Question

Find the values of $x$ and $y$, if $y^x=x^y$ and $x=2y$

Answer 1

Given equations
$y^x=x^y$ ....(1)
and $x=2y$ ...(2)
Taking log on both sides of eqn (1), we get
$x\log y=y\log x$ ($\because \log x^m=m\log x$)
$2y\log y=y\log 2y$ (by (2))
$\Rightarrow 2y\log y=y\log 2+y\log y$ ($\because \log mn=\log m+\log n$
$\Rightarrow y\log y-y\log 2=0$
$\Rightarrow y(\log y-\log 2)=0$
$\Rightarrow y=0$ or $\log y=\log 2$
$\Rightarrow y=0$ or $y=2$
But $y=2$ is only possible as $y>0$ for log to be defined.
Hence, $x=2\left(2\right)=4$

Hence, $x=4,y=2$