We are tasked with evaluating the following expression:

$\log_2(3) \cdot \log_3(4) \cdot \log_6(16) \cdot \log_4(6)$

Step-by-Step Evaluation:

1. Change of Base Formula: Recall that for any logarithms $\log_b(a)$, we can apply the change of base formula:

   $\log_b(a) = \frac{\log_c(a)}{\log_c(b)}$

   We will simplify each logarithmic term.

2. Simplifying $\log_6(16)$: Using the change of base formula, we can express this as:

   $\log_6(16) = \frac{\log_2(16)}{\log_2(6)}$

   Since $16 = 2^4$, we know that:

   $\log_2(16) = 4$

   Therefore,

   $\log_6(16) = \frac{4}{\log_2(6)}$

3. Simplifying $\log_4(6)$: Similarly,

   $\log_4(6) = \frac{\log_2(6)}{\log_2(4)}$

   Since $4 = 2^2$, we know that:

   $\log_2(4) = 2$

   Therefore,

   $\log_4(6) = \frac{\log_2(6)}{2}$

4. Substituting into the Original Expression: The original expression now becomes:

   $\log_2(3) \cdot \log_3(4) \cdot \left( \frac{4}{\log_2(6)} \right) \cdot \left( \frac{\log_2(6)}{2} \right)$

   Notice that $\log_2(6)$ cancels out, leaving us with:

   $\log_2(3) \cdot \log_3(4) \cdot \frac{4}{2}$

   Simplifying further:

   $\log_2(3) \cdot \log_3(4) \cdot 2$

5. Simplifying $\log_2(3) \cdot \log_3(4)$: We again use the change of base formula to simplify $\log_3(4)$:

   $\log_3(4) = \frac{\log_2(4)}{\log_2(3)}$

   Since $\log_2(4) = 2$, we have:

   $\log_3(4) = \frac{2}{\log_2(3)}$

   Now, the expression becomes:

   $\log_2(3) \cdot \frac{2}{\log_2(3)} \cdot 2$

   The $\log_2(3)$ terms cancel, leaving:

   $2 \cdot 2 = 4$

Final Answer:

The value of the given expression is:

$\boxed{4}$

Would you like more details or have any questions?

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Here are some related questions:

1. How does the change of base formula help in simplifying logarithmic expressions?
2. Why does $\log_2(6)$ cancel out in the simplification process?
3. What is the importance of understanding logarithmic identities in algebra?
4. How can you evaluate more complex logarithmic expressions without a calculator?
5. What role do powers of two play in simplifying logarithms?

Tip: The change of base formula is a powerful tool for converting logarithms into more manageable forms, especially when dealing with non-standard bases.