Chapter 6.1, Problem 69E

(a)

To determine

To compare: the values of $a^n\text{ and }{a}^{-n}$ when $n<0,n=0\text{ and }n>0$ for $a>1$ .

Answer to Problem 69E

For $a>1$

When $n<0$ , then $a^n\text{ < }{a}^{-n}$

When $n=0$ , then $a^n\text{ = }{a}^{-n}=1$

When $n>0$ , then $a^n\text{ > }{a}^{-n}$

Explanation of Solution

Since $a>1$ and $n<0$ then $a^n$ will be always less than 1 and $a^{-n}$ will be always greater than 1

And hence $a^n\text{ < }{a}^{-n}$

When $a>1$ and $n=0$ then $a^n{\text{=a}}^0\text{=1}$ and $a^{-n}=a^0=1$

And hence $a^n\text{ = }{a}^{-n}=1$

When $a>1$ and $n>0$ then $a^n$ will be always greater than 1 and $a^{-n}$ will be always less than 1

And hence $a^n\text{ > }{a}^{-n}$

(b)

To determine

To compare: the values of $a^n\text{ and }{a}^{-n}$ when $n<0,n=0\text{ and }n>0$ for $0<a<1$ .

Answer to Problem 69E

For $0<a<1$

When $n<0$ , then $a^n\text{ > }{a}^{-n}$

When $n=0$ , then $a^n\text{ = }{a}^{-n}=1$

When $n>0$ , then $a^n\text{ < }{a}^{-n}$

Explanation of Solution

Since $0<a<1$ and $n<0$ then $a^n$ will be always greater than 1 and $a^{-n}$ will be always less than 1

And hence $a^n\text{ > }{a}^{-n}$

When $0<a<1$ and $n=0$ then $a^n{\text{=a}}^0\text{=1}$ and $a^{-n}=a^0=1$

And hence $a^n\text{ = }{a}^{-n}=1$

When $0<a<1$ and $n>0$ then $a^n$ will be always less than 1 and $a^{-n}$ will be always greater than 1

And hence $a^n\text{ < }{a}^{-n}$