Question

The domain of the function $f\left(x\right)={\log }_{10}{\log }_{10}{\log }_{10}......{\log }_{10}x\left(n\text{times}\right)$ is

• A. $[{10}^n,\infty )$
• B. $\left({10}^{n-1},\infty \right)$
• C. $\left({10}^{n-2},\infty \right)$
• D. None of these

Answer 1

The correct option is D

None of these

Explanation for correct answer Option(s)

Finding the domain of the function $f\left(x\right)$

The given function is defined as:

$$\begin{gathered} {\log }_{10}{\log }_{10}{\log }_{10}......{\log }_{10}x>0\left[\text{for}n-1\right] \\ {\log }_{10}{\log }_{10}{\log }_{10}......{\log }_{10}x>1\left[\text{for}n-2\right] \end{gathered}$$

We know that

$$\begin{gathered} {\log }_{a}x=b \\ \therefore x=a^b \end{gathered}$$

Then,

$$\begin{gathered} \Rightarrow {\log }_{10}{\log }_{10}{\log }_{10}......{\log }_{10}x>{10}^1\left[\text{for}n-3\right] \\ \Rightarrow {\log }_{10}{\log }_{10}{\log }_{10}......{\log }_{10}x>10 \\ \Rightarrow {\log }_{10}{\log }_{10}{\log }_{10}......{\log }_{10}x>{10}^{n-3}\left[\text{for}n-4\right] \\ \Rightarrow {\log }_{10}{\log }_{10}{\log }_{10}......{\log }_{10}x>{\left({10}^{n-3}\right)}^{n-4}\left[\text{for}n-5\right] \\ \Rightarrow {\log }_{10}{\log }_{10}{\log }_{10}......{\log }_{10}x>{\left({10}^{\left(n-3\right)\left(n-4\right)}\right)}^{n-5}\left[\text{for}n-6\right] \\ \Rightarrow x>{10}^{\left(n-3\right)\left(n-4\right)\left(n-5\right)......1} \end{gathered}$$

Since,

$\frac{n!}{n\left(n-1\right)\left(n-2\right)}=\left(n-3\right)\left(n-4\right)\left(n-5\right)......1$

Hence, the domain of the function is $x>{10}^{\frac{n!}{n\left(n-1\right)\left(n-2\right)}}$

Therefore, the correct answer is Option D.