Question

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

• A. $4$
• B. $3$
• C. $-2$
• D. $1$

Answer 1

The correct option is D $1$

$\because a^{{\log }_{b}x}=x^{{\log }_{b}a}$
$\therefore 2^{{\log }_2\sqrt{(9^{x-1}+7)}}={\left(\sqrt{9^{x-1}+7}\right)}^{{\log }_{2}2}=\sqrt{9^{x-1}+7}$
$2^{\frac{1}{5}{\log }_2(3^{x-1}+1)}=(3^{x-1}+1{)}^{1/5}$

$T_{r+1}={}^7{C}_r{\left(\sqrt{9^{x-1}+7}\right)}^{7-r}\times \left(\right(3^{x-1}+1{)}^{-1/5}{)}^r$

$\because T_6=84$
$\Rightarrow {}^7{C}_5{\left(\sqrt{9^{x-1}+7}\right)}^{7-5}\times \left(\right(3^{x-1}+1{)}^{-1/5}{)}^5=84$

$\Rightarrow 21\times \frac{(3^{2x-2}+7)}{3^{x-1}+1}=84$
$\Rightarrow \frac{(3^x{)}^2}{9}+7=\frac{4}{3}\left(3^x\right)+4$

Let $3^x=t$
$\Rightarrow t^2-12t+27=0$
$\Rightarrow (t-9)(t-3)=0\Rightarrow t=9,3$
$\Rightarrow 3^x=9,3$
or, $x=2,1$