Chapter 8, Problem 10CA

To determine

To find:The expressions which are equivalent to ${(b^{-5})}^{-4}$

Answer to Problem 10CA

The expressions which are equivalent to ${(b^{-5})}^{-4}$ are ${(b^{-4})}^{-5}$ , ${(b^{10})}^2$ , $b^{20}$ , $b^{12}{b}^8$ .

Explanation of Solution

Given information:The expressions which may be equivalent to ${(b^{-5})}^{-4}$ are,

  $b^{-20}$ , $b^{-6}{b}^{-14}$ , ${(b^{-4})}^{-5}$ , $b^{-9}$ , ${(b^{-2})}^{-7}$ , $b^{-5}{b}^{-4}$ , ${(b^{10})}^2$ , $b^{20}$ , ${(b^{-12})}^3$ , $b^{12}{b}^8$

Calculation:Simplifying the expressions,

1. ${(b^{-5})}^{-4}$

  $\begin{array}{l}$ =b^{(-5)\times (-4)}[\because {(a^x)}^y=a^{xy}]$=b^{20}\end{array}$

2. $b^{-20}$

  $=\frac{1}{b^{20}}$

3. $b^{-6}{b}^{-14}$

  $\begin{array}{l}=b^{-6-14}[\because a^x{a}^y=a^{x+y}]\\ =b^{-20}\\ =\frac{1}{b^{20}}\end{array}$

4. ${(b^{-4})}^{-5}$

  $\begin{array}{l}$ =b^{(-4)\times (-5)}[\because {(a^x)}^y=a^{xy}]$=b^{20}\end{array}$

5. $b^{-9}$

  $=\frac{1}{b^9}$

6. ${(b^{-2})}^{-7}$

  $\begin{array}{l}$ =b^{(-2)\times (-7)}[\because {(a^x)}^y=a^{xy}]$=b^{14}\end{array}$

7. $b^{-5}{b}^{-4}$

  $\begin{array}{l}=b^{-5-4}[\because a^x{a}^y=a^{x+y}]\\ =b^{-9}\\ =\frac{1}{b^9}\end{array}$

8. ${(b^{10})}^2$

  $\begin{array}{l}$ =b^{10\times 2}[\because {(a^x)}^y=a^{xy}]$=b^{20}\end{array}$

9. $b^{20}$
10. ${(b^{-12})}^3$

  $\begin{array}{l}$ =b^{(-12)\times 3}[\because {(a^x)}^y=a^{xy}]$=b^{-36}\\ =\frac{1}{b^{36}}\end{array}$

11. $b^{12}{b}^8$

  $\begin{array}{l}$ =b^{12+8}[\because a^x{a}^y=a^{x+y}]$=b^{20}\end{array}$

Thus, the expressions which are equivalent to ${(b^{-5})}^{-4}$ are ${(b^{-4})}^{-5}$ , ${(b^{10})}^2$ , $b^{20}$ , $b^{12}{b}^8$ .