The domain of definition of `f(x)=log_(10) log_(10)…..log_(10)x` n times, is

Options

A. `(10^(n),oo)`
B. `(10^(n-1),oo)`
C. `(10^(n-2),oo)`
D. none of these

Answer

To find the domain of the function \( f(x) = \log_{10}(\log_{10}(\log_{10}(\ldots \log_{10}(x) \ldots))) \) taken \( n \) times, we need to ensure that the argument of each logarithm is positive.

### Step-by-Step Solution:

1. **Understanding the Function**:
The function \( f(x) \) involves taking the logarithm \( n \) times. Therefore, we need to ensure that each logarithm is defined and positive.

2. **First Logarithm**:
The innermost logarithm is \( \log_{10}(x) \). For this to be defined, we need:
\[ x > 0 \]
Additionally, for \( \log_{10}(x) \) to be positive:
\[ \log_{10}(x) > 0 \implies x > 10^0 = 1 \]

3. **Second Logarithm**:
Now consider the second logarithm \( \log_{10}(\log_{10}(x)) \). For this to be defined and positive:
\[ \log_{10}(x) > 1 \implies x > 10^1 = 10 \]

4. **Third Logarithm**:
Next, we consider the third logarithm \( \log_{10}(\log_{10}(\log_{10}(x))) \). For this to be defined and positive:
\[ \log_{10}(\log_{10}(x)) > 1 \implies \log_{10}(x) > 10 \implies x > 10^{10} \]

5. **Generalizing for n Logarithms**:
Continuing this process, we can see a pattern. For the \( n \)-th logarithm:
\[ \log_{10}(\log_{10}(\ldots \log_{10}(x) \ldots)) > 1 \implies \log_{10}(\log_{10}(\ldots)) > 10^{(n-1)} \]
Therefore, we can generalize that:
\[ x > 10^{10^{(n-1)}} \]

6. **Final Domain**:
Thus, the domain of the function \( f(x) \) is:
\[ x > 10^{10^{(n-1)}} \]

### Conclusion:
The domain of the function \( f(x) = \log_{10}(\log_{10}(\ldots \log_{10}(x) \ldots)) \) taken \( n \) times is \( x > 10^{10^{(n-1)}} \).