Question

If $m\ne n$ and $(m+n{)}^{-1}(m^{-1}+n^{-1})=m^x{n}^y;$, then $x+y+2$ is equal to:

Answer 1

Given equation is $(m+n{)}^{-1}(m^{-1}+n^{-1})=m^x{n}^y$
By using law $a^{-m}=\frac{1}{a^m}$
$=>\frac{1}{m+n}(\frac{1}{m}+\frac{1}{n})=m^x{n}^y$
$=>\frac{1}{m+n}\left(\frac{m+n}{mn}\right)=m^x{n}^y$
$=>\frac{1}{mn}=m^x{n}^y$
$=>m^x{n}^y\times mn=1$
$=>m^{x+1}{n}^{y+1}=1$
$=>x+1+y+1=0$
$=>x+y+2=0$
Hence proved.