7.196. $3^{\log _{3} x+\log _{3} x^{2}+\log _{3} x^{3}+\ldots+\log _{3} x^{8}}=27 x^{30}$.
Solution.

Domain of definition: $x>0$.

Rewrite the equation as $3^{\log _{3} x+2 \log _{3} x+3 \log _{3} x+\ldots+8 \log _{3} x}=27 x^{30} \Leftrightarrow$ $\Leftrightarrow\left(3^{\log _{3} x}\right)^{(1+2+3+\ldots+8)}=27 x^{30} \Leftrightarrow x^{1+2+3+\ldots+8}=27 x^{30} \Leftrightarrow x^{36}=27 x^{30} \Leftrightarrow$ $\Leftrightarrow x^{6}=27$, from which $x=\sqrt[6]{27}=\sqrt{3}$.

Answer: $\sqrt{3}$.