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Chapter 6: Problem 169

Use the Product Property for Exponents to explain why \(x \cdot x=x^{2}\).

Short Answer

Expert verified
Using the Product Property for Exponents: \[ x \times x = x^2 \]

Step by step solution

01

Understand the problem

The problem asks to use the Product Property for Exponents to explain why multiplying the same base, x, by itself results in the base raised to the power of 2.
02

Recall the Product Property for Exponents

The Product Property for Exponents states that when you multiply two expressions with the same base, you add their exponents: \[ a^m \times a^n = a^{m+n} \]
03

Apply the Property to the Problem

In the given problem, x does not explicitly show exponents, which implies that each x has an exponent of 1. Write this as: \[ x^1 \times x^1 \]
04

Combine the Exponents

Using the Product Property for Exponents, add the exponents together: \[ x^1 \times x^1 = x^{1+1} = x^2 \]
05

Conclude the Explanation

Thus, we have explained why \( x \times x = x^2 \). The multiplication of the base by itself results in the base raised to the sum of the exponents, which in this case is 2.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Understanding Multiplication of Exponents
When dealing with exponents, the **Multiplication of Exponents** property is fundamental. Imagine you have a base, let's say **x**. If you multiply **x** by itself, it鈥檚 similar to adding one more instance of **x**.
For example, if you start with \( x= x^1 \) and you multiply it by another instance of **x**, you get \( x^1 \times x^1 \). According to the multiplication property, we add the exponents together while keeping the base the same:
\[ x^1 \times x^1 = x^{1+1} = x^2 \]
This means when you multiply **x** by itself, you raise it to the power of 2.
Base and Exponents Explained
Understanding the **Base and Exponents** concept makes the multiplication of exponents easier.
The **Base** is the number being multiplied, while the **Exponent** tells us how many times to multiply the base by itself.
In the case of \( x^2 \), **x** is the base, and 2 is the exponent. This implies:
\[ x^2 = x \times x \]
Essentially, **x** is multiplied by itself twice. Applying this idea to our example, \( x^1 \times x^1 \) combines both bases (both are the same base) and adds their exponents, resulting in \( x^{1+1}= x^2 \). This property holds for any base and exponent scenario.
Algebraic Properties Simplified
The Product Property of Exponents is one of the key **Algebraic Properties**. This property guides us on how to handle multiplication when dealing with exponents.
The general rule states that for any base **a** with exponents **m** and **n**, if you multiply them, you add the exponents:
\[ a^m \times a^n = a^{m+n} \]
This turns complex multiplications into simple addition problems, keeping calculations manageable.
For our exercise, using the same principle, we demonstrate that \( x^1 \times x^1 = x^{1+1} = x^2 \). Recognizing and applying these algebraic properties correctly allows for the simplification and solving of more complex mathematical expressions.

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Most popular questions from this chapter

Mixed Practice $$\frac{27 a^{7}}{3 a^{3}}+\frac{54 a^{9}}{9 a^{5}}$$

Divide the monomials. $$\frac{\left(10 m^{5} n^{4}\right)\left(5 m^{3} n^{6}\right)}{25 m^{7} n^{5}}$$

Maurice simplifies the quotient \(\frac{d^{7}}{d}\) by writing \(\frac{d^{7}}{d}=7 .\) What is wrong with his reasoning?

Mixed Practice (a) \(\frac{z^{6}}{z^{5}}\) (b) \(\frac{z^{5}}{z^{6}}\)

Divide each polynomial by the binomial. $$\left(125 y^{3}-64\right) \div(5 y-4)$$
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