Chapter 7.9, Problem 49WE

To determine

To calculate: The simplified form of the expression ${\left(a^2\cdot a^{-5}\right)}^2$ .

Answer to Problem 49WE

The simplified form of the expression ${\left(a^2\cdot a^{-5}\right)}^2$ is $\frac{1}{a^6}$ .

Explanation of Solution

Given information:

The expression ${\left(a^2\cdot a^{-5}\right)}^2$ .

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.

Calculation:

Consider the expression ${\left(a^2\cdot a^{-5}\right)}^2$ .

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,

  $\begin{array}{c}{\left(a^2\cdot a^{-5}\right)}^2={\left(a^{2+\left(-5\right)}\right)}^2\\ ={\left(a^{-3}\right)}^2\end{array}$

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,

  $\begin{array}{c}{\left(a^2\cdot a^{-5}\right)}^2={\left(a^{2+\left(-5\right)}\right)}^2\\ ={\left(a^{-3}\right)}^2\\ =a^{-6}\end{array}$

Recall that expression $a^{-n}$ is expressed as $\frac{1}{a^n}$ where a is a nonzero real number and n is a positive integer.

Apply it,

  $\begin{array}{c}{\left(a^2\cdot a^{-5}\right)}^2={\left(a^{2+\left(-5\right)}\right)}^2\\ ={\left(a^{-3}\right)}^2\\ =a^{-6}\\ =\frac{1}{a^6}\end{array}$

Thus, the simplified form of the expression ${\left(a^2\cdot a^{-5}\right)}^2$ is $\frac{1}{a^6}$ .