The value of `3^(-1/2 log_3 9)` is

Options

A. `3`
B. `1/3`
C. `2/3`
D. none of these

Answer

To solve the expression \(3^{-\frac{1}{2} \log_3 9}\), we will follow these steps:

### Step 1: Simplify the logarithm
We start with the expression:
\[ -\frac{1}{2} \log_3 9 \]
We know that \(9\) can be expressed as \(3^2\). Thus, we can rewrite the logarithm:
\[ \log_3 9 = \log_3 (3^2) \]
Using the logarithmic identity \(\log_b (a^m) = m \cdot \log_b a\), we have:
\[ \log_3 (3^2) = 2 \cdot \log_3 3 \]
Since \(\log_3 3 = 1\), we find:
\[ \log_3 9 = 2 \cdot 1 = 2 \]

### Step 2: Substitute the logarithm back into the expression
Now we can substitute this value back into our expression:
\[ -\frac{1}{2} \log_3 9 = -\frac{1}{2} \cdot 2 = -1 \]

### Step 3: Substitute into the power of 3
Now we substitute \(-1\) back into the expression for \(3\):
\[ 3^{-\frac{1}{2} \log_3 9} = 3^{-1} \]

### Step 4: Calculate the final value
We know that:
\[ 3^{-1} = \frac{1}{3} \]

Thus, the final value of \(3^{-\frac{1}{2} \log_3 9}\) is:
\[ \frac{1}{3} \]

### Final Answer:
\(\frac{1}{3}\)

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