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The value of $x$, for which the 6th term in the expansion of ${\left\{ {{2^{{{\log }_2}\sqrt {({9^{x - 1}} + 7)} }} + \frac{1}{{{2^{(1/5){{\log }_2}({3^{x - 1}} + 1)}}}}} \right\}^7}$ is $84$, is equal to
$4$
$1$
$2$
$b$ or $c$ both
Solution
(d) We have ${\left[ {{2^{{{\log }_2}\sqrt {{9^{x – 1}} + 7} }} + \frac{1}{{{2^{(1/5){{\log }_2}({3^{x – 1}} + 1)}}}}} \right]^7}$
$ = {\left[ {\sqrt {{9^{x – 1}} + 7} + \frac{1}{{{{({3^{x – 1}} + 1)}^{1/5}}}}} \right]^7}$
$\therefore \,\,\,{T_6} = {\,^7}{C_5}{\left( {\sqrt {{9^{x – 1}} + 7} } \right)^{7 – 5}}{\left[ {\frac{1}{{{{({3^{x – 1}} + 1)}^{1/5}}}}} \right]^5}$
$ = {\,^7}{C_5}({9^{x – 1}} + 7)\frac{1}{{({3^{x – 1}} + 1)}}$
Now ${T_6} = 84 \Rightarrow {\,^7}{C_5}\frac{{({9^{x – 1}} + 7)}}{{({3^{x – 1}} + 1)}} = 84$
==> ${9^{x – 1}} + 7 = 4({3^{x – 1}} + 1)$
==> ${3^{2x}} – 12({3^x}) + 27 = 0\, \Rightarrow {y^2} – 12y + 27 = 0$ (Where $y = {3^x}$)
==> $y = 3,\,9 \Rightarrow {3^x} = 3,9 \Rightarrow x = 1,2$
