Answer :
Sure! Let's work through the given logarithmic expressions step-by-step:
### Part 1: Expand \(\ln \sqrt{\frac{x}{y}}\)
First, consider the given expression:
[tex]\[ \ln \sqrt{\frac{x}{y}} \][/tex]
1. Simplify the square root:
Use the property that \(\sqrt{a} = a^{1/2}\):
[tex]\[ \ln \sqrt{\frac{x}{y}} = \ln \left( \left( \frac{x}{y} \right)^{1/2} \right) \][/tex]
2. Power rule of logarithms:
Use the rule \(\ln (a^b) = b \ln (a)\):
[tex]\[ \ln \left( \left( \frac{x}{y} \right)^{1/2} \right) = \frac{1}{2} \ln \left( \frac{x}{y} \right) \][/tex]
3. Quotient rule of logarithms:
Use the rule \(\ln \left( \frac{a}{b} \right) = \ln (a) - \ln (b)\):
[tex]\[ \frac{1}{2} \ln \left( \frac{x}{y} \right) = \frac{1}{2} (\ln (x) - \ln (y)) \][/tex]
Therefore, the expanded form of \(\ln \sqrt{\frac{x}{y}}\) is:
[tex]\[ \ln \sqrt{\frac{x}{y}} = \frac{1}{2} (\ln (x) - \ln (y)) \][/tex]
### Part 2: Simplify \(\frac{1}{-} \ln x - \frac{1}{\ln x}\)
Consider the given expression:
[tex]\[ \frac{1}{- \ln x} - \frac{1}{\ln x} \][/tex]
1. Simplify first term:
The first term can be interpreted as:
[tex]\[ \frac{1}{- \ln x} = -\frac{1}{\ln x} \][/tex]
2. Combine the terms:
Combine the simplified first term with the second term:
[tex]\[ -\frac{1}{\ln x} - \frac{1}{\ln x} = -\frac{1}{\ln x} - \frac{1}{\ln x} \][/tex]
3. Addition of like terms:
[tex]\[ -\frac{1}{\ln x} - \frac{1}{\ln x} = -2 \left( \frac{1}{\ln x} \right) \][/tex]
4. We can approach it slightly differently by adding only rather than explicitly expanding −2:
Combining these identically as:
[tex]\[ -1.0 \cdot \ln (x) - \frac{1}{\ln (x)} \][/tex]
Therefore, \(-\frac{1}{\ln x} - \frac{1}{\ln x}\) simplifies to:
[tex]\[ -\frac{1}{\ln x} - \frac{1}{\ln x} = -1.0 \ln (x) - \frac{1}{\ln (x)} \][/tex]
### Summary
1. The expanded form of \(\ln \sqrt{\frac{x}{y}}\) is:
[tex]\[ \frac{1}{2} (\ln (x) - \ln (y)) \][/tex]
2. The simplified form of \(\frac{1}{- \ln x} - \frac{1}{\ln x}\) is:
[tex]\[ -1.0 \ln (x) - \frac{1}{\ln (x)} \][/tex]
### Part 1: Expand \(\ln \sqrt{\frac{x}{y}}\)
First, consider the given expression:
[tex]\[ \ln \sqrt{\frac{x}{y}} \][/tex]
1. Simplify the square root:
Use the property that \(\sqrt{a} = a^{1/2}\):
[tex]\[ \ln \sqrt{\frac{x}{y}} = \ln \left( \left( \frac{x}{y} \right)^{1/2} \right) \][/tex]
2. Power rule of logarithms:
Use the rule \(\ln (a^b) = b \ln (a)\):
[tex]\[ \ln \left( \left( \frac{x}{y} \right)^{1/2} \right) = \frac{1}{2} \ln \left( \frac{x}{y} \right) \][/tex]
3. Quotient rule of logarithms:
Use the rule \(\ln \left( \frac{a}{b} \right) = \ln (a) - \ln (b)\):
[tex]\[ \frac{1}{2} \ln \left( \frac{x}{y} \right) = \frac{1}{2} (\ln (x) - \ln (y)) \][/tex]
Therefore, the expanded form of \(\ln \sqrt{\frac{x}{y}}\) is:
[tex]\[ \ln \sqrt{\frac{x}{y}} = \frac{1}{2} (\ln (x) - \ln (y)) \][/tex]
### Part 2: Simplify \(\frac{1}{-} \ln x - \frac{1}{\ln x}\)
Consider the given expression:
[tex]\[ \frac{1}{- \ln x} - \frac{1}{\ln x} \][/tex]
1. Simplify first term:
The first term can be interpreted as:
[tex]\[ \frac{1}{- \ln x} = -\frac{1}{\ln x} \][/tex]
2. Combine the terms:
Combine the simplified first term with the second term:
[tex]\[ -\frac{1}{\ln x} - \frac{1}{\ln x} = -\frac{1}{\ln x} - \frac{1}{\ln x} \][/tex]
3. Addition of like terms:
[tex]\[ -\frac{1}{\ln x} - \frac{1}{\ln x} = -2 \left( \frac{1}{\ln x} \right) \][/tex]
4. We can approach it slightly differently by adding only rather than explicitly expanding −2:
Combining these identically as:
[tex]\[ -1.0 \cdot \ln (x) - \frac{1}{\ln (x)} \][/tex]
Therefore, \(-\frac{1}{\ln x} - \frac{1}{\ln x}\) simplifies to:
[tex]\[ -\frac{1}{\ln x} - \frac{1}{\ln x} = -1.0 \ln (x) - \frac{1}{\ln (x)} \][/tex]
### Summary
1. The expanded form of \(\ln \sqrt{\frac{x}{y}}\) is:
[tex]\[ \frac{1}{2} (\ln (x) - \ln (y)) \][/tex]
2. The simplified form of \(\frac{1}{- \ln x} - \frac{1}{\ln x}\) is:
[tex]\[ -1.0 \ln (x) - \frac{1}{\ln (x)} \][/tex]
