- Apply the quotient rule: \(\log \frac{1}{9k} = \log 1 - \log(9k)\).
- Simplify the expression: \(0 - (\log 9 + \log k) = -\log 9 - \log k\).
- Analyze the options and identify that \(\log \frac{1}{9} - \log k = -\log 9 - \log k\).
- Conclude that the equivalent expression is \(\boxed{\log \frac{1}{9} - \log k}\).
Explanation
- Understanding the Logarithm Properties
We are asked to identify which expression is equivalent to \(\log \frac{1}{9k}\). To solve this, we rely on the fundamental properties of logarithms, specifically the quotient rule and the product rule. The quotient rule states that \(\log \left(\frac{A}{B}\right) = \log A - \log B\) and the product rule states that \(\log (AB) = \log A + \log B\). These properties allow us to break down complex logarithmic expressions into simpler components.
- Simplifying the Expression \(\log \frac{1}{9k}\)
Starting with the expression \(\log \left(\frac{1}{9k}\right)\), we can apply the quotient rule:
\[\log \left(\frac{1}{9k}\right) = \log 1 - \log(9k)\]
Since \(\log 1 = 0\), the expression simplifies further to:
\[0 - \log(9k) = -\log(9k)\]
Using the product rule on \(\log(9k)\), we have \(\log(9k) = \log 9 + \log k\). Substituting this back, we get:
\[-\log(9k) = -(\log 9 + \log k) = -\log 9 - \log k\]
- Comparing with the Given Options
Now, let us examine the provided options to see which one matches our derived expression of \(-\log 9 - \log k\). Let's evaluate the first option: \(\log \frac{1}{9} - \log k\).
Using the quotient rule again on \(\log \frac{1}{9}\):
\[\log \frac{1}{9} = \log 1 - \log 9 = 0 - \log 9 = -\log 9\]
Therefore, the first option becomes:
\[\log \frac{1}{9} - \log k = -\log 9 - \log k\]
This matches our simplified result perfectly.
- Conclusion
By simplifying both the target expression and the first choice, we have shown that they are equivalent. The other choices do not lead to the same result. Thus, the correct equivalent expression is \(\log \frac{1}{9} - \log k\), which is the first option. The final answer is \(\boxed{\log \frac{1}{9} - \log k}\).
Examples
Logarithms are used extensively in fields like acoustics to measure sound levels in decibels. The formula involves logarithmic ratios. For example, if you are comparing two sound intensities, you might encounter expressions similar to \(\log(\frac{I_1}{I_2 \times I_3})\). Understanding how to expand these using log properties allows engineers and scientists to easily analyze and manipulate these measurements to determine the relative intensity or power of signals in a system.