Question

Using the change of base formula, the following expression was created. log base 2 128/ log base 2 32. Identify the base of the original expression before the change of base formula was used

Answer 1

The expression given is \(\frac{\log_2 128}{\log_2 32}\). This can be simplified using the change of base formula, which states:

\[
\log_a b = \frac{\log_c b}{\log_c a}
\]

Where \(c\) is any valid base for the logarithm.

Notice that \(128\) can be expressed as \(2^7\) and \(32\) as \(2^5\). Thus:

\[
\log_2 128 = 7 \quad \text{and} \quad \log_2 32 = 5
\]

So,

\[
\frac{\log_2 128}{\log_2 32} = \frac{7}{5}
\]

Now, if we denote the original logarithm as \(\log_{b} 128\) and \(\log_{b} 32 = x\), we can express the change of base formula applied to this base \(b\):

Using the change of base formula, it can be written as:

\[
\log_{b} 128 = \frac{\log_2 128}{\log_2 b} \quad \text{and} \quad \log_{b} 32 = \frac{\log_2 32}{\log_2 b}
\]

So the fraction \(\frac{\log_b 128}{\log_b 32}\) can be rewritten as:

\[
\frac{\frac{\log_2 128}{\log_2 b}}{\frac{\log_2 32}{\log_2 b}} = \frac{\log_2 128}{\log_2 32}
\]

This implies that the logarithms in the numerator and denominator share the same base \(b\).

Thus, the original base \(b\) of the logarithm can be inferred directly from the values of \(128\) and \(32\). The original expression before the change of base might typically be in terms of \(b\) such as:

\[
\log_{32} 128
\]

Based on our evaluations, we established that:

\[
x = \frac{7}{5}, \text{ reverts back to } \log_{32} 128
\]

Therefore, the original base of the logarithm before the change of base formula was \(32\).