Question

What is the smallest integer base we can change in log25 125

Answer 1

To determine the smallest integer base in which we can evaluate the logarithm \(\log_{25} 125\), we can express both 25 and 125 as powers of a common base.

We start by rewriting both numbers:

\[
25 = 5^2
\]
\[
125 = 5^3
\]

Now we can substitute these expressions into the logarithm:

\[
\log_{25} 125 = \log_{5^2} 5^3
\]

We can use the change of base formula or properties of logarithms to simplify this further. Using the change of base formula:

\[
\log_{a^b} c = \frac{1}{b} \log_a c
\]

We get:

\[
\log_{5^2} 5^3 = \frac{1}{2} \log_5 5^3
\]

Since \(\log_5 5^3 = 3\):

\[
\log_{5^2} 5^3 = \frac{1}{2} \cdot 3 = \frac{3}{2}
\]

This indicates that we can evaluate \(\log_{25} 125\) in base \(5\) and that the computation is valid since \(5\) is an integer base.

To find the smallest integer base possible, recall that the logarithm is defined if the base is greater than 1 and is not equal to 1. The two numbers, 25 and 125, must also be valid in that base.

The smallest integer base we can use while keeping both numbers valid (and greater than 1) is base \(5\). So, we conclude that:

\[
\text{The smallest integer base we can change in } \log_{25} 125 \text{ is } 5.
\]