Answer

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**Hint:**First, we have to know that modulus is. After applying the condition, the value in the log cannot be less than 0. Since the value inside the modulus function returns a positive value. Then again apply the condition, the value in the log cannot be less than 0. Then, apply the condition for which the equation satisfies.

**Complete step-by-step solution:**

Given:- $f\left( x \right) = \log \left| {\log x} \right|$

Since \[\left| {\log x} \right|\] is in the log and we know that the value in the log cannot be less than 0. Then,

$ \Rightarrow \left| {\log x} \right| > 0$

A modulus function is a function that gives an absolute value to a number or variable. This results in the magnitude of the number of variables. It is often referred to as the function of absolute value. The output of the function is always positive irrespective of the data.

Since the value of $\log x$ can be either positive or negative. Then,

For the positive value of $\log x$, the value of $x$ will be,

$ \Rightarrow x \in \left[ {1,\infty } \right)$

For the negative value of $\log x$, the value of $x$ will be,

$ \Rightarrow x \in \left( {0,1} \right)$

Now, add both domains to get the actual domain of $x$,

$ \Rightarrow x \in \left( {0,1} \right) \cup \left[ {1,\infty } \right)$

Take the union of the domain on the right side,

$\therefore x \in \left( {0,\infty } \right)$

Thus, the domain of the function is $\left( {0,\infty } \right)$.

**Hence, option (A) is the correct answer.**

**Note:**The students might make mistakes in the log value as the log value can be less than 1 as the modulus function will give a positive value for the outer log.

Logarithms are the opposite of exponentials, just as the opposite of addition is subtraction and the opposite of multiplication is division.

In other words, a logarithm is essentially an exponent that is written in a particular manner.

Logarithms can make multiplication and division of large numbers easier, because adding logarithms is the same as multiplying, and subtracting logarithms is the same as dividing.

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