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

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?

---

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.

Learn how to evaluate the logarithmic expression $$ \log_2(3) \cdot \log_3(4) \cdot \log_6(16) \cdot \log_4(6) $$ using the change of base formula. This problem covers key algebraic concepts, including logarithmic simplification techniques and important logarithmic identities. Suitable for students in Grades 10-12.

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