Q.97 - Intermediate Algebra

Question

Use the Definition of a Negative Exponent

In the following exercises, simplify each expression.

$$\begin{gathered} \left(a\right)a^{-2} \\ \left(b\right){10}^{-3} \\ \left(c\right)\frac{1}{c^{-5}} \\ \left(d\right)\frac{1}{3^{-2}} \end{gathered}$$

Step-by-Step Solution

Answer

The simplified form of the expressions are:

$\begin{array}{rcl}\left(a\right)a^{-2}& =& \frac{1}{a^2}\\ \left(b\right){10}^{-3}& =& \frac{1}{{10}^3}\\ \left(c\right)\frac{1}{c^{-5}}& =& c^5\\ \left(d\right)\frac{1}{3^{-2}}& =& 3^2\end{array}$

Step 1

The properties of negative exponents:

$\begin{array}{rcl}{a}^{-n}& =& \frac{1}{a^n}\\ \frac{1}{a^{-n}}& =& a^n\end{array}$

Step 2

Use the negative property $\begin{array}{rcl}{a}^{-n}& =& \frac{1}{a^n}\end{array}$, to simplify the expressions:

$\begin{array}{rcl}\left(a\right)a^{-2}& =& \frac{1}{a^2}\\ \left(b\right){10}^{-3}& =& \frac{1}{{10}^3}\end{array}$

Step 3

Use the negative exponent property $\frac{1}{a^{-n}}=a^n$, to simplify the expressions:

$\begin{array}{rcl}\left(c\right)\frac{1}{c^{-5}}& =& c^5\\ \left(d\right)\frac{1}{3^{-2}}& =& 3^2\end{array}$