(c) $4.\,{9^{x – 1}} = 3\,.\,\sqrt {({2^{2x + 1}})} $$ \Rightarrow $${3^{2x – 2 – 1}} = {2^{{{2x + 1} \over 2} – 2}}$

$ \Rightarrow $ ${3^{2x – 3}} = {2^{{{2x – 3} \over 2}}}$ $ \Rightarrow $ ${2^{{{2x – 3} \over 2}}} = {\left( {{3^{{{2x – 3} \over 2}}}} \right)^2}$
 $ \Rightarrow $$2x – 3 = 0$,

$\therefore x = {3 \over 2}$.