Precalculus (6th Edition) Blitzer

As we know, $\log_a1=0$, $\log_aa=1$, and $\log_ab^c=c\log_ab$. So we have$$\log_4 [ \log_3 ( \log_28) ]=\log_4 [ \log_3 ( \log_22^3 ) ]=\log_4 [ ( \log_3 ( 3 \log_22) ] = \log_4 [ ( \log_3 ((3)(1) ) ] = \log_4 [ \log_33 ]= \log_41=0$$