The problem asks to evaluate several natural logarithm expressions. The expressions are: 1. $ln(\frac{1}{e^9})$

1. Problem Description

The problem asks to evaluate several natural logarithm expressions. The expressions are:

1. $ln(\frac{1}{e^9})$

2. $ln(\sqrt[3]{e^2})$

3. $ln(e)$

4. $ln(\frac{1}{e})$

5. $ln(e^4)$

6. $ln(\sqrt{e})$

7. $ln(\frac{1}{\sqrt{e^9}})$

2. Solution Steps

We will use the following properties of logarithms:

* $ln(e^x) = x$

* $ln(\frac{1}{x}) = -ln(x)$

* $ln(x^a) = a \cdot ln(x)$

* $\sqrt[n]{x} = x^{\frac{1}{n}}$

1. $ln(\frac{1}{e^9}) = ln(e^{-9}) = -9$

2. $ln(\sqrt[3]{e^2}) = ln((e^2)^{\frac{1}{3}}) = ln(e^{\frac{2}{3}}) = \frac{2}{3}$

3. $ln(e) = ln(e^1) = 1$

4. $ln(\frac{1}{e}) = ln(e^{-1}) = -1$

5. $ln(e^4) = 4$

6. $ln(\sqrt{e}) = ln(e^{\frac{1}{2}}) = \frac{1}{2}$

7. $ln(\frac{1}{\sqrt{e^9}}) = ln(\frac{1}{(e^9)^{\frac{1}{2}}}) = ln(\frac{1}{e^{\frac{9}{2}}}) = ln(e^{-\frac{9}{2}}) = -\frac{9}{2}$

3. Final Answer

1. $ln(\frac{1}{e^9}) = -9$

2. $ln(\sqrt[3]{e^2}) = \frac{2}{3}$

3. $ln(e) = 1$

4. $ln(\frac{1}{e}) = -1$

5. $ln(e^4) = 4$

6. $ln(\sqrt{e}) = \frac{1}{2}$

7. $ln(\frac{1}{\sqrt{e^9}}) = -\frac{9}{2}$