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

Options

A. `4`
B. `5`
C. `0`
D. `6`

Answer

To solve the problem, we need to evaluate the expression:

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

Let's denote the repeating part as \( t \):

\[ t = \frac{1}{3\sqrt{2}} \sqrt{4 - t} \]

### Step 1: Solve for \( t \)

To eliminate the square root, we square both sides:

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

This simplifies to:

\[ t^2 = \frac{1}{18}(4 - t) \]

### Step 2: Rearranging the equation

Multiply both sides by 18 to eliminate the fraction:

\[ 18t^2 = 4 - t \]

Rearranging gives us:

\[ 18t^2 + t - 4 = 0 \]

### Step 3: Solve the quadratic equation

Using the quadratic formula \( t = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \):

Here, \( a = 18 \), \( b = 1 \), and \( c = -4 \).

Calculating the discriminant:

\[ b^2 - 4ac = 1^2 - 4 \cdot 18 \cdot (-4) = 1 + 288 = 289 \]

Now substituting back into the quadratic formula:

\[ t = \frac{-1 \pm \sqrt{289}}{2 \cdot 18} = \frac{-1 \pm 17}{36} \]

Calculating the two possible values:

1. \( t = \frac{16}{36} = \frac{4}{9} \)
2. \( t = \frac{-18}{36} = -\frac{1}{2} \) (not valid since \( t \) must be positive)

Thus, we have:

\[ t = \frac{4}{9} \]

### Step 4: Substitute \( t \) back into the logarithm

Now we substitute \( t \) back into the logarithmic expression:

\[ 6 + \log_{\frac{3}{2}}\left(\frac{4}{9}\right) \]

### Step 5: Simplifying the logarithm

Using the property of logarithms:

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

### Step 6: Final calculation

Now, we can finalize the expression:

\[ 6 + (-2) = 4 \]

Thus, the final answer is:

\[ \boxed{4} \]