Chapter 1.2, Problem 52E

(a)

To determine

To simplify: given expression $\left(\frac{s^2{t}^{-4}}{5s^{-1}t}\right)$ .

Answer to Problem 52E

  $\left(\frac{s^2{t}^{-4}}{5s^{-1}t}\right)=\frac{s^3}{5t^5}$

Explanation of Solution

Given information:

Given expression

  $\left(\frac{s^2{t}^{-4}}{5s^{-1}t}\right)$

Calculation:

Consider the given expression.

  $\left(\frac{s^2{t}^{-4}}{5s^{-1}t}\right)$

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

  $\begin{array}{c}\left(\frac{s^2{t}^{-4}}{5s^{-1}t}\right)=\frac{1}{5}{s}^{2-\left(-1\right)}{t}^{-4-1}\\ =\frac{1}{5}{s}^3{t}^{-5}\\ =\frac{s^3}{5t^5}\end{array}$

Hence,

  $\left(\frac{s^2{t}^{-4}}{5s^{-1}t}\right)=\frac{s^3}{5t^5}$

(b)

To determine

To evaluate: the given expression ${\left(\frac{x{y}^{-2}{z}^{-3}}{x^2{y}^3{z}^{-4}}\right)}^{-3}$ .

Answer to Problem 52E

  ${\left(\frac{x{y}^{-2}{z}^{-3}}{x^2{y}^3{z}^{-4}}\right)}^{-3}=\frac{x^3{y}^{15}}{z^3}$

Explanation of Solution

Given information:

Given expression

  ${\left(\frac{x{y}^{-2}{z}^{-3}}{x^2{y}^3{z}^{-4}}\right)}^{-3}$

Calculation:

Consider the given expression.

  ${\left(\frac{x{y}^{-2}{z}^{-3}}{x^2{y}^3{z}^{-4}}\right)}^{-3}$

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

  $\begin{array}{c}{\left(\frac{x{y}^{-2}{z}^{-3}}{x^2{y}^3{z}^{-4}}\right)}^{-3}=\frac{{\left(x\right)}^{-3}{\left(y^{-2}\right)}^{-3}{\left(z^{-3}\right)}^{-3}}{{\left(x^2\right)}^{-3}{\left(y^3\right)}^{-3}{\left(z^{-4}\right)}^{-3}}\\ =\frac{x^{-3}{y}^6{z}^9}{x^{-6}{y}^{-9}{z}^{12}}\end{array}$

Now, use the second law of exponents, which states

  $\frac{a^m}{a^n}=a^{m-n}$

And combine like terms

  $\begin{array}{c}\frac{x^{-3}{y}^6{z}^9}{x^{-6}{y}^{-9}{z}^{12}}=x^{-3-\left(-6\right)}{y}^{6-\left(-9\right)}{z}^{9-12}\\ =x^{-3+6}{y}^{6+9}{z}^{-3}\\ =x^3{y}^{15}{z}^{-3}\\ =\frac{x^3{y}^{15}}{z^3}\end{array}$

Hence,

  ${\left(\frac{x{y}^{-2}{z}^{-3}}{x^2{y}^3{z}^{-4}}\right)}^{-3}=\frac{x^3{y}^{15}}{z^3}$