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8. Sequences and Series
medium
If ${\log _3}2,\;{\log _3}({2^x} - 5)$ and ${\log _3}\left( {{2^x} - \frac{7}{2}} \right)$ are in $A.P.$, then $x$ is equal to
A
$1,\;\frac{1}{2}$
B
$1,\;\frac{1}{3}$
C
$1,\;\frac{3}{2}$
D
None of these
(IIT-1990)
Solution
(d) ${\log _3}2,\;{\log _3}({2^x} – 5)$ and ${\log _3}\left( {{2^x} – \frac{7}{2}} \right)$ are in $A.P.$
$ \Rightarrow $$2{\log _3}({2^x} – 5) = {\log _3}\left[ {(2)\,\left( {{2^x} – \frac{7}{2}} \right)} \right]$
$ \Rightarrow $ ${({2^x} – 5)^2} = {2^{x + 1}} – 7$
$ \Rightarrow $${2^{2x}} – 12\;.\;{2^x} – 32 = 0$
$ \Rightarrow $ $x = 2,\;3$
But $x = 2$ does not hold, hence $x = 3$.
Standard 11
Mathematics
