Q. 128 - Intermediate Algebra

Question

In the following exercises, simplify each expression by applying several properties.

$$\begin{gathered} \left(a\right). {\left(10k^4\right)}^3{\left(5k^6\right)}^2 \\ \left(b\right). \frac{{\left(q^3\right)}^6{\left(q^{-2}\right)}^3}{{\left(q^4\right)}^8} \end{gathered}$$

Step-by-Step Solution

Answer

The simplified form of the expressions

$$\begin{gathered} \left(a\right). {\left(10k^4\right)}^3{\left(5k^6\right)}^2=25000k^{24} \\ \left(b\right). \frac{{\left(q^3\right)}^6{\left(q^{-2}\right)}^3}{{\left(q^4\right)}^8}=\frac{1}{q^{20}} \end{gathered}$$

Part (a) Step 1. Given information.

The given expression is ${\left(10k^4\right)}^3{\left(5k^6\right)}^2$.

Part (a) Step 2. Calculate the expression.

Apply product to a power property

We get, ${10}^3.k^{12} \times 5^2.k^{12}$

$25000k^{24}$

Part (b) Step 1. Given information.

The given expression is $\frac{{\left(q^3\right)}^6{\left(q^{-2}\right)}^3}{{\left(q^4\right)}^8}$.

Part (b) Step 2. Calculate the expression.

First of all, apply power property

We get,

$\frac{q^{18} \times q^{-6}}{q^{32}}$

Now, apply the property of the negative exponent and cancel out the common terms

We get,

$\frac{q^{12}}{q^{32}}=\frac{1}{q^{20}}.$