Question
20
Using the change of base formula, the following expression was created: log_2 128/ log_2 32
2
128
2
32
log
2
32
log
2
128
Identify the base of the original expression before the change of base formula was used.
Using the change of base formula, the following expression was created: log_2 128/ log_2 32
2
128
2
32
log
2
32
log
2
128
Identify the base of the original expression before the change of base formula was used.
Answers
GPT-4o mini
To identify the base of the original expression before the change of base formula was used, we can analyze the given expression:
\[
\frac{\log_2 128}{\log_2 32}
\]
Using the change of base formula, which states that \(\log_b a = \frac{\log_k a}{\log_k b}\) for any logarithmic base \(k\), we can rewrite the expression with a new base.
In this case, we can let the original base be denoted as \(b\). The change of base formula applied to both numerator and denominator can be expressed as:
\[
\log_2 128 = \frac{\log_b 128}{\log_b 2}
\]
\[
\log_2 32 = \frac{\log_b 32}{\log_b 2}
\]
Substituting these into the original expression, we get:
\[
\frac{\frac{\log_b 128}{\log_b 2}}{\frac{\log_b 32}{\log_b 2}} = \frac{\log_b 128}{\log_b 32}
\]
This simplifies to \( \log_{32} 128 \) if we work out the division.
Now, to find the base \(b\) originally used in the logarithm (from which \( \log_2 128 \) and \( \log_2 32 \) were derived), notice that:
To revert from 32 and 128 to powers of a common base, we observe:
- \(32 = 2^5\)
- \(128 = 2^7\)
This implies:
\[
\log_2 128 = 7 \quad \text{and} \quad \log_2 32 = 5 \implies \frac{\log_2 128}{\log_2 32} = \frac{7}{5}
\]
This expression can also be rewritten as \( \log_{32} 128 \), and the base of the logarithm comes from the positive exponent in the power of 2 representation.
From further analysis, taking into account this reduction using powers of the same base indicates that the implicit base of our original expression was indeed \(2\).
Thus, **the base of the original expression before the change of base formula was**:
\[
\textbf{2}
\]
\[
\frac{\log_2 128}{\log_2 32}
\]
Using the change of base formula, which states that \(\log_b a = \frac{\log_k a}{\log_k b}\) for any logarithmic base \(k\), we can rewrite the expression with a new base.
In this case, we can let the original base be denoted as \(b\). The change of base formula applied to both numerator and denominator can be expressed as:
\[
\log_2 128 = \frac{\log_b 128}{\log_b 2}
\]
\[
\log_2 32 = \frac{\log_b 32}{\log_b 2}
\]
Substituting these into the original expression, we get:
\[
\frac{\frac{\log_b 128}{\log_b 2}}{\frac{\log_b 32}{\log_b 2}} = \frac{\log_b 128}{\log_b 32}
\]
This simplifies to \( \log_{32} 128 \) if we work out the division.
Now, to find the base \(b\) originally used in the logarithm (from which \( \log_2 128 \) and \( \log_2 32 \) were derived), notice that:
To revert from 32 and 128 to powers of a common base, we observe:
- \(32 = 2^5\)
- \(128 = 2^7\)
This implies:
\[
\log_2 128 = 7 \quad \text{and} \quad \log_2 32 = 5 \implies \frac{\log_2 128}{\log_2 32} = \frac{7}{5}
\]
This expression can also be rewritten as \( \log_{32} 128 \), and the base of the logarithm comes from the positive exponent in the power of 2 representation.
From further analysis, taking into account this reduction using powers of the same base indicates that the implicit base of our original expression was indeed \(2\).
Thus, **the base of the original expression before the change of base formula was**:
\[
\textbf{2}
\]