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8. Sequences and Series
medium
If $1,\,\,{\log _9}({3^{1 - x}} + 2),\,\,{\log _3}({4.3^x} - 1)$ are in $A.P.$ then $x$ equals
A
${\log _3}4$
B
$1 - {\log _3}4$
C
$1 - {\log _4}3$
D
${\log _4}3$
(AIEEE-2002)
Solution
(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$.
Standard 11
Mathematics
