6. (3 points) Find the total length of intervals on the number line where the inequalities $|x|<1$ and $\operatorname{tg} \log _{5}|x|<0$ are satisfied.
Answer:

$$
\frac{2 \cdot 5^{\frac{\pi}{2}}}{1+5^{\frac{\pi}{2}}}
$$

Solution:

It is sufficient to solve the problem for positive $x$, and then multiply the result by 2.

If $\operatorname{tg} \log _{5} x<0$, then $\pi\left(k-\frac{1}{2}\right)<\log _{5} x<\pi k$ for some integer $k$, which means $5^{\pi\left(k-\frac{1}{2}\right)}<x<5^{\pi k}$. Since $x<1$, we are only interested in non-positive $k$. Let $n=-k$. The number $n$ takes non-negative values, then the length of the interval of negativity of our function, having the number $n$ when numbered from the left (numbering starts from 0), is $5^{-\pi n}-5^{-\pi\left(n+\frac{1}{2}\right)}=5^{-\pi n}\left(1-5^{-\frac{\pi}{2}}\right)$.

Therefore, the lengths of all intervals form an infinitely decreasing geometric progression with a common ratio of $5^{-\pi}$, and their sum is $\sum_{n=0}^{\infty} 5^{-\pi n}\left(1-5^{-\frac{\pi}{2}}\right)=\frac{1-5^{-\frac{\pi}{2}}}{1-5^{-\pi}}=\frac{1}{1+5^{-\frac{\pi}{2}}}=\frac{5^{\frac{\pi}{2}}}{1+5^{\frac{\pi}{2}}}$.

Multiplying this number by 2, we get the answer.