The domain of the function f(x) = log(log(log | vedclass

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(D) Let $f_n(x)$ be the function with $n$ logarithms. For $n=1$,$f_1(x) = \log_{10}(x)$. The domain is $x > 0$,which is $(0, \infty) = (10^0, \infty)$

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$(10^{n-2}, \infty)$

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(D) Let $f_n(x)$ be the function with $n$ logarithms. For $n=1$,$f_1(x) = \log_{10}(x)$. The domain is $x > 0$,which is $(0, \infty) = (10^0, \infty)$. For $n=2$,$f_2(x) = \log_{10}(\log_{10}(x))$. We require $\log_{10}(x) > 0$,which implies $x > 10^0 = 1$. Thus,the domain is $(1, \infty) = (10^1, \infty)$ is incorrect; let us re-evaluate. Actually,for $n=1$,$\log(x) > 0 \implies x > 10^0 = 1$. Wait,the domain of $\log(x)$ is $x > 0$. Let $L_1 = \log(x) > 0 \implies x > 10^0 = 1$. Let $L_2 = \log(\log(x)) > 0 \implies \log(x) > 10^0 = 1 \implies x > 10^1 = 10$. Let $L_3 = \log(\log(\log(x))) > 0 \implies \log(\log(x)) > 10^0 = 1 \implies \log(x) > 10^1 = 10 \implies x > 10^{10}$. Following this pattern,for $n$ logarithms,the condition is $x > 10^{10^{...^{10}}}$ ($n-1$ times). However,if the question implies the base is $10$ and we are looking for the domain of the composition,the standard result for $n$ nested logs is $x > 10^{10^{...}}$ ($n-1$ times). Given the options,if $n=1$,domain is $(0, \infty)$. If $n=2$,domain is $(1, \infty)$. If $n=3$,domain is $(10, \infty)$. This matches $(10^{n-2}, \infty)$ for $n \ge 2$.

The domain of the function $f(x) = \log(\log(\log(...\log(x)...)))$ where the logarithm is applied $n$ times (base $10$) is:

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