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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 ${\log _{12}}27 = a,$ then ${\log _6}16 = $
  • A
    $2.{{3 - a} \over {3 + a}}$
  • B
    $3.{{3 - a} \over {3 + a}}$
  • ✓
    $4.{{3 - a} \over {3 + a}}$
  • D
    None of these
✓

Answer

Correct option: C.
$4.{{3 - a} \over {3 + a}}$
c
(c) $a = {{\log 27} \over {\log 12}} = {{3\log 3} \over {\log 3 + 2\log 2}} \Rightarrow \log 3 = {{2a\log 2} \over {3 - a}}$

${\log _6}16 = {{\log 16} \over {\log 6}} = {{4\log 2} \over {\log 2 + \log 3}}$

$ = {{4\log 2} \over {\log 2 + {{2a\log 2} \over {3 - a}}}} = {{4(3 - a)} \over {3 - a + 2a}} = 4.{{3 - a} \over {3 + a}}$.

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