Question

The value of $x$, for which the $6th$ term in the expansion
${\{2^{{\log }_2\sqrt{(9^{x-1}+7)}}+\frac{1}{2^{\left(\frac{1}{5}\right)}{\log }_2(3^{x-1}+1)}\}}^7$ is $84$, is equal to

• A. $4$
• B. $3$
• C. $2$
• D. $0$

Answer 1

The correct option is C $2$

Given, expression
$={[\sqrt{9^{x-1}+7}+\frac{1}{(3^{x-1}+1{)}^{1/5}}]}^7$
$\therefore T_6=T_{5+1}$
${=}^7{C}_5(\sqrt{9^{x-1}+7}{)}^{7-5}{\left\{\frac{1}{(3^{x-1}+1{)}^{1/5}}\right\}}^5$
$=21(9^{x-1}+7)\cdot \frac{1}{3^{x-1}+1}$
$=21\cdot \frac{(3^{2x-2}+7)}{3^{x-1}+1}=84$ (given)
$\Rightarrow 3^{2x-2}+7=4(3^{x-1}+1)$
$\Rightarrow y^2+7-4y-4=0$
where $y=3^{x-1}$
$\therefore y=3$
$\Rightarrow 3^{x-1}=3$
$\Rightarrow x-1=1$
$\Rightarrow x=2$ and $y=1$
$\Rightarrow 3^{x-1}=3^0$
$\Rightarrow x-1=0\Rightarrow x=1$.