\section*{Problem 3A - 101043A}

Determine all positive real numbers \(c\) for which \(\left[\log _{12} c\right] \leq\left[\log _{4} c\right]\) holds.

Here, \([x]\) denotes the greatest integer not greater than \(x\).
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It is \(\ln 12 > \ln 4\), so \(\frac{1}{\ln 12} < \frac{1}{\ln 4}\).

In order for \(\left[\log _{12} c\right] \leq \left[\log _{4} c\right]\) to hold, both logarithms must round down to the same integer \(n\), i.e., there must be a negative integer \(n\) such that \(n \leq \log _{4} c < \log _{12} c < n+1\). Therefore, \(4^{n} \leq c < 12^{n+1}\).

For \(n = -1\), this yields \(c \in \left[\frac{1}{4} ; 1\right)\) and for \(n = -2\) the statement \(c \in \left[\frac{1}{16} ; \frac{1}{12}\right)\).

For \(n \leq -3\), \(3^{n} \leq \frac{1}{27} < \frac{1}{12}\), so \(12^{n+1} = 12 \cdot 12^{n} = 12 \cdot 3^{n} \cdot 4^{n} < 4^{n}\), which means the upper interval boundary would be smaller than the lower one, and thus no further solutions would arise.

The equation is therefore satisfied for all \(c\) with \(\frac{1}{16} \leq c < \frac{1}{12}\) or \(\frac{1}{4} \leq c\).