If `3^(2n-1)= (1)/(27^(n-3))`, then the value of n is

Options

A. 5
B. 3
C. 6
D. 2

Answer

To solve the equation \( 3^{2n-1} = \frac{1}{27^{n-3}} \), we can follow these steps:

### Step 1: Rewrite \( 27 \) in terms of base \( 3 \)
We know that \( 27 = 3^3 \). Therefore, we can rewrite \( 27^{n-3} \) as:
\[ 27^{n-3} = (3^3)^{n-3} = 3^{3(n-3)} = 3^{3n - 9} \]

### Step 2: Rewrite the equation
Now, substituting this back into the original equation gives us:
\[ 3^{2n-1} = \frac{1}{3^{3n - 9}} \]

### Step 3: Rewrite the right-hand side
The right-hand side can be rewritten using the property of exponents:
\[ \frac{1}{3^{3n - 9}} = 3^{-(3n - 9)} = 3^{9 - 3n} \]

### Step 4: Set the exponents equal
Now we have:
\[ 3^{2n-1} = 3^{9 - 3n} \]
Since the bases are the same, we can set the exponents equal to each other:
\[ 2n - 1 = 9 - 3n \]

### Step 5: Solve for \( n \)
Now, we solve for \( n \):
1. Add \( 3n \) to both sides:
\[ 2n + 3n - 1 = 9 \]
\[ 5n - 1 = 9 \]

2. Add \( 1 \) to both sides:
\[ 5n = 10 \]

3. Divide by \( 5 \):
\[ n = 2 \]

### Final Answer
Thus, the value of \( n \) is \( 2 \).

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