Chapter 1.2, Problem 39E

(a)

To determine

To simplify: given expression ${\left(a^2{a}^4\right)}^3$ .

Answer to Problem 39E

  ${\left(a^2{a}^4\right)}^3=a^{18}$ .

Explanation of Solution

Given information:

Given expression

  ${\left(a^2{a}^4\right)}^3$

Calculation:

Consider the given expression.

  ${\left(a^2{a}^4\right)}^3$

To simplify this, use the rule of product of exponents, which states

  $\begin{array}{c}{a}^m{a}^n=a^{m+n}\\ {\left(a^2{a}^4\right)}^3={\left(a^{2+4}\right)}^3\\ ={\left(a^6\right)}^3\end{array}$

Now, use ${\left(a^m\right)}^n=a^{mn}$ ,

  $\begin{array}{c}{\left(a^2{a}^4\right)}^3={\left(a^6\right)}^3\\ =a^{6\times 3}\\ =a^{18}\end{array}$

(b)

To determine

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

Answer to Problem 39E

  ${\left(\frac{a^2}{4}\right)}^3=\frac{a^6}{64}$

Explanation of Solution

Given information:

Given expression

  ${\left(\frac{a^2}{4}\right)}^3$ .

Calculation:

Consider the given expression.

  ${\left(\frac{a^2}{4}\right)}^3$

To simplify this, use the fifth law of exponents, which states

  $\begin{array}{c}{\left(\frac{a}{b}\right)}^n=\frac{a^n}{b^n}\\ {\left(\frac{a^2}{4}\right)}^3=\frac{{\left(a^2\right)}^3}{4^3}\\ =\frac{{\left(a^2\right)}^3}{64}\end{array}$

Now, use the third law of exponents, which states

  $\begin{array}{c}{\left(a^m\right)}^n=a^{mn}\\ {\left(\frac{a^2}{4}\right)}^3=\frac{{\left(a^2\right)}^3}{64}\\ =\frac{a^6}{64}\end{array}$

(c)

To determine

To evaluate: given expression ${\left(3z\right)}^2{\left(6z^2\right)}^{-3}$ .

Answer to Problem 39E

  ${\left(3z\right)}^2{\left(6z^2\right)}^{-3}=\frac{1}{24z^4}$

Explanation of Solution

Given information:

Given expression

  ${\left(3z\right)}^2{\left(6z^2\right)}^{-3}$

Calculation:

Consider the given expression.

  ${\left(3z\right)}^2{\left(6z^2\right)}^{-3}$

To simplify this, use the fourth law of exponents, which states

  $\begin{array}{c}{\left(ab\right)}^n=a^n{b}^n\\ {\left(3z\right)}^2{\left(6z^2\right)}^{-3}={\left(3\right)}^2{\left(z\right)}^2{\left(6\right)}^{-3}{\left(z^2\right)}^{-3}\\ =9z^2\cdot \frac{1}{6\cdot 6\cdot 6}{z}^{-6}\left[\text{use}{\left(a^m\right)}^n=a^{mn}\right]\\ =\frac{z^{2-6}}{24}\left[\text{use}{a}^m\times a^n=a^{m+n}\right]\\ =\frac{1}{24z^4}\left[\text{use}{a}^{-m}=\frac{1}{a^m}\right]\end{array}$