Chapter 7.9, Problem 23WE

To determine

To calculate: The simplified form of the expression $\frac{{\left(2^{-3}\cdot 4^2\right)}^{-1}}{2^{-1}}$ .

Answer to Problem 23WE

The simplified form of the expression $\frac{{\left(2^{-3}\cdot 4^2\right)}^{-1}}{2^{-1}}$ is $1$ .

Explanation of Solution

Given information:

The expression $\frac{{\left(2^{-3}\cdot 4^2\right)}^{-1}}{2^{-1}}$ .

Formula used:

Law of exponent for product of power is $a^m\cdot a^n=a^{m+n}$ where a andb are nonzero integers and n and m are positive integers.

Law of exponent for power of power is ${\left(a^m\right)}^n=a^{mn}$ where a is a nonzero integers and n and m are positive integers.

Calculation:

Consider the expression $\frac{{\left(2^{-3}\cdot 4^2\right)}^{-1}}{2^{-1}}$ .

Recall that law of exponent for product of power is $a^m\cdot a^n=a^{m+n}$ where a andb are nonzero integers and n and m are positive integers.

Apply it,

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

Recall that law of exponent for power of power is ${\left(a^m\right)}^n=a^{mn}$ where a is a nonzero integers and n and m are positive integers.

Apply it,

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

Thus, the simplified form of the expression $\frac{{\left(2^{-3}\cdot 4^2\right)}^{-1}}{2^{-1}}$ is $1$ .