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I do know how to find domain for most of the functions I find but this one's a bit off. A step-by-step response would be appreciated. $$f(x)=\log_2(\log_3(\log_2(\log_3(\log_2x))))$$ Thanks in advance!

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    $\begingroup$ MathJax hint: if you put a backslash before common functions you get the proper font and spacing, so \log_2 x gives $\log_2 x$ as contrasted with log_2 x giving $log_2 x$ It is better to make sure the body of the question has the whole question in it, rather than relying on the title. $\endgroup$ Jun 30 '17 at 2:24
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Using the fact $\log_bx>a\iff x>b^a$ (when $b>1$) we can reason as follows:

$$\begin{array}{ll} & \log_2\log_3\log_2\log_3\log_2 x \quad \textrm{exists} \\ \iff & \phantom{\log_2}\log_3\log_2\log_3\log_2x>0 \\ \iff & \phantom{\log_2\log_3}\log_2\log_3\log_2x>3^0=1 \\ \iff & \phantom{\log_2\log_3\log_2}\log_3\log_2 x>2^1=2 \\ \iff & \phantom{\log_2\log_3\log_2\log_3}\log_2 x>3^2=9 \\ \iff & \phantom{\log_2\log_3\log_2\log_3\log_2}x>2^9=512. \end{array} $$

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  • $\begingroup$ Well done. Would be better if the fact is supported by $b>1$. $\endgroup$
    – farruhota
    Jun 30 '17 at 6:30
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Clearly you have no trouble if $x$ is very large, so you are just trying to find the minimum value of $x$ that makes sense. The problem comes if the argument of the outer $\log_2$ is negative or zero. What is the minimum argument of the first $\log 3$ to avoid that? Keep going in layer by layer to find the minimum value of $x$.

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  • $\begingroup$ Minor nitpick: the problem is if the argument is nonpositive, not just negative. $\endgroup$
    – Ian
    Jun 30 '17 at 2:21
  • $\begingroup$ @Ian: Good point. I have updated the answer. Thanks. $\endgroup$ Jun 30 '17 at 2:22
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You can make changes: $$x=2^a \Rightarrow \log_2 x = a$$ $$a=3^b \Rightarrow \log_3 a = b$$ $$b=2^c \Rightarrow \log_2 b = c$$ $$c=3^d \Rightarrow \log_3 c = d$$ $$d>0 \Rightarrow c>1 \Rightarrow b>2 \Rightarrow a>3^2=9 \Rightarrow x>2^9.$$

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