Q.92 - Intermediate Algebra

Question

In the following exercises, simplify each expression using the properties for exponents.

$$\begin{gathered} \left(a\right)\frac{u^{24}}{u^3} \\ \left(b\right)\frac{9^{15}}{9^5} \\ \left(c\right)\frac{x}{x^7} \\ \left(d\right)\frac{10}{{10}^3} \end{gathered}$$

Step-by-Step Solution

Answer

The simplified form is:

$$\begin{gathered} \left(a\right)\frac{u^{24}}{u^3}=u^{21} \\ \left(b\right)\frac{9^{15}}{9^5}=9^{10} \\ \left(c\right)\frac{x}{x^7}=\frac{1}{x^6} \\ \left(d\right)\frac{10}{{10}^3}=\frac{1}{100} \end{gathered}$$

Step 1

Some properties of exponents are:

$$\begin{gathered} \frac{a^m}{a^n}=a^{m-n},m>n \\ \frac{a^m}{a^n}=\frac{1}{a^{n-m}},n>m \end{gathered}$$

Step 2

Use the property $\frac{a^m}{a^n}=a^{m-n},m>n$to simplify the expressions:

$\begin{array}{rcl}\left(a\right)\frac{u^{24}}{u^3}& =& u^{24-3}\\ & =& u^{21}\\ \left(b\right)\frac{9^{15}}{9^5}& =& 9^{15-5}\\ & =& 9^{10}\end{array}$

Step 3

Use the property that $\frac{a^m}{a^n}=\frac{1}{a^{n-m}},n>m$to simplify the expressions:

$\begin{array}{rcl}\left(c\right)\frac{x}{x^7}& =& \frac{x^1}{x^7}\\ & =& \frac{1}{x^{7-1}}\\ & =& \frac{1}{x^6}\\ \left(d\right)\frac{10}{{10}^3}& =& \frac{{10}^1}{{10}^3}\\ & =& \frac{1}{{10}^{3-1}}\\ & =& \frac{1}{{10}^2}\\ & =& \frac{1}{100}\end{array}$