To solve this, we need to look at the expression you've provided:
\[ \frac{\log_2{128}}{\log_2{32}} \]
This expression is derived from the change of base formula. The change of base formula states that:
\[ \log_b{a} = \frac{\log_k{a}}{\log_k{b}} \]
for any positive base \(k\). In this case, the expression can be rewritten by choosing \(k = 2\):
This means that if we want to rewrite \(\log_b{a}\) using this formula, we could express:
\[ \log_b{128} = \frac{\log_2{128}}{\log_2{b}} \]
and
\[ \log_b{32} = \frac{\log_2{32}}{\log_2{b}} \]
If we take the ratio of these two logarithms using the change of base formula, we get:
\[ \frac{\log_b{128}}{\log_b{32}} = \frac{\frac{\log_2{128}}{\log_2{b}}}{\frac{\log_2{32}}{\log_2{b}}} = \frac{\log_2{128}}{\log_2{32}} \]
This indicates that the ratio simplifies to the expression you provided.
Now, since we started with expressions that followed the structure of the change of base formula, we can identify that the original base \(b\) before applying the change of base formula is:
The original base of the logarithm was 2.