(b) ${9^x} – {2^{x + (1/2)}} = {2^{x + (3/2)}} – {3^{2x – 1}}$

==> ${3^{2x}} + {1 \over 3}{.3^{2x}} = {2.2^{x + {1 \over 2}}} + {2^{x + {1 \over 2} – 2}}$

==> $4\,.\,{3^{2x – 1}} = 3.\,{2^{x + {1 \over 2}}}$ ==> ${3^{2x – 2}} = {2^{x + {1 \over 2} – 2}}$

==> ${3^{2x – 2}} = {2^{x – {3 \over 2}}}$ ==> ${\left( {{9 \over 2}} \right)^{x – 1}} = {2^{ – 1/2}}$

==> $(x – 1)\,{\log _{9/2}}9/2 = – {1 \over 2}{\log _{9/2}}2$

==> $x – 1 = – {1 \over 2}{\log _{9/2}}2$

==> $x = 1 – {\log _{9/2}}\sqrt 2 = {\log _{9/2}}9/2 – {\log _{9/2}}\sqrt 2 $

==> $x = {\log _{9/2}}(9/2\sqrt 2 )$;

$\therefore x = {\log _{9/2}}(9/\sqrt 8 )$.