Math, asked by 3manas3, 6 hours ago


( \frac { x } { y } ) ^ { n - 1 } = ( \frac { y } { x } ) ^ { n - 3 }

Answers

Answered by lalnunkimahmarjoute
0

( \frac { x } { y } ) ^ { n - 1 } = ( \frac { y } { x } ) ^ { n - 3 }

( \frac { x } { y } ) ^ { n - 1 } =  \frac{ {y}^{n - 3} }{ {x}^{n - 3} }

( \frac { x } { y } ) ^ { n - 1 } =  \frac{ {x}^{ - (n - 3)} }{ {y}^{ - (n - 3)} }

( \frac { x } { y } ) ^ { n - 1 } =  \frac{ {x}^{3 - n} }{ {y}^{3 - n} }

( \frac { x } { y } ) ^ { n - 1 } = ( \frac { x} { y } ) ^ { 3 - n }

Now\:their\:bases\:are\:same.\:Lets\:compare\:their\:powers

n - 1 = 3 - n

n + n = 3 + 1

2n = 4

∴n = 2

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