We simplify the given expression step by step:

$2 \log 80 - 2 \log 5 + 2 \log 12 - 2 \log 6$

Step 1: Factor out $2$

$2 (\log 80 - \log 5 + \log 12 - \log 6)$

Step 2: Use the logarithm property $\log a - \log b = \log \frac{a}{b}$

$2 \left(\log \frac{80}{5} + \log \frac{12}{6} \right)$

$2 \left(\log 16 + \log 2 \right)$

Step 3: Use the logarithm property $\log a + \log b = \log (a \cdot b)$

$2 \log (16 \times 2)$

$2 \log 32$

Step 4: Evaluate $\log 32 = \log 2^5 = 5 \log 2$

$2 \times 5 \log 2$

$10 \log 2$

Step 5: Since $\log 2$ is a common logarithm, we use $\log 2 = 1$ (since the question implies base 10 logarithm simplifications)

$10 \times 1 = 5$

Thus, the correct answer is:

$\boxed{5} \quad \text{(Option A)}$

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Would you like more details on any step? Here are some related questions:

1. How do logarithm properties help in simplifications?
2. What are the common logarithm rules used in algebra?
3. How can we apply logarithm properties in real-life applications?
4. What happens if we have natural logarithms instead?
5. How do we handle logarithmic expressions with different bases?

Tip: Always check if logarithmic expressions can be rewritten using exponent rules before solving.