Find the value of `log_(1//3)root(4)(729*root(3)(9^(-1)*27^(-4//3)))`.

Explanation

To find the value of \( \log_{(1/3)} \left( \sqrt[4]{729} \cdot \sqrt[3]{9^{-1} \cdot 27^{-4/3}} \right) \), we will break down the expression step by step.

### Step 1: Simplify the expression inside the logarithm
We start with the expression:
\[ \sqrt[4]{729} \cdot \sqrt[3]{9^{-1} \cdot 27^{-4/3}} \]

### Step 2: Simplify \( \sqrt[4]{729} \)
We know that:
\[ 729 = 3^6 \quad \text{(since \( 729 = 27^2 = (3^3)^2 = 3^6 \))} \]
Thus,
\[ \sqrt[4]{729} = \sqrt[4]{3^6} = 3^{6/4} = 3^{3/2} \]

### Step 3: Simplify \( \sqrt[3]{9^{-1} \cdot 27^{-4/3}} \)
First, we simplify \( 9^{-1} \) and \( 27^{-4/3} \):
\[ 9 = 3^2 \quad \Rightarrow \quad 9^{-1} = (3^2)^{-1} = 3^{-2} \]
\[ 27 = 3^3 \quad \Rightarrow \quad 27^{-4/3} = (3^3)^{-4/3} = 3^{-4} \]
Now, we can combine these:
\[ 9^{-1} \cdot 27^{-4/3} = 3^{-2} \cdot 3^{-4} = 3^{-2 - 4} = 3^{-6} \]
Now, take the cube root:
\[ \sqrt[3]{3^{-6}} = 3^{-6/3} = 3^{-2} \]

### Step 4: Combine the results
Now we can combine the results from Steps 2 and 3:
\[ \sqrt[4]{729} \cdot \sqrt[3]{9^{-1} \cdot 27^{-4/3}} = 3^{3/2} \cdot 3^{-2} = 3^{3/2 - 2} = 3^{3/2 - 4/2} = 3^{-1/2} \]

### Step 5: Substitute back into the logarithm
Now we substitute back into the logarithm:
\[ \log_{(1/3)}(3^{-1/2}) \]

### Step 6: Use the logarithm property
Using the property \( \log_a(b^c) = c \cdot \log_a(b) \):
\[ \log_{(1/3)}(3^{-1/2}) = -\frac{1}{2} \cdot \log_{(1/3)}(3) \]

### Step 7: Calculate \( \log_{(1/3)}(3) \)
Using the change of base formula:
\[ \log_{(1/3)}(3) = \frac{\log(3)}{\log(1/3)} = \frac{\log(3)}{\log(3^{-1})} = \frac{\log(3)}{-\log(3)} = -1 \]

### Step 8: Final calculation
Substituting back, we get:
\[ -\frac{1}{2} \cdot (-1) = \frac{1}{2} \]

### Final Answer
Thus, the value of \( \log_{(1/3)} \left( \sqrt[4]{729} \cdot \sqrt[3]{9^{-1} \cdot 27^{-4/3}} \right) \) is:
\[ \frac{1}{2} \]