b
(b) \({a^x} = {b^y} = {(ab)^{xy}}\)

\( \Rightarrow \)\(x\ln a = y\ln b = xy\ln (ab) = k\,{\rm{(say)}}\)

\(\ln a = {k \over x},\,\,\ln b = {k \over y}\)

\(\ln (a\,b) = {k \over {xy}}\)\( \Rightarrow \) \(\ln a + \ln b = {k \over {xy}}\)

\( \Rightarrow\) \({k \over x} + {k \over y} = {k \over {xy}}\)

\( \Rightarrow \) \({1 \over x} + {1 \over y} = {1 \over {xy}}\) \( \Rightarrow \) \({{x + y} \over {xy}} = {1 \over {xy}}\);

\(\therefore x + y = 1\).