Find the value of:
`[(3^(-7)div3^(-10))xx3^(-5)]//[((-2)/3)^(-2)]^2`

Options

A. `(742)/(16)`
B. `(729)/(16)`
C. `(16)/(729)`
D. `(16)/(829)`

Answer

To solve the expression \(\frac{(3^{-7} \div 3^{-10}) \times 3^{-5}}{((-2/3)^{-2})^2}\), we will follow the properties of exponents step by step.

### Step 1: Simplify the numerator
We start with the numerator:
\[ 3^{-7} \div 3^{-10} \]
Using the property of exponents that states \(a^m \div a^n = a^{m-n}\):
\[ 3^{-7} \div 3^{-10} = 3^{-7 - (-10)} = 3^{-7 + 10} = 3^{3} \]

### Step 2: Multiply by \(3^{-5}\)
Now we multiply \(3^{3}\) by \(3^{-5}\):
\[ 3^{3} \times 3^{-5} = 3^{3 + (-5)} = 3^{-2} \]

### Step 3: Simplify the denominator
Next, we simplify the denominator:
\[ ((-2/3)^{-2})^2 \]
Using the property \((a^m)^n = a^{m \cdot n}\):
\[ ((-2/3)^{-2})^2 = (-2/3)^{-4} \]
Now, applying the property of negative exponents:
\[ (-2/3)^{-4} = \frac{1}{(-2/3)^{4}} = \frac{1}{\left(\frac{16}{81}\right)} = \frac{81}{16} \]

### Step 4: Combine the results
Now we can combine the results from the numerator and denominator:
\[ \frac{3^{-2}}{(-2/3)^{-4}} = \frac{3^{-2}}{\frac{81}{16}} = 3^{-2} \times \frac{16}{81} \]
Using the property \(a^{-m} = \frac{1}{a^m}\):
\[ 3^{-2} = \frac{1}{3^2} = \frac{1}{9} \]
Thus, we have:
\[ \frac{1}{9} \times \frac{16}{81} = \frac{16}{729} \]

### Final Answer
The final value of the expression is:
\[ \frac{16}{729} \]