Q.90 - Intermediate Algebra

Question

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

$$\begin{gathered} \left(a\right)\frac{y^{20}}{y^{10}} \\ \left(b\right)\frac{7^{16}}{7^2} \\ \left(c\right)\frac{t^{10}}{t^{40}} \\ \left(d\right)\frac{8^3}{8^5} \end{gathered}$$

Step-by-Step Solution

Answer

The solution is:

$$\begin{gathered} \left(a\right)\frac{y^{20}}{y^{10}}=y^{10} \\ \left(b\right)\frac{7^{16}}{7^2}=7^{14} \\ \left(c\right)\frac{t^{10}}{t^{40}}=\frac{1}{t^{30}} \\ \left(d\right)\frac{8^3}{8^5}=\frac{1}{64} \end{gathered}$$

Part(a) Step 1. Given information

The given expression is $\frac{y^{20}}{y^{10}}$.

The main objective is to simplify the expression.

Some of the properties of exponents are:

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

Part(a) Step 2. Simplify

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

$\begin{array}{rcl}\frac{y^{20}}{y^{10}}& =& y^{20-10}\\ & =& y^{10}\end{array}$

So, $\frac{y^{20}}{y^{10}}=y^{10}$.

Part(b)-Step 3:Given information

The given expression is $\frac{7^{16}}{7^2}$.

The main objective is to simplify the expression.

Part(b)-Step 2. Simplify

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

$\begin{array}{rcl}\frac{7^{16}}{7^2}& =& 7^{16-2}\\ & =& 7^{14}\end{array}$

So, $\frac{7^{16}}{7^2}=7^{14}$.

Part (c)-Step 1. Given information

The given expression is$\frac{t^{10}}{t^{40}}$.

The main objective is to simplify the expression.

Part(c)-Step 2. Simplify

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

$\begin{array}{rcl}\frac{t^{10}}{t^{40}}& =& \frac{1}{t^{40-10}}\\ & =& \frac{1}{t^{30}}\end{array}$

So, $\frac{t^{10}}{t^{40}}=\frac{1}{t^{30}}$.

Part(d). Step 1. Given information

The given expression is $\frac{8^3}{8^5}$.

The main objective is to simplify the expression.

Part(d)-Step 2. Simplify

Use the property $\frac{a^m}{a^n}=\frac{1}{a^{n-m}},n>m$, to simplify the expression.

$\begin{array}{rcl}\frac{8^3}{8^5}& =& \frac{1}{8^{5-3}}\\ & =& \frac{1}{8^2}\end{array}$

So, $\frac{8^3}{8^5}=\frac{1}{8^2}$.