The maximum domain of the function $f(x)=\log_2(\log_3(\log_2(\log_3(\log_2x))))$ is? 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!
 A: Using the fact $\log_bx>a\iff x>b^a$ (when $b>1$) we can reason as follows:
$$\begin{array}{ll} &  \log_2\log_3\log_2\log_3\log_2 x \quad \textrm{exists} \\ \iff & \phantom{\log_2}\log_3\log_2\log_3\log_2x>0 \\ \iff & \phantom{\log_2\log_3}\log_2\log_3\log_2x>3^0=1 \\ \iff & \phantom{\log_2\log_3\log_2}\log_3\log_2 x>2^1=2 \\ \iff & \phantom{\log_2\log_3\log_2\log_3}\log_2 x>3^2=9 \\ \iff &  \phantom{\log_2\log_3\log_2\log_3\log_2}x>2^9=512. \end{array} $$
A: 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 $x$.
A: You can make changes:
$$x=2^a \Rightarrow \log_2 x = a$$
$$a=3^b \Rightarrow \log_3 a = b$$
$$b=2^c \Rightarrow \log_2 b = c$$
$$c=3^d \Rightarrow \log_3 c = d$$
$$d>0 \Rightarrow c>1 \Rightarrow b>2 \Rightarrow a>3^2=9 \Rightarrow x>2^9.$$
