smc

Let $k$ be an arbitrary integer. For which $x$ do we have $\lfloor {\log }_{10}4x\rfloor =\lfloor {\log }_{10}x\rfloor =k$? The equation $\lfloor {\log }_{10}x\rfloor =k$ can be rewritten as $10^k\le x<10^{k+1}$. The second one gives us $10^k\le 4x<10^{k+1}$. Combining these, we get that both hold at the same time if and only if $10^k\le x<\frac{10^{k+1}}{4}$. Hence for each integer $k$ we get an interval of values for which $\lfloor {\log }_{10}4x\rfloor -\lfloor {\log }_{10}x\rfloor =0$. These intervals are obviously pairwise disjoint. For any $k\ge 0$ the corresponding interval is disjoint with $(0,1)$, so it does not contribute to our answer. On the other hand, for any $k<0$ the entire interval is inside $(0,1)$. Hence our answer is the sum of the lengths of the intervals for $k<0$. For a fixed $k$ the length of the interval $\left[10^k,\frac{10^{k+1}}{4}\right)$ is $\frac{3}{2}\cdot 10^k$. This means that our result is $\frac{3}{2}\left(10^{-1}+10^{-2}+\cdots \right)=\frac{3}{2}\cdot \frac{1}{9}=\boxed{\frac{1}{6}}$.