Suppose that `beta` is a real number such that `3^(2x +3) = (3^(x))/(2^(x))` , then the value of `2^(-(1+log_(2)3)beta)` is equal to (A) `6 (B) `9 (C) `18 (D) `27

Options

A. 6
B. 9
C. 18
D. 27

Explanation

To solve the equation \( 3^{2x + 3} = \frac{3^x}{2^x} \) and find the value of \( 2^{-(1 + \log_2 3) \beta} \), we can follow these steps:

### Step 1: Rewrite the equation
Start with the given equation:
\[ 3^{2x + 3} = \frac{3^x}{2^x} \]
We can rewrite the right side:
\[ 3^{2x + 3} = 3^x \cdot 2^{-x} \]

### Step 2: Simplify the equation
Now, we can express the left side:
\[ 3^{2x} \cdot 3^3 = 3^x \cdot 2^{-x} \]
This simplifies to:
\[ 27 \cdot 3^{2x} = \frac{3^x}{2^x} \]
Now, we can multiply both sides by \( 2^x \):
\[ 27 \cdot 3^{2x} \cdot 2^x = 3^x \]

### Step 3: Rearranging the terms
Rearranging gives:
\[ 27 \cdot 3^{2x} \cdot 2^x - 3^x = 0 \]
Factoring out \( 3^x \):
\[ 3^x (27 \cdot 3^x \cdot 2^x - 1) = 0 \]
Since \( 3^x \neq 0 \), we have:
\[ 27 \cdot 3^x \cdot 2^x = 1 \]

### Step 4: Solving for \( 3^x \)
From the equation:
\[ 3^x \cdot 2^x = \frac{1}{27} \]
This implies:
\[ (3 \cdot 2)^x = \frac{1}{27} \]
Thus:
\[ 6^x = 27^{-1} \]
We can express \( 27 \) as \( 3^3 \):
\[ 6^x = 3^{-3} \]

### Step 5: Taking logarithms
Taking logarithms on both sides:
\[ x \log 6 = -3 \log 3 \]
Thus:
\[ x = -\frac{3 \log 3}{\log 6} \]

### Step 6: Finding \( \beta \)
Now, we can express \( \beta \) in terms of \( x \):
\[ \beta = -x = \frac{3 \log 3}{\log 6} \]

### Step 7: Substitute \( \beta \) into the expression
Now we need to find:
\[ 2^{-(1 + \log_2 3) \beta} \]
Substituting \( \beta \):
\[ 2^{-(1 + \log_2 3) \cdot \frac{3 \log 3}{\log 6}} \]

### Step 8: Simplifying the exponent
Using the change of base formula:
\[ \log_2 3 = \frac{\log 3}{\log 2} \]
Thus:
\[ 1 + \log_2 3 = 1 + \frac{\log 3}{\log 2} = \frac{\log 2 + \log 3}{\log 2} = \frac{\log 6}{\log 2} \]
Now substituting back:
\[ 2^{-\left(\frac{\log 6}{\log 2}\right) \cdot \frac{3 \log 3}{\log 6}} = 2^{-3 \frac{\log 3}{\log 2}} = \left(2^{-\log_2 3}\right)^3 = \left(\frac{1}{3}\right)^3 = \frac{1}{27} \]

### Step 9: Final result
Thus, the value of \( 2^{-(1 + \log_2 3) \beta} \) is:
\[ 27 \]

### Final Answer
The value of \( 2^{-(1 + \log_2 3) \beta} \) is \( 27 \) (Option D).