Question

If $x^y=y^x$ and $x=2y$, then the values of $x$ and $y$ are $(x,y>0)$

• A. $x=4,y=2$
• B. $x=3,y=2$
• C. $x=1,y=1$
• D. $noneofthese$

Answer 1

Answer: (A)

$x=4,y=2$

Consider the given equation,

$x^y=y^x$ ……(1)

$x=2y..........\left(2\right)$

From equation (1),

$y\log x=x\log y$ taking log both sides,

$\frac{\log x}{x}=\frac{\log y}{y}$

$\frac{\log 2y}{2y}=\frac{\log y}{y}$

$\frac{1}{2}\log 2y=\log y$

$\log {\left(2y\right)}^{\frac{1}{2}}=\log y$

$\sqrt{2}{y}^{\frac{1}{2}}=y$

$\sqrt{2}{y}^{\frac{1}{2}}-y=0$

$y^{\frac{1}{2}}(\sqrt{2}-y^{\frac{1}{2}})=0$

When ,

$y^{\frac{1}{2}}=0$

$y=0$

When,

$\sqrt{2}-y^{\frac{1}{2}}=0$

$y^{\frac{1}{2}}=2^{\frac{1}{2}}$

Taking, square both sides,

$y=2$

Now, put the value of $y=0$ and 2 in equation (2), we get

$X=0,4$

Hence , $(0,0)$ and $(4,2)$ are the answes.