The value of `"log"_(sqrt(2)) sqrt(2sqrt(2sqrt(2sqrt(2))))`, is

Options

A. `(15)/(16)`
B. `(7)/(16)`
C. `(15)/(8)`
D. `(31)/(32)`

Answer

To solve the expression \( \log_{\sqrt{2}} \sqrt{2\sqrt{2\sqrt{2\sqrt{2}}}} \), we will follow these steps:

### Step 1: Simplify the expression inside the logarithm
We start with the expression inside the logarithm:
\[ \sqrt{2\sqrt{2\sqrt{2\sqrt{2}}}} \]
We can denote this entire expression as \( x \):
\[ x = \sqrt{2\sqrt{2\sqrt{2\sqrt{2}}}} \]

### Step 2: Break down the nested square roots
We can simplify \( x \) by working from the innermost square root outward. Let's denote:
\[ y = \sqrt{2\sqrt{2\sqrt{2}}} \]
Then we can express \( x \) as:
\[ x = \sqrt{2y} \]

### Step 3: Continue simplifying \( y \)
Now we simplify \( y \):
\[ y = \sqrt{2\sqrt{2\sqrt{2}}} \]
Let \( z = \sqrt{2\sqrt{2}} \):
\[ y = \sqrt{2z} \]

### Step 4: Simplify \( z \)
Now we simplify \( z \):
\[ z = \sqrt{2\sqrt{2}} = \sqrt{2 \cdot 2^{1/2}} = \sqrt{2^{1 + 1/2}} = \sqrt{2^{3/2}} = 2^{3/4} \]
Thus,
\[ y = \sqrt{2 \cdot 2^{3/4}} = \sqrt{2^{1 + 3/4}} = \sqrt{2^{7/4}} = 2^{7/8} \]

### Step 5: Substitute back to find \( x \)
Now substituting \( y \) back into the equation for \( x \):
\[ x = \sqrt{2 \cdot 2^{7/8}} = \sqrt{2^{1 + 7/8}} = \sqrt{2^{15/8}} = 2^{15/16} \]

### Step 6: Substitute \( x \) back into the logarithm
Now we substitute \( x \) back into the logarithm:
\[ \log_{\sqrt{2}}(2^{15/16}) \]

### Step 7: Use logarithm properties
Using the property of logarithms \( \log_b(a^c) = c \cdot \log_b(a) \):
\[ \log_{\sqrt{2}}(2^{15/16}) = \frac{15}{16} \cdot \log_{\sqrt{2}}(2) \]

### Step 8: Calculate \( \log_{\sqrt{2}}(2) \)
Now we calculate \( \log_{\sqrt{2}}(2) \):
\[ \log_{\sqrt{2}}(2) = \frac{1}{\frac{1}{2}} = 2 \]

### Step 9: Final calculation
Now substituting back:
\[ \log_{\sqrt{2}}(2^{15/16}) = \frac{15}{16} \cdot 2 = \frac{15}{8} \]

### Final Answer
Thus, the value of \( \log_{\sqrt{2}} \sqrt{2\sqrt{2\sqrt{2\sqrt{2}}}} \) is:
\[ \boxed{\frac{15}{8}} \]