Q 119. - Intermediate Algebra

Question

In the following exercises, simplify each expression using the Product to a Power Property. 

$$\begin{gathered} \left(a\right) {\left(-3xy\right)}^2 \\ {\left(b\right) \left(6a\right)}^0 \\ \left(c\right) {\left(5x^2\right)}^{-2} \\ \left(d\right) {\left(-4y^{-3}\right)}^2 \end{gathered}$$

Step-by-Step Solution

Answer

$$\begin{gathered} \mathrm{Part} \left(\mathrm{a}\right) 9x^2{y}^2 \\ \mathrm{Part} \left(\mathrm{b}\right) 1 \\ \mathrm{Part} \left(\mathrm{c}\right) \frac{1}{25x^4} \\ \mathrm{Part} \left(\mathrm{d}\right) \frac{16}{y^6} \end{gathered}$$

Part (a) Step 1. Simplify the given expression.

In the given expression, we will use the Product to a Power Property,

${\left(-3xy\right)}^2={\left(-3\right)}^2{x}^2{y}^2$

Simplify,

$9x^2{y}^2$

Part (b) Step 1. Simplify the given expression.

Use the Product to a Power Property,

$$\begin{gathered} {\left(6a\right)}^0=6^0{a}^0 \\ =1·1 \\ =1 \end{gathered}$$

Part (c) Step 1. Simplify the given expression.

Use the Product to a Power Property and simplify.

$$\begin{gathered} {\left(5x^2\right)}^{-2}={\left(5\right)}^{-2}{\left(x^2\right)}^{-2} \\ =\frac{1}{5^2}·\frac{1}{{\left(x^2\right)}^2} \\ =\frac{1}{25x^4} \end{gathered}$$

Part (d) Step 1. Simplify the given expression.

Apply the concept of power product.

$$\begin{gathered} {\left(-4y^{-3}\right)}^2={\left(-4\right)}^2{\left(y^{-3}\right)}^2 \\ =16y^{-3·2} \\ =16y^{-6} \\ =16·\frac{1}{y^6} \\ =\frac{16}{y^6} \end{gathered}$$