The domain of the function $f (x) = \underset{n times}{\underset{⏟}{\log_{2} \log_{2} \log_{2} \dots \log_{2} x}} is$

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$(2^{n - 1}, \infty)$

$[2^{n}, \infty)$

$(2^{n - 2}, \infty)$

None of these

answer is D.

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Detailed Solution

f (x) is defined if
$(1) x > 0 (2) \log_{2} x > 0 \Rightarrow x > 2^{0} = 1 (3) \log_{2} \log_{2} x > 0 \Rightarrow \log_{2} x > 2^{0} = 1 \Rightarrow x > 2^{1} = 2 (4) \log_{2} \log_{2} \log_{2} x > 0 \Rightarrow \log_{2} \log_{2} x > 2^{0} = 1 \begin{array}{l} \Rightarrow \log_{2} x > 2^{1} = 2 \\ \Rightarrow x > 2^{2} \end{array} (5) \log_{2} \log_{2} \log_{2} \log_{2} x > 0 \Rightarrow \log_{2} \log_{2} \log_{2} x > 2^{0} = 1 \begin{array}{l} \Rightarrow \log_{2} \log_{2} x > 2^{1} \\ \Rightarrow \log_{2} x > 2^{2} \Rightarrow x > 2^{2^{2}} \end{array}$
Continuing like this, we get $x > 2^{2^{2 (n - 1) times}}$
$∴ Domain of f = (2^{2^{2 . (n - 1)} iimes}, \infty)$