Answer
1. The number of digits in 1000 is 4, and $log1000=3$,
2. For 10,000, the number of digits in 10,000 is 5, and $log10000=4$,
3. Any number between 1000 and 10,000 have a number of digit= $4$.
4. The common logarithm of such a number must lie between 3 and 4.
5. $m=[[log(x)]]+1$ see prove below.
6. The number $2^{100}$ have 31 digits.
Work Step by Step
1. The number of digits in 1000 is 4, and $log1000=3(=4-1)$,
2. For 10,000, the number of digits in 10,000 is 5, and $log10000=4(=5-1)$,
3. Any number between 1000 and 10,000 have a number of digit= $4$.
4. The common logarithm of such a number must lie between 3 and 4.
5. Assume the number of digits of the positive integer $x$ is $m$,
based on the above observations, the common logarithm of $x$
must lie between $m-1$ and $m$, which means that $[[log(x)]]=m-1$
hence we proved that $[[log(x)]]+1=m$ the number of digits in $x$.
6. For $x=2^100$, $m=[[log(2^{100})]]+1=[[100log2]]+1=[[30.1]]+1=31$