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STD 11 - 8. sequence and series — JEE STD 11 Science — Question

Gujarat BoardEnglish MediumSTD 11 ScienceJEESTD 11 - 8. sequence and series4 Marks
MCQ
If $1,\,\,{\log _9}({3^{1 - x}} + 2),\,\,{\log _3}({4.3^x} - 1)$ are in $A.P.$ then $x$ equals
  • A
    ${\log _3}4$
  • ✓
    $1 - {\log _3}4$
  • C
    $1 - {\log _4}3$
  • D
    ${\log _4}3$
✓

Answer

Correct option: B.
$1 - {\log _3}4$
b
(b) The given number are in $A.P.$

$\therefore 2{\log _9}({3^{1 - x}} + 2) = {\log _3}({4.3^x} - 1) + 1$

$⇒$  $2{\log _{{3^2}}}({3^{1 - x}} + 2) = {\log _3}({4.3^x} - 1) + {\log _3}3$

$⇒$  $\frac{2}{2}{\log _3}({3^{1 - x}} + 2) = {\log _3}[3({4.3^x} - 1)]$

$⇒$  ${3^{1 - x}} + 2 = 3\,({4.3^x} - 1)$

$⇒$  $\frac{3}{y} + 2 = 12y - 3,$ where $y = {3^x}$

$⇒$  $12{y^2} - 5y - 3 = 0$

$y = \frac{{ - 1}}{3}$ or $\frac{3}{4}$

$\Rightarrow {3^x} = \frac{{ - 1}}{3}\,\,$or ${3^x} = \frac{3}{4}$

$x = {\log _3}\,(3/4)$

$ \Rightarrow x = 1 - {\log _3}4$.

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