Explanation

(d)
We have, $2^{x + 2}3^{3x/(x - 1)} = 9$
Taking log on both sides, we get
$(x + 2)\log{2 + \left( \ \frac{3x}{x - 1} \right)\log{3 = 2\log 3}}$
$\Rightarrow \ (x + 2)\left( \log{2 + \frac{1}{x - 1}\log 3} \right) = 0$
$\Rightarrow x = - 2\ or\ \frac{1}{1 - x} = \frac{\log 2}{\log 3}$
$\Rightarrow 1 - x = \frac{\log 3}{\log 2}$
$\Rightarrow x = 1 - \frac{\log 3}{\log 2}$