The value of `((1)/(sqrt(27)))^(2-((log_(5)16)/(2log_(5)9)))` equal to :

Options

A. `(5sqrt(2))/(27)`
B. `(sqrt(2))/(27)`
C. `(4sqrt(2))/(27)`
D. `(2sqrt(2))/(27)`

Answer

To solve the expression \(\left(\frac{1}{\sqrt{27}}\right)^{2 - \frac{\log_{5} 16}{2 \log_{5} 9}}\), we will follow these steps:

### Step 1: Simplify the base
The base \(\frac{1}{\sqrt{27}}\) can be rewritten as:
\[ \frac{1}{\sqrt{27}} = \frac{1}{27^{1/2}} = 27^{-1/2} \]

### Step 2: Rewrite the exponent
Next, we simplify the exponent \(2 - \frac{\log_{5} 16}{2 \log_{5} 9}\). We can rewrite \(\log_{5} 16\) and \(\log_{5} 9\) using properties of logarithms:
\[ \log_{5} 16 = \log_{5} (2^4) = 4 \log_{5} 2 \]
\[ \log_{5} 9 = \log_{5} (3^2) = 2 \log_{5} 3 \]
Thus, we can substitute these into the exponent:
\[ 2 - \frac{4 \log_{5} 2}{2 \cdot 2 \log_{5} 3} = 2 - \frac{4 \log_{5} 2}{4 \log_{5} 3} = 2 - \frac{\log_{5} 2}{\log_{5} 3} \]

### Step 3: Combine the exponent
Now we can express the exponent as:
\[ 2 - \frac{\log_{5} 2}{\log_{5} 3} = 2 - \log_{3} 2 \]
(using the change of base formula, \(\log_{5} 2 / \log_{5} 3 = \log_{3} 2\))

### Step 4: Substitute back into the expression
Now we can substitute back into our expression:
\[ (27^{-1/2})^{2 - \log_{3} 2} \]
Using the property of exponents, we can simplify this to:
\[ 27^{-\frac{1}{2}(2 - \log_{3} 2)} = 27^{-\frac{1}{2} \cdot 2 + \frac{1}{2} \log_{3} 2} = 27^{-1 + \frac{1}{2} \log_{3} 2} \]

### Step 5: Rewrite 27 in terms of base 3
Since \(27 = 3^3\), we can rewrite the expression as:
\[ (3^3)^{-1 + \frac{1}{2} \log_{3} 2} = 3^{-3 + \frac{3}{2} \log_{3} 2} \]

### Step 6: Simplify the exponent
Now we simplify the exponent:
\[ -3 + \frac{3}{2} \log_{3} 2 = -3 + \log_{3} 2^{3/2} = \log_{3} \left(\frac{2^{3/2}}{27}\right) \]

### Step 7: Final expression
Thus, we can express our final answer as:
\[ 3^{-3 + \frac{3}{2} \log_{3} 2} = \frac{2^{3/2}}{27} \]

### Conclusion
The value of \(\left(\frac{1}{\sqrt{27}}\right)^{2 - \frac{\log_{5} 16}{2 \log_{5} 9}}\) is:
\[ \frac{2^{3/2}}{27} \]