, from which $x=10$.

Answer: $x=10$.