find the value of n

$\begin{aligned} \log_b(\log_b(\log_b x)) & = \log_c(\log_c(\log_c x)) \\ \log_b \left(\log_b \left(\frac {\log_2 x}8 \right) \right) & = \log_c \left(\log_c \left(\frac {\log_2 x}{16} \right) \right) \\ \log_b \left(\frac {\log_2(\log_2 x) - 3}8 \right) & = \log_c \left(\frac {\log_2(\log_2 x)-4}{16} \right) \\ \frac {\log_2(\log_2(\log_2 x) - 3)-3}8 & = \frac {\log_2(\log_2(\log_2 x)-4)-4}{16} \\ 2\log_2(\log_2(\log_2 x) - 3)-6 & = \log_2(\log_2(\log_2 x)-4)-4 \\ 2\log_2 \left(\log_2 \left(\frac {\log_2 x}8 \right)\right) & = \log_2 \left(\log_2 \left(\frac {\log_2 x}{16}\right) \right) +2 \\ \log_2 \left(\log_2 \left(\frac {\log_2 x}8 \right)^2 \right) & = \log_2 \left(4\log_2 \left(\frac {\log_2 x}{16}\right) \right) \\ \log_2 \left(\frac {\log_2 x}8 \right)^2 & = 4 \log_2 \left(\frac {\log_2 x}{16}\right) \\ \implies 2^2 & = 4 \cdot 1 & \small \blue{\implies \frac {\log_2 x}8 = 4 \implies \frac {\log_2 x}{16} = 2} \\ \implies \log_2 x & = 32 \\ \implies n & = \boxed{32} \end{aligned}$