Find the value of m - n, if
`(9^(n)xx3^(2)xx(3^((-n)/2))^(-2)-(27)^(n))/(3^(3m)xx2^(3))=1/27`

Options

A. 1
B. -2
C. -1
D. 2

Answer

To solve the equation and find the value of \( m - n \), we start with the given expression:

\[ \frac{9^n \cdot 3^2 \cdot (3^{-\frac{n}{2}})^{-2} - 27^n}{3^{3m} \cdot 2^3} = \frac{1}{27} \]

### Step 1: Rewrite the terms in base 3
First, we rewrite \( 9^n \) and \( 27^n \) in terms of base 3:
- \( 9^n = (3^2)^n = 3^{2n} \)
- \( 27^n = (3^3)^n = 3^{3n} \)

Now, substitute these into the equation:

\[ \frac{3^{2n} \cdot 3^2 \cdot (3^{-\frac{n}{2}})^{-2} - 3^{3n}}{3^{3m} \cdot 2^3} = \frac{1}{27} \]

### Step 2: Simplify the expression
Next, simplify \( (3^{-\frac{n}{2}})^{-2} \):
\[ (3^{-\frac{n}{2}})^{-2} = 3^{n} \]

Now substitute this back into the equation:

\[ \frac{3^{2n} \cdot 3^2 \cdot 3^n - 3^{3n}}{3^{3m} \cdot 2^3} = \frac{1}{27} \]

Combine the powers of 3 in the numerator:

\[ 3^{2n} \cdot 3^2 \cdot 3^n = 3^{2n + 2 + n} = 3^{3n + 2} \]

So the equation now looks like:

\[ \frac{3^{3n + 2} - 3^{3n}}{3^{3m} \cdot 2^3} = \frac{1}{27} \]

### Step 3: Factor out \( 3^{3n} \)
Factor \( 3^{3n} \) from the numerator:

\[ \frac{3^{3n}(3^2 - 1)}{3^{3m} \cdot 2^3} = \frac{1}{27} \]

Calculate \( 3^2 - 1 = 9 - 1 = 8 \):

\[ \frac{3^{3n} \cdot 8}{3^{3m} \cdot 2^3} = \frac{1}{27} \]

### Step 4: Simplify further
Since \( 2^3 = 8 \), the equation simplifies to:

\[ \frac{3^{3n} \cdot 8}{3^{3m} \cdot 8} = \frac{1}{27} \]

Cancel \( 8 \):

\[ \frac{3^{3n}}{3^{3m}} = \frac{1}{27} \]

### Step 5: Rewrite \( \frac{1}{27} \)
Rewrite \( \frac{1}{27} \) as \( 3^{-3} \):

\[ 3^{3n - 3m} = 3^{-3} \]

### Step 6: Set the exponents equal
Since the bases are the same, we can set the exponents equal to each other:

\[ 3n - 3m = -3 \]

### Step 7: Solve for \( n - m \)
Divide the entire equation by 3:

\[ n - m = -1 \]

### Step 8: Rearranging gives us \( m - n \)
Rearranging gives:

\[ m - n = 1 \]

Thus, the final answer is:

\[ \boxed{1} \]