The $x^\text{th}$ root of $x$

$Let\quad x=\frac { p }{ q } \quad and\quad y=\frac { r }{ s } ,\quad where\quad p,q\quad are\quad coprime\quad positive\quad integers,\quad and\quad similarily\\ for\quad r\quad and\quad s.\\ Now\quad \sqrt [ \frac { p }{ q } ]{ \frac { p }{ q } } =\sqrt [ \frac { r }{ s } ]{ \frac { r }{ s } } \Rightarrow { \frac { p }{ q } }^{ \frac { q }{ p } }={ \frac { r }{ s } }^{ \frac { s }{ r } }.\\ Taking\quad logarithms\quad w.r.t.\quad \frac { p }{ q } \quad on\quad both\quad sides\quad we\quad get\\ \frac { q }{ p } =\frac { s }{ r } \log _{ \frac { p }{ q } }{ \frac { r }{ s } } .\\ It\quad follows\quad that\quad \log _{ \frac { p }{ q } }{ \frac { r }{ s } } \quad must\quad be\quad a\quad rational\quad number,\quad i.e.\quad \frac { p }{ q } \\ is\quad a\quad rational\quad power\quad of\quad \frac { r }{ s } .\\ Let\quad p={ r }^{ n },\quad where\quad n\quad is\quad a\quad rational\quad number\quad not\quad equal\quad to\quad 1\quad (since\quad x\neq y).\quad \\ Then\quad { r }^{ \frac { n{ s }^{ n } }{ { r }^{ n } } }={ \quad r }^{ \frac { s }{ r } }(as\quad r\quad and\quad s\quad are\quad coprime).\\ Simplifying\quad we\quad obtain\quad n={ (\frac { r }{ s } ) }^{ n-1 }.\quad Therefore\quad \sqrt [ n-1 ]{ n } \quad is\quad rational.\quad \\ Letting\quad n=\frac { t }{ u } ,\quad where\quad t,\quad u\quad are\quad coprime\quad integers\quad we\quad see\quad that\quad t\quad and\quad u\quad \\ must\quad both\quad be\quad rational\quad numbers\quad raised\quad to\quad the\quad powers\quad of\quad t-u.\quad \\ The\quad only\quad solution\quad for\quad this\quad is\quad t=2\quad and\quad u=1\quad (or\quad vice\quad versa\quad but\quad this\quad is\quad impossible\quad as\quad x<y).\quad \\ Hence\quad x=2\quad and\quad y=4\quad is\quad the\quad only\quad solution.\quad So\quad the\quad answer\quad is\quad \boxed { 2 } .$