6. (3 points) Find the total length of intervals on the number line where the inequalities $x<1$ and $\sin \log _{2} x<0$ are satisfied.
Answer:

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

## Solution:

Note that we are only interested in positive $x$.

If $\sin \log _{2} x<0$, then $2(k-1) \pi<\log _{2} x<2 k \pi$ for some integer $k$, which means $2^{2(k-1) \pi}<x<2^{2 k \pi}$. 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 number $n$ when numbered from the left (numbering starts from 0), is $2^{-2 n \pi}-2^{-(2 n+1) \pi}=2^{-2 n \pi}\left(1-2^{-\pi}\right)$.

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