In the expansion of ;$\left[2^{\log_2 {\sqrt{9^{x-1}+7}}}+\displaystyle \frac{1}{2^{\frac{1}{5}\log_2 \displaystyle {(3^{x-1}+1})}}\right]^{7}$,& ;6th term is $84$. Then $x =$

The given expression $\displaystyle ={ \left[ \sqrt { { 9 }^{ x-1 }+7 } +\frac { 1 }{ { \left( { 3 }^{ x-1 }+1 \right) }^{ \frac { 1 }{ 5 } } } \right] }^{ 7 }$

Given, ${ T }_{ 6 }=84$

$\Rightarrow\displaystyle {}^{ 7 }{ { C }_{ 5 } }{ \left( \sqrt { { 9 }^{ x-1 }+7 } \right) }^{ 7-5 }{ \left( \frac { 1 }{ { \left( { 3 }^{ x-1 }+1 \right) }^{ \frac { 1 }{ 5 } } } \right) }^{ 5 }=84$

$\displaystyle\Rightarrow {}^{ 7 }{ { C }_{ 5 } }\left( { 9 }^{ x-1 }+7 \right) .\frac { 1 }{ \left( { 3 }^{ x-1 }+1 \right) } =84$

$\Rightarrow { 9 }^{ x-1 }+7=4\left( { 3 }^{ x-1 }+1 \right)$

$\Rightarrow { 3 }^{ 2x }-12.{ 3 }^{ x }+27=0$$\Rightarrow \left( { 3 }^{ x }-3 \right) \left( { 3 }^{ x }-9 \right) =0$

$\Rightarrow { 3 }^{ x }=3,9$

$\Rightarrow x=1,2.$