The problem asks to simplify the following logarithmic expression:

$\ln\left( \frac{1}{e^{-2}} \right) + 4\ln(e^3) + \ln(e^{-2})$

Let's break this down step by step:

Step 1: Simplify $\ln\left( \frac{1}{e^{-2}} \right)$

$\ln\left( \frac{1}{e^{-2}} \right) = \ln\left(e^{2}\right)$ Since $\ln(e^a) = a$, we get: $\ln(e^2) = 2$

Step 2: Simplify $4\ln(e^3)$

Using the property $\ln(e^a) = a$, we have: $4\ln(e^3) = 4 \times 3 = 12$

Step 3: Simplify $\ln(e^{-2})$

Again using the property $\ln(e^a) = a$, we get: $\ln(e^{-2}) = -2$

Step 4: Combine the results

Now, adding all the simplified terms: $2 + 12 + (-2) = 12$

Final Answer:

The simplified form of the given expression is: $12$

Would you like further details or have any questions?

Here are some related questions to explore:

1. How do logarithmic properties work when combining sums and differences?
2. Can you explain the logarithmic identity $\ln(e^x) = x$ in more detail?
3. What are the steps to simplify more complex logarithmic expressions?
4. What other properties of logarithms are useful in similar problems?
5. How would the expression change if different bases (not $e$) were involved?

Tip: Always apply logarithmic properties one at a time to avoid confusion in complex expressions!