Anticipated 100 day streak problem

Let $\{x\}$ denote the fractional part of x, and $\lfloor x \rfloor$ denote the floor function of x.

So, the first digit of $1729^a$ in base 9 is equal to $\lfloor 9^{\{a log_9 1729\}} \rfloor$

Now, let us concentrate on $\{a log_9 1729\}$ . Since $a$ ranges from 1 to infinity (all powers of 1729 means that a ranges from one to infinity), $\{a log_9 1729\}$ has equal probability of being any number between $[0,1)$ .

Now, since $probability = \frac{number of appearances of case}{number of cases}$ and $mean = \frac{(number)*(number of appearances of number)}{number of cases}$ , we can conclude that $mean = (number)*(probability of that no. being selected)$

So, if $0 \leq \{a log_9 1729\} < log _9 2$ , then the first digit is 1. So the probability that the first digit is 1 is $log _9 2 - 0 = log_9 2$

Similarly, if $log _9 2 \leq \{a log_9 1729\} < log _9 3$ , then the first digit is 2, So, the probability that the first digit is 2 is $log_9 3- log _9 2 = log_9 \frac{3}{2}$

Continuing on, we get that the probability of $n$ as the first digit is $log_9 \frac{n+1}{n}$

By the formula for the mean stated above, we get our probability as $\sum_{i=1}^8 (i* log_9 \frac{i+1}{i})$

By a calculator, we get our answer as $\boxed{3.173}$

This is basically an extension of Benford's law