Question

Identify the smallest integer base that the given logarithm can be changed to in order to easily evaluate the expression: log25 125 .

Answer 1

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:

1. A prime number.
2. 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:

\[
\boxed{5}
\]