Chapter 5, Problem 4CT

To determine

To simplify the answer only positive-exponent of given expression : ${\left(\frac{u^2}{v}\right)}^{-2}{\left(\frac{v^3}{u}\right)}^{-1}$

Answer to Problem 4CT

The answer is $\frac{1}{u^{3}v}$ .

Explanation of Solution

Given that: : ${\left(\frac{u^2}{v}\right)}^{-2}{\left(\frac{v^3}{u}\right)}^{-1}$

Apply law of Exponents ${\left(\frac{a}{b}\right)}^n=\frac{a^n}{b^n}$

  ${\left(\frac{u^2}{v}\right)}^{-2}{\left(\frac{v^3}{u}\right)}^{-1}=\frac{{\left(u^2\right)}^{-2}}{v^{-2}}.\frac{{\left(v^3\right)}^{-1}}{u^{-1}}$

Now, apply law of Exponent: ${\left(a^n\right)}^m=a^{nm}$

  $\frac{\left(u^{-4}\right)}{v^{-2}}.\frac{\left(v^{-3}\right)}{u^{-1}}$

Rearrange in positive exponent

  $\begin{array}{l}\frac{\left(v^2\right)}{u^4}.\frac{\left(u^1\right)}{v^3}\\ =u^{(1-4)}{v}^{(2-3)}[a^{-n}=\frac{1}{a^n}]\\ =u^{-3}{v}^{-1}=\frac{1}{u^{3}v}\end{array}$

Hence, the answer is $\frac{1}{u^{3}v}$ .