Question
If $A = {\log _2}{\log _2}{\log _4}256 + 2{\log _{\sqrt 2 \,}}\,2,$ then $A$ is equal to
✓
Answer
c
(c) $A = {\log _2}{\log _2}{\log _4}256$ + $2{\log _2}_{^{1/2}}\,2$
(c) $A = {\log _2}{\log _2}{\log _4}256$ + $2{\log _2}_{^{1/2}}\,2$
$ = {\log _2}{\log _2}{\log _4}{4^4} + 2 \times {1 \over {(1/2)}}{\log _2}2$
$ = {\log _2}{\log _2}4 + 4 = {\log _2}{\log _2}{2^2} + 4$
$ = {\log _2}2 + 4 = 1 + 4 = 5$.
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