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

$\therefore x + y = 1$.