The value of `((27)^(n//3) xx (8)^(-n//6))/((162)^(-n//2))` is equal to

Options

A. `2^(n)`
B. `3^(n)`
C. `3^(3n)`
D. `35^(n)`

Answer

To solve the expression \(\frac{(27)^{(n/3)} \times (8)^{(-n/6)}}{(162)^{(-n/2)}}\), we can simplify it step by step.

### Step 1: Rewrite the bases in terms of prime factors
- \(27 = 3^3\)
- \(8 = 2^3\)
- \(162 = 2 \times 81 = 2 \times 3^4\)

### Step 2: Substitute the prime factor forms into the expression
The expression becomes:
\[ \frac{(3^3)^{(n/3)} \times (2^3)^{(-n/6)}}{(2 \times 3^4)^{(-n/2)}} \]

### Step 3: Apply the power of a power property
Using the property \((a^m)^n = a^{m \cdot n}\), we simplify:
\[ (3^3)^{(n/3)} = 3^{(3 \cdot n/3)} = 3^n \]
\[ (2^3)^{(-n/6)} = 2^{(3 \cdot -n/6)} = 2^{-n/2} \]
\[ (2 \times 3^4)^{(-n/2)} = 2^{-n/2} \times (3^4)^{-n/2} = 2^{-n/2} \times 3^{-2n} \]

### Step 4: Substitute back into the expression
Now, substituting these results back into the expression gives:
\[ \frac{3^n \times 2^{-n/2}}{2^{-n/2} \times 3^{-2n}} \]

### Step 5: Simplify the expression
The \(2^{-n/2}\) terms cancel out:
\[ \frac{3^n}{3^{-2n}} = 3^n \times 3^{2n} = 3^{n + 2n} = 3^{3n} \]

### Final Result
Thus, the value of the expression is:
\[ 3^{3n} \]