To solve this expression step by step:

$\frac{\log_3(12) - \log_9(4)}{\log_{1/3}(6)}$

Step 1: Simplify the numerator using logarithmic properties

The numerator is: $\log_3(12) - \log_9(4)$

Step 1.1: Rewrite $\log_9(4)$ in terms of base 3

Since $9 = 3^2$, we use the change of base property: $\log_9(4) = \frac{\log_3(4)}{\log_3(9)} = \frac{\log_3(4)}{2}$

Now, substitute into the numerator: $\log_3(12) - \frac{\log_3(4)}{2}$

Step 1.2: Combine terms under base 3

The first term, $\log_3(12)$, can be written as: $\log_3(12) = \log_3(4 \cdot 3) = \log_3(4) + \log_3(3)$

Thus, the numerator becomes: $\log_3(4) + \log_3(3) - \frac{\log_3(4)}{2}$

Combine like terms involving $\log_3(4)$: $\log_3(4) - \frac{\log_3(4)}{2} = \frac{2\log_3(4)}{2} - \frac{\log_3(4)}{2} = \frac{\log_3(4)}{2}$

So the numerator simplifies to: $\frac{\log_3(4)}{2} + \log_3(3)$

Step 2: Simplify the denominator

The denominator is: $\log_{1/3}(6)$

Using the change of base formula: $\log_{1/3}(6) = \frac{\log_3(6)}{\log_3(1/3)}$

Recall that $\log_3(1/3) = -1$, so: $\log_{1/3}(6) = \frac{\log_3(6)}{-1} = -\log_3(6)$

Step 3: Combine numerator and denominator

The expression becomes: $\frac{\frac{\log_3(4)}{2} + \log_3(3)}{-\log_3(6)}$

Factor out the negative sign in the denominator: $-\left(\frac{\frac{\log_3(4)}{2} + \log_3(3)}{\log_3(6)}\right)$

Step 4: Simplify further

Use the property $\log_3(6) = \log_3(2 \cdot 3) = \log_3(2) + \log_3(3)$ in the denominator: $\frac{\frac{\log_3(4)}{2} + \log_3(3)}{\log_3(2) + \log_3(3)}$

Final Answer:

$-\frac{\frac{\log_3(4)}{2} + \log_3(3)}{\log_3(2) + \log_3(3)}$

Let me know if you would like numerical approximations or more clarifications.

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Expanding questions:

1. What is the significance of the change of base property in logarithms?
2. How does the choice of base affect the calculation of logarithms?
3. Why does $\log_3(1/3) = -1$?
4. How can $\log_3(6)$ be broken into simpler components?
5. What are the implications of combining logarithms with different bases?

Tip:

Always simplify logarithmic expressions step by step, and watch