7.62 $\log _{\sqrt{3}} x+\log _{\sqrt[4]{3}} x+\log _{\sqrt[6]{3}} x+\ldots+\log _{\sqrt[6]{3}} x=36$.

Translate the above text into English, please retain the original text's line breaks and format, and output the translation result directly.

7.62 $\log _{\sqrt{3}} x+\log _{\sqrt[4]{3}} x+\log _{\sqrt[6]{3}} x+\ldots+\log _{\sqrt[6]{3}} x=36$.
7.62 Applying formulas (7.8) and (7.6), we get

$$
\begin{aligned}
& \frac{1}{\log _{x} \sqrt{3}}+\frac{1}{\log _{x} \sqrt[4]{3}}+\frac{1}{\log _{x} \sqrt[6]{3}}+\ldots+\frac{1}{\log _{x} \sqrt[16]{3}}=36 \\
& \frac{2}{\log _{x} 3}+\frac{4}{\log _{x} 3}+\frac{6}{\log _{x} 3}+\ldots+\frac{16}{\log _{x} 3}=36 \\
& \frac{1}{\log _{x} 3}(2+4+6+\ldots+16)=36
\end{aligned}
$$

The expression in parentheses represents an arithmetic progression, where $a_{1}=2, a_{n}=16$, $d=2$. Using formula (6.1), i.e., $a_{n}=a_{1}+d(n-1)$, we get $16=2+2(n-1)$, from which $n=8$. Then, according to formula (6.2), we find

$$
2+4+6+\ldots+16=S_{8}=\frac{a_{1}+a_{8}}{2} \cdot 8=\frac{2+16}{2} \cdot 8=72
$$

Substituting this value into equation (1), we have

$$
\frac{1}{\log _{x} 3} \cdot 72=36, \text { or } 72 \log _{3} x=36
$$

from which $x=3^{\frac{1}{2}}=\sqrt{3}$.

Answer: $x=\sqrt{3}$.