5000 Floors, Ceilings, Fracs, Logs, Powers

Hint: $\lceil \log (5000^{5000})\rceil=18495$ just means that there are 18495 digits in the integer $5000^{5000}$ (in base 10).

Can we say a similar thing about $f(x)$ ?

Nice problem! There is a typo though! It should be $f({k}_{i})$ than $g({k}_{i})$ ! Got me once!

To find which base it is, $\lceil 5000log(5000) \rceil = 18495$ is given from where we find that base is $10$ .

$f(x) = \lfloor {10}^{xlog(5000) - \lfloor (xlog(5000)) \rfloor} \rfloor$

We know that $5000 = \frac{{10}^{4}}{2}$ (not a necessary step, just cleans it a little bit)

$f(x) = \lfloor \frac{{5000}^{x}}{{10}^{\lfloor 4x - xlog(2) \rfloor}} \rfloor$

$f(x) = \lfloor \frac{{10}^{4x}({10}^{\lceil xlog(2) \rceil})}{{10}^{4x}({2}^{x})} \rfloor$

$f(x) = \lfloor \frac{{10}^{\lceil xlog(2) \rceil}}{{2}^{x}} \rfloor$

Well, now it can be seen that $log(2) \approx \frac{3}{10}$ (This is a wrong step though but quite useful)

$f(x) = \lfloor \frac{{10}^{\lceil\frac{3x}{10} \rceil}}{{2}^{x}} \rfloor$

Now it can easily be seen that $x = 3k$ will do our job at least for a few values.

Hence, after a little bashing,

$f(x) = \lfloor {\frac{5}{4}}^{k} \rfloor$ and hence it can be seen that it is valid for $k = 1,2,3$

and $k=4$ makes it $2$ , so $x = 13$ will be our fourth solution.

As a result,

this will go on as after an increment of 1 here after 3 values, there is a decrement of 1 in $\lfloor {\frac{5}{4}}^{k} \rfloor$ , therefore, again we have to increase 1 after 3 values and so on.

Therefore, our set of solutions becomes $[3,6,9,13,16,19,23..... 4996,4999]$ . I mean an increment of 3 for 3 times and then an increment of 4 and again(starting from 3). As a result of counting, our answer becomes $\boxed{1505}$