6. (3 points) Find the total length of intervals on the number line where the inequalities $|x|<1$ and $|x|<3^{-\pi}$ are satisfied.

Answer:

$$
\frac{2}{1+3^{\pi}}
$$
Solution:

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

If $\sin \log _{3} x>0$, then $2 k \pi<\log _{3} x<(2 k+1) \pi$ for some integer $k$, which means $3^{2 k \pi}<x<3^{(2 k+1) \pi}$. Since $x<1$, we are only interested in negative $k$. Let $n=-k$. The number $n$ takes positive values, then the length of the interval of positivity of our function, having the number $n$ when numbered from the left, is $3^{-(2 n-1) \pi}-3^{-2 n \pi}=2^{-2(n-1) \pi}\left(3^{-\pi}-3^{-2 \pi}\right)$.

Therefore, the lengths of all intervals form an infinitely decreasing geometric progression with a common ratio of $3^{-2 \pi}$, and their sum is $\sum_{n=1}^{\infty} 3^{-(2 n-1) \pi}=\frac{3^{-\pi}-3^{-2 \pi}}{1-3^{-2 \pi}}=\frac{3^{-\pi}}{1+3^{-\pi}}=\frac{1}{1+3^{\pi}}$.

Multiplying this number by 2, we get the answer.