- Apply the quotient rule: \(\ln \left(\frac{2e}{x}\right) = \ln(2e) - \ln x\).
- Apply the product rule: \(\ln(2e) = \ln 2 + \ln e\).
- Use the identity \(\ln e = 1\).
- The equivalent expression is \(\boxed{1 + \ln 2 - \ln x}\).
Explanation
- Analysis and logarithmic properties
The problem asks us to find an expression equivalent to \(\ln \left(\frac{2e}{x}\right)\). To do this, we will use fundamental properties of the natural logarithm, specifically the quotient rule, product rule, and the value of \(\ln(e)\). The rules are as follows:
- The quotient rule: \(\ln \left(\frac{a}{b}\right) = \ln a - \ln b\)
- The product rule: \(\ln(ab) = \ln a + \ln b\)
- The identity: \(\ln e = 1\)
- Simplifying the expression
Starting with the expression \(\ln \left(\frac{2e}{x}\right)\), we apply the quotient rule to separate the terms:
\[\ln \left(\frac{2e}{x}\right) = \ln(2e) - \ln x\]
Next, we apply the product rule to the term \(\ln(2e)\):
\[\ln(2e) = \ln 2 + \ln e\]
Now, we substitute these into our expression:
\[\ln \left(\frac{2e}{x}\right) = (\ln 2 + \ln e) - \ln x\]
Since \(\ln e = 1\), we replace \(\ln e\) with 1:
\[\ln \left(\frac{2e}{x}\right) = \ln 2 + 1 - \ln x = 1 + \ln 2 - \ln x\]
This matches the third option provided.
- Conclusion and final answer
Comparing our result \(1 + \ln 2 - \ln x\) with the given choices:
- \(\ln 2 - \ln x\)
- \(\ln 1 + \ln 2 - \ln x\)
- \(1 + \ln 2 - \ln x\)
- \(\ln 2 + \ln x\)
The third option is the correct equivalent expression. Thus, the final answer is \(1 + \ln 2 - \ln x\).
Examples
Logarithmic properties like these are essential in fields like finance and science for modeling exponential growth or decay. For instance, if you want to calculate how long it takes for an investment to grow by a factor involving both a growth constant and the natural constant \(e\), you might encounter expressions similar to this one. By simplifying them, you can isolate variables and perform easier calculations for decision-making or predicting future outcomes.