Question

If $x^y=y^x$ and $x^2=y,$ where $x$ and $y$ are distinct Positive real number, then the value of $x+2y$ is :

• A. $3$
• B. $8$
• C. $10$
• D. $12$

Answer 1

Answer: (C)

$10$

Given, $x^2=y$
$\therefore x^y=y^x\Rightarrow x^{x^2}=(x^2{)}^x$
$\Rightarrow x^{x^2}=x^{2x}$
$\Rightarrow x^2=2x$
$\Rightarrow x^2-2x=0$
$\Rightarrow x=0,2$
But, $x=0$ is not possible because $x$ and $y$ are distinct.
Now, $y=x^2=4$
$\Rightarrow x+2y=2+2\times 4=10$