The value of `log_((9)/(4))((1)/(2sqrt(3))sqrt(6-(1)/(2sqrt(3))sqrt(6-(1)/(2sqrt(3))sqrt(6-(1)/(2sqrt(3)))))...oo)` is

Options

A. `-2`
B. `-1`
C. `-1//2`
D. none of these

Explanation

To solve the problem, we need to evaluate the logarithm given by:

\[ \log_{\frac{9}{4}} \left( \frac{1}{2\sqrt{3}} \sqrt{6 - \frac{1}{2\sqrt{3}} \sqrt{6 - \frac{1}{2\sqrt{3}} \sqrt{6 - \frac{1}{2\sqrt{3}} \cdots}}} \right) \]

Let us denote the infinite nested radical by \( x \):

\[ x = \frac{1}{2\sqrt{3}} \sqrt{6 - x} \]

### Step 1: Square both sides

First, we square both sides to eliminate the square root:

\[ x^2 = \left(\frac{1}{2\sqrt{3}}\right)^2 (6 - x) \]

### Step 2: Simplify the equation

Calculating the left side, we have:

\[ x^2 = \frac{1}{4 \cdot 3} (6 - x) = \frac{1}{12} (6 - x) \]

Multiplying through by 12 to eliminate the fraction gives:

\[ 12x^2 = 6 - x \]

### Step 3: Rearranging the equation

Rearranging this equation leads to:

\[ 12x^2 + x - 6 = 0 \]

### Step 4: Factor the quadratic equation

Next, we can factor the quadratic equation. We need two numbers that multiply to \( 12 \times -6 = -72 \) and add to \( 1 \). The numbers \( 9 \) and \( -8 \) work:

\[ 12x^2 + 9x - 8x - 6 = 0 \]

Grouping gives:

\[ 3x(4x + 3) - 2(4x + 3) = 0 \]

Factoring out \( (4x + 3) \):

\[ (4x + 3)(3x - 2) = 0 \]

### Step 5: Solve for \( x \)

Setting each factor to zero gives:

1. \( 4x + 3 = 0 \) → \( x = -\frac{3}{4} \) (not valid since \( x \) must be positive)
2. \( 3x - 2 = 0 \) → \( x = \frac{2}{3} \)

### Step 6: Substitute back into the logarithm

Now we substitute \( x = \frac{2}{3} \) back into the logarithm:

\[ \log_{\frac{9}{4}} \left( \frac{2}{3} \right) \]

### Step 7: Use logarithm properties

Using the property of logarithms, we can rewrite this as:

\[ \log_{\frac{9}{4}} \left( \frac{2}{3} \right) = \log_{\frac{9}{4}} \left( \left(\frac{9}{4}\right)^{-\frac{1}{2}} \right) \]

This simplifies to:

\[ -\frac{1}{2} \log_{\frac{9}{4}} \left( \frac{9}{4} \right) = -\frac{1}{2} \cdot 1 = -\frac{1}{2} \]

### Final Answer

Thus, the value of the logarithm is:

\[ \boxed{-\frac{1}{2}} \]