7.090. $27^{\lg x}-7 \cdot 9^{\lg x}-21 \cdot 3^{\lg x}+27=0$.
## Solution.

Domain of definition: $x>0$.

We have

$$
\begin{aligned}
& 3^{3 \lg x}-7 \cdot 3^{2 \lg x}-21 \cdot 3^{\lg x}+27=0,\left(3^{3 \lg x}+27\right)-7 \cdot 3^{\lg x}\left(3^{\lg x}+3\right)=0 \\
& \left(3^{\lg x}+3\right)\left(3^{2 \lg x}-3 \cdot 3^{\lg x}+9\right)-7 \cdot 3^{\lg x}\left(3^{\lg x}+3\right)=0 \\
& \left(3^{\lg x}+3\right)\left(3^{2 \lg x}-10 \cdot 3^{\lg x}+9=0\right)
\end{aligned}
$$

from which $3^{2 \lg x}-10 \cdot 3^{\lg x}+9=0$. Solving this equation as a quadratic equation in terms of $3^{\lg x}$, we get $\left(3^{\lg x}\right)_1=1$ or $\left(3^{\lg x}\right)_2=9$, from which $(\lg x)_1=0$ or $(\lg x)_2=2$. Hence, $x_1=1, x_2=100$.

Answer: $1 ; 100$.