Chapter 1.2, Problem 50E

(a)

To determine

To simplify: given expression $\frac{5x{y}^{-2}}{x^{-1}{y}^{-3}}$ .

Answer to Problem 50E

  $\frac{5x{y}^{-2}}{x^{-1}{y}^{-3}}=5x^{2}y$

Explanation of Solution

Given information:

Given expression

  $\frac{5x{y}^{-2}}{x^{-1}{y}^{-3}}$

Calculation:

Consider the given expression.

  $\frac{5x{y}^{-2}}{x^{-1}{y}^{-3}}$

Now, use the first and second law of exponents, which states

  $a^m\cdot a^n=a^{m+n},\frac{a^m}{a^n}=a^{m-n}$

And combine like terms

  $\begin{array}{c}\frac{5x{y}^{-2}}{x^{-1}{y}^{-3}}=5\cdot x^{1-\left(-1\right)}{y}^{-2-\left(-3\right)}\\ =5x^{1+1}{y}^{-2+3}\\ =5x^2{y}^1\\ =5x^{2}y\end{array}$

Hence,

  $\frac{5x{y}^{-2}}{x^{-1}{y}^{-3}}=5x^{2}y$

(b)

To determine

To evaluate: the given expression ${\left(\frac{2a^{-1}b}{a^2{b}^{-3}}\right)}^{-3}$ .

Answer to Problem 50E

  ${\left(\frac{2a^{-1}b}{a^2{b}^{-3}}\right)}^{-3}=\frac{a^9}{8b^{12}}$

Explanation of Solution

Given information:

Given expression

  ${\left(\frac{2a^{-1}b}{a^2{b}^{-3}}\right)}^{-3}$

Calculation:

Consider the given expression.

  ${\left(\frac{2a^{-1}b}{a^2{b}^{-3}}\right)}^{-3}$

Now. Simplify using law of exponents ${\left(ab\right)}^n=a^n{b}^n$ and ${\left(a^m\right)}^n=a^{mn}$ .

  $\begin{array}{c}{\left(\frac{2a^{-1}b}{a^2{b}^{-3}}\right)}^{-3}=\frac{2^{-3}{\left(a^{-1}\right)}^{-3}{b}^{-3}}{{\left(a^2\right)}^{-3}{\left(b^{-3}\right)}^{-3}}\\ =\frac{a^3{b}^{-3}}{2^3{a}^{-6}{b}^9}\end{array}$

Now, use the second law of exponents, which states

  $\frac{a^m}{a^n}=a^{m-n}$

And combine like terms

  $\begin{array}{c}\frac{a^3{b}^{-3}}{2^3{a}^{-6}{b}^9}=\frac{1}{8}{a}^{3-\left(-6\right)}{b}^{-3-9}\\ =\frac{1}{8}{a}^{3+6}{b}^{-12}\\ =\frac{a^9}{8b^{12}}\end{array}$

Hence,

  ${\left(\frac{2a^{-1}b}{a^2{b}^{-3}}\right)}^{-3}=\frac{a^9}{8b^{12}}$