Question

${\left(\frac{a^{-1}{b}^2}{a^2{b}^{-4}}\right)}^7÷{\left(\frac{a^3{b}^{-5}}{a^{-2}{b}^3}\right)}^{-5}$ is equal to:

• A. $a^4{b}^2$
• B. $a^2{b}^4$
• C. $a^3{b}^2$
• D. $a^2{b}^3$

Answer 1

The correct option is A $a^4{b}^2$

${\left(\frac{a^{-1}{b}^2}{a^2{b}^{-4}}\right)}^7÷{\left(\frac{a^3{b}^{-5}}{a^{-2}{b}^3}\right)}^{-5}=\frac{a^{-7}{b}^{14}}{a^{14}{b}^{-28}}÷\frac{a^{-15}{b}^{25}}{a^{10}{b}^{-15}}$

$=[a^{-7-14}\times b^{14-(-28)}]÷[a^{-15-10}\times b^{25-(-15)}]$

$=\frac{a^{-21}{b}^{42}}{a^{-25}{b}^{40}}$

$=a^{-21-(-25)}\times b^{42-40}$

$=a^4{b}^2$