7.191. $\log _{10} x+\log _{\sqrt{10}} x+\log _{\sqrt[3]{10}} x+\ldots+\log _{\sqrt[1]{10}} x=5.5$.
Solution.

Domain of definition: $x>0$.

Let's switch to base 10. We have

$$
\lg x+2 \lg x+3 \lg x+\ldots+10 \lg x=5.5, (1+2+3+\ldots+10) \lg x=5.5
$$

In the parentheses, we have the sum of the terms of an arithmetic progression $S_{n}$ with $a_{1}=1$, $d=1, a_{n}=10, n=10: S_{n}=\frac{a_{1}+a_{n}}{2} n=\frac{1+10}{2} \cdot 10=55$. Then $55 \lg x=5.5 \Leftrightarrow$ $\Leftrightarrow \lg x=\frac{1}{10}$, from which $x=\sqrt[10]{10}$.

Answer: $\sqrt[10]{10}$.