Q. 120. - Intermediate Algebra

Question

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

(a) (-4ab)² 

(b) (5x)⁰ 

(c) (4y³)^-3 

(d) (-7y^-3)²

Step-by-Step Solution

Answer

The simplified form of the given expression is :-

$$\begin{gathered} \left(a\right) 16a^2{b}^2 \\ \left(b\right) 1 \\ \left(c\right) \frac{1}{64y^9} \\ \left(d\right) 49y^{-6} \end{gathered}$$

Part (a) Step 1 . Given information.

The given expression is ${(-4ab)}^2$.

Part (a). Step 2 . Apply the power rule.

Here we use the concept of power rule,

${\left(ab\right)}^n=a^n b^n$

So,

${(-4ab)}^2=16a^2{b}^2.$

Part (b) Step 1 . Given information.

The given expression is ${\left(5x\right)}^0.$

part (b). Step 2 . Apply the concept.

Here we apply the concept of anything raised to zero is $1$.

So, 

${\left(5x\right)}^0=1.$

Part (c) Step 1 . Given information.

The given expression is ${\left(4y^3\right)}^{-3}.$

Part (c) Step 2 . We apply the concept.

In this expression we use negative exponents rule that is-

$b^{-n}=\frac{1}{b^n}.$

 Our final expression is $\frac{1}{64y^9}.$

Part (d) step 1 . Given information.

The given expression is ${(-7y^{-3})}^2.$

Part (d) Step 2 . Apply the concept.

We use the concept of negative exponents.

${(-ab)}^2=a^2{b}^2.$

So,

${(-7y^{-3})}^2 = 49y^{-6}$