As in the problem statement, “doing something repeatedly until the condition is met” can be implemented by a while statement.

As the upper bound of the answer is \(\log_2 10^9 < 30\), the answer can be found fast enough by a simulation. Be careful enough of overflows. (The product of \(A\) and \(K\) may not fit in a 32-bit integer.)

Sample code (C++):

#include <bits/stdc++.h>
using namespace std;

int main() {
    long long a, b, k;
    cin >> a >> b >> k;

    int res = 0;
    while (a < b) {
        a *= k;
        res++;
    }

    cout << res << endl;
}