Question

If $x=2$ and $y=4$, then

${\left(\frac{x}{y}\right)}^{x-y}+{\left(\frac{y}{x}\right)}^{y-x}=$ ?

• A. 4
• B. 8
• C. 12
• D. 2

Answer 1

Answer: (B)

Given:

$x=2$ and $y=4$

${\left(\frac{x}{y}\right)}^{x-y}+{\left(\frac{y}{x}\right)}^{y-x}$

$={\left(\frac{2}{4}\right)}^{2-4}+{\left(\frac{4}{2}\right)}^{4-2}$

$={\left(\frac{1}{2}\right)}^{-2}+(2{)}^2$

$=2^2+2^2$ [Since, $(\frac{1}{a}{)}^{-m}=a^m,a\ne 0$]

$=4+4$

$=8$

$\therefore ,{\left(\frac{x}{y}\right)}^{x-y}+{\left(\frac{y}{x}\right)}^{y-x}=8$ when $x=2$ and $y=4$