`log_2log_2(sqrt(sqrt(...sqrtsqrt2)))/(nTimes)` is equal to

Options

A. 0
B. 1
C. n
D. `-n`

Answer

To solve the problem \( \log_2 \log_2(\sqrt{\sqrt{\ldots \sqrt{2}}}) \) where the square root operation is applied \( n \) times, we can follow these steps:

### Step 1: Express the nested square roots
The expression \( \sqrt{\sqrt{\ldots \sqrt{2}}} \) (with \( n \) square roots) can be rewritten in exponential form. We know that:
\[ \sqrt{2} = 2^{1/2} \]
Thus, applying the square root \( n \) times gives us:
\[ \sqrt{\sqrt{\ldots \sqrt{2}}} = 2^{1/2^n} \]

### Step 2: Apply the logarithm
Now we need to find \( \log_2(\sqrt{\sqrt{\ldots \sqrt{2}}}) \):
\[ \log_2(2^{1/2^n}) = \frac{1}{2^n} \cdot \log_2(2) \]
Since \( \log_2(2) = 1 \), we have:
\[ \log_2(\sqrt{\sqrt{\ldots \sqrt{2}}}) = \frac{1}{2^n} \]

### Step 3: Apply the logarithm again
Next, we need to find \( \log_2(\log_2(\sqrt{\sqrt{\ldots \sqrt{2}}})) \):
\[ \log_2\left(\frac{1}{2^n}\right) = \log_2(2^{-n}) = -n \]

### Conclusion
Thus, the final result is:
\[ \log_2 \log_2(\sqrt{\sqrt{\ldots \sqrt{2}}}) = -n \]