Chapter 7.3, Problem 51PPE

To determine

To complete: The equation ${\left(m^2{n}^3\right)}^{\_}=\frac{1}{m^6{n}^9}$

Answer to Problem 51PPE

The required answer is: ${\left(m^2{n}^3\right)}^{-3}=\frac{1}{m^6{n}^9}$

Explanation of Solution

Given information:

The given equation:

  ${\left(m^2{n}^3\right)}^{\_}=\frac{1}{m^6{n}^9}$

Calculation:

  $\begin{array}{l}\Rightarrow {\left(m^2{n}^3\right)}^{\_}=\frac{1}{m^6{n}^9}\\ \Rightarrow {\left(m^2{n}^3\right)}^{-3}=\frac{1}{m^6{n}^9}\end{array}$

(Multiplying exponents when raising a power to a power)

  $\begin{array}{l}\Rightarrow \left(m^{-6}{n}^{-9}\right)=\frac{1}{m^6{n}^9}\\ \Rightarrow \frac{1}{m^6{n}^9}=\frac{1}{m^6{n}^9}\end{array}$

(Simplify)

Therefore, ${\left(m^2{n}^3\right)}^{-3}=\frac{1}{m^6{n}^9}$