Question

In the expansion of ${[2^{{\log }_2\sqrt{9^{x-1}+7}}+\frac{1}{2^{\frac{1}{5}{\log }_2(3^{x-1}+1)}}]}^7$, 6th term is $84$. Then $x=$

• A. $1$
• B. $2$
• C. $1$ or $2$
• D. $2$ or $4$

Answer 1

The correct option is C $1$ or $2$

The given expression $={[\sqrt{9^{x-1}+7}+\frac{1}{{(3^{x-1}+1)}^{\frac{1}{5}}}]}^7$
Given, $T_6=84$
$\Rightarrow {}^7{C}_5{\left(\sqrt{9^{x-1}+7}\right)}^{7-5}{\left(\frac{1}{{(3^{x-1}+1)}^{\frac{1}{5}}}\right)}^5=84$
$\Rightarrow {}^7{C}_5(9^{x-1}+7).\frac{1}{(3^{x-1}+1)}=84$

$\Rightarrow 9^{x-1}+7=4(3^{x-1}+1)$
$\Rightarrow 3^{2x}-12.3^x+27=0$$\Rightarrow (3^x-3)(3^x-9)=0$
$\Rightarrow 3^x=3,9$

$\Rightarrow x=1,2.$