Explanation

(b)
Since, $2\log_{3}\left( 3^{1 - x} + 2 \right)^{\frac{1}{2}} = 1 + \log_{3}{({4.3}^{x} - 1)}$
$\Rightarrow \ \log_{3}{(3^{1 - x} + 2)} = \log_{3}3 + \log_{3}{({4.3}^{x} - 1)}$
$\Rightarrow 3^{1 - x} + 2 = 3({4.3}^{x} - 1)$
$\Rightarrow {3.3}^{- x} + 2 = {12.3}^{x} - 3$
Let $3^{x} = t$
$\therefore\frac{3}{t} + 2 = 12t - 3$
$\Rightarrow 12t^{2} - 5t - 3 = 0$
$\Rightarrow \ t = - \frac{1}{3},\frac{3}{4}$
$\therefore\ 3^{x} = \frac{3}{4}$ [$\because 3^{x}$ cannot be negative]
$\Rightarrow \ x = \log_{3}\left( \frac{3}{4} \right)\ \Rightarrow x = 1 - \log_{3}4\$