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STD 11 - basic of algoritham — JEE STD 11 Science — Question

Gujarat BoardEnglish MediumSTD 11 ScienceJEESTD 11 - basic of algoritham4 Marks
MCQ
If $A = {\log _2}{\log _2}{\log _4}256 + 2{\log _{\sqrt 2 \,}}\,2,$ then $A$ is equal to
  • A
    $2$
  • B
    $3$
  • ✓
    $5$
  • D
    $7$
✓

Answer

Correct option: C.
$5$
c
(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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