b
(b) \({9^x} - {2^{x + (1/2)}} = {2^{x + (3/2)}} - {3^{2x - 1}}\)

 ==> \({3^{2x}} + {1 \over 3}{.3^{2x}} = {2.2^{x + {1 \over 2}}} + {2^{x + {1 \over 2} - 2}}\)

==> \(4\,.\,{3^{2x - 1}} = 3.\,{2^{x + {1 \over 2}}}\) ==> \({3^{2x - 2}} = {2^{x + {1 \over 2} - 2}}\)

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

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

==> \(x - 1 = - {1 \over 2}{\log _{9/2}}2\)

==> \(x = 1 - {\log _{9/2}}\sqrt 2 = {\log _{9/2}}9/2 - {\log _{9/2}}\sqrt 2 \)

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

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