Explanation

(c)
We have,
$\left\lbrack 2^{\log_{2}\sqrt{9^{x - 1} + 7}} + \frac{1}{2^{(1/5)\log_{2}\left( 3^{x - 1} + 1 \right)}} \right\rbrack^{7}$
$= \left\lbrack \sqrt{9^{x - 1} + 7} + \frac{1}{\left( 3^{x - 1} + 1 \right)^{1/5}} \right\rbrack^{7}$
$\therefore T_{6} = \ ^{7}C_{5}\left( \sqrt{9^{x - 1} + 7} \right)^{7 - 5}\left\lbrack \frac{1}{\left( 3^{x - 1} + 1 \right)^{\frac{1}{5}}} \right\rbrack^{5}$
$= \ ^{7}C_{5}(9^{x - 1} + 7)\frac{1}{(3^{x - 1} + 1)}$
$\Rightarrow \ 84 = \ ^{7}C_{5}\frac{(9^{x - 1} + 7)}{(3^{x - 1} + 1)}$
$\Rightarrow \ 9^{x - 1} + 7 = 4(3^{x - 1} + 1)$
$\Rightarrow \ \frac{3^{2x}}{9} + 7 = 4\left( \frac{3^{x}}{3} + 1 \right)$
$\Rightarrow \ 3^{2x} - 12\left( 3^{x} \right) + 27 = 0$
$\Rightarrow \ y^{2} - 12y + 27 = 0$ $(put\ y = 3^{x})$
$\Rightarrow \ (y - 3)(y - 9) = 0$
$\Rightarrow \ y = 3,\ 9\ \Rightarrow \ 3^{x} = 3,\ 9\ \Rightarrow \ x = 1,\ 2$