To simplify the expression

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

we can start by breaking it down into smaller parts using properties of logarithms. Here are the steps:

Step 1: Simplify each logarithmic term separately

We have two terms:

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

Term 1: Simplify $\ln\left(\frac{3e^2}{e^{-3}}\right)$

Using the property of logarithms, $\ln\left(\frac{a}{b}\right) = \ln(a) - \ln(b)$: $\ln\left(\frac{3e^2}{e^{-3}}\right) = \ln(3e^2) - \ln(e^{-3}).$ Now, apply the property $\ln(ab) = \ln(a) + \ln(b)$: $= \ln(3) + \ln(e^2) - \ln(e^{-3}).$ Since $\ln(e^x) = x$, we get: $= \ln(3) + 2 + 3 = \ln(3) + 5.$

Term 2: Simplify $\ln\left(\frac{4e^{-1}}{e^{-4}}\right)$

Similarly, using the same logarithmic properties: $\ln\left(\frac{4e^{-1}}{e^{-4}}\right) = \ln(4e^{-1}) - \ln(e^{-4}).$ Expanding $\ln(4e^{-1})$ as $\ln(4) + \ln(e^{-1})$: $= \ln(4) - 1 + 4 = \ln(4) + 3.$

Step 2: Combine the simplified terms

Now, we have: $\ln\left(\frac{3e^2}{e^{-3}}\right) + \ln\left(\frac{4e^{-1}}{e^{-4}}\right) = (\ln(3) + 5) + (\ln(4) + 3).$ Simplify by adding the constants: $= \ln(3) + \ln(4) + 8.$ Using the property $\ln(a) + \ln(b) = \ln(ab)$, we combine $\ln(3) + \ln(4)$ into $\ln(12)$: $= \ln(12) + 8.$

Final Answer

The correct answer is: $\boxed{\text{(c) } \ln(12) + 8}.$

Would you like a further breakdown of any steps?

Here are five related questions to deepen understanding:

1. How do the properties of logarithms help simplify complex expressions?
2. Can you expand and simplify $\ln\left(\frac{6e^3}{2e^{-2}}\right)$?
3. What happens to the logarithmic expression if the base is not $e$?
4. How would you handle expressions involving $\ln(x^2 + y^2)$?
5. How do logarithmic properties apply to division and multiplication inside the log function?

Tip: Remember, logarithmic properties are powerful tools for simplifying expressions—mastering them can make complex expressions more manageable.