Question

Solve $(\frac{x^{-1}{y}^2}{x^2{y}^{-4}}{)}^7÷(\frac{x^3{y}^{-5}}{x^{-2}{y}^3}{)}^{-5}=x^p.y^q$ , then $p+q$

Answer 1

We have,

${\left(\frac{x^{-1}{y}^2}{x^2{y}^{-4}}\right)}^7÷{\left(\frac{x^3{y}^{-5}}{x^{-2}{y}^3}\right)}^{-5}=x^p.y^q$

${\left(x^{-3}{y}^6\right)}^7÷{\left(x^5{y}^{-8}\right)}^{-5}=x^p.y^q$

$\left(x^{-21}{y}^{42}\right)÷\left(x^{-25}{y}^{40}\right)=x^p.y^q$

$x^4.y^2=x^p.y^q$

So,

$p=4,q=2$

Therefore,

$p+q=4+2=6$

Hence, this is the answer.