Chapter 7.3, Problem 33PPE

To determine

To simplify: The expression ${\left(2j^2{k}^4\right)}^{-5}{\left(k^{-1}{j}^7\right)}^6$ .

Answer to Problem 33PPE

The required answer is $\frac{j^{32}}{32k^{26}}$ .

Explanation of Solution

Given information:

The given expression ${\left(2j^2{k}^4\right)}^{-5}{\left(k^{-1}{j}^7\right)}^6$ .

Calculation:

  $\begin{array}{l}={\left(2j^2{k}^4\right)}^{-5}{\left(k^{-1}{j}^7\right)}^6\\ =2^{-5}{\left(j^2\right)}^{-5}{\left(k^4\right)}^{-5}{\left(k^{-1}\right)}^6{\left(j^7\right)}^6\end{array}$

(Raise each factor to the second power)

  $\begin{array}{l}=2^{-5}{j}^{2\cdot \left(-5\right)}{k}^{4\cdot \left(-5\right)}{k}^{-1\cdot 6}{j}^{7\cdot 6}\\ =2^{-5}{j}^{-10}{k}^{-20}{k}^{-6}{j}^{42}\end{array}$

(Multiply the exponents of a power raised to a power)

  $\begin{array}{l}=2^{-5}{j}^{-10}{j}^{42}{k}^{-20}{k}^{-6}\\ =2^{-5}{j}^{-10+42}{k}^{-20+\left(-6\right)}\end{array}$

(Add exponents when multiplying powers with same base)

  $\begin{array}{l}=\frac{1}{32}\cdot j^{32}{k}^{-26}\\ =\frac{j^{32}}{32k^{26}}\end{array}$

(Simplify and the expression using only positive exponents)