To evaluate the expression \(\log_{25} 125\), we want to rewrite it in a way that uses simpler bases. We start by expressing both 25 and 125 in terms of powers:
- \(25\) can be rewritten as \(5^2\).
- \(125\) can be rewritten as \(5^3\).
Now, substituting these values into the logarithm gives us:
\[ \log_{25} 125 = \log_{5^2} 5^3 \]
We can use the change of base property for logarithms, which states:
\[ \log_{a^b} c^d = \frac{d}{b} \log_a c \]
In our case, \(a = 5\), \(b = 2\), \(c = 5\), and \(d = 3\). Using the property, we have:
\[ \log_{5^2} 5^3 = \frac{3}{2} \log_5 5 \]
Since \(\log_5 5 = 1\), we simplify this to:
\[ \log_{5^2} 5^3 = \frac{3}{2} \cdot 1 = \frac{3}{2} \]
Thus, \(\log_{25} 125 = \frac{3}{2}\).
Next, we identify the smallest integer base that makes it easier to evaluate logarithmic expressions. The base we used here is 5, which is:
- A prime number.
- The smallest integer base that simplifies our logarithmic expression.
Since the base of our logarithm simplifies nicely to compute, the smallest integer base that can be used to evaluate \(\log_{25} 125\) efficiently is:
\[ \boxed{5} \]