Problem 77 - University Calculus: Early Transcendentals

Question

Starting with the equation \(e^{x_{1}} e^{x_{2}}=e^{x_{1}+x_{2}},\) derived in the text, show that \(e^{-x}=1 / e^{x}\) for any real number \(x .\) Then show that \(e^{x_{1}} / e^{x_{2}}=e^{x_{1}-x_{2}}\) for any numbers \(x_{1}\) and \(x_{2}\).

Step-by-Step Solution

Answer

\( e^{-x} = \frac{1}{e^x} \) and \( \frac{e^{x_1}}{e^{x_2}} = e^{x_1-x_2} \).

Step 1: Rewrite the Equation Using Negative Exponents

Start with the identity \( e^{x_1} e^{x_2} = e^{x_1+x_2} \). Specifically, consider \( x_2 = -x_1 \). Substituting gives: \( e^{x_1} e^{-x_1} = e^{x_1-x_1} = e^{0} = 1 \). This confirms that \( e^{x_1} e^{-x_1} = 1 \). Rearranging gives \( e^{-x_1} = \frac{1}{e^{x_1}} \). Thus, we have shown that \( e^{-x} = \frac{1}{e^x} \).

Step 2: Derive the Division Rule for Exponentials

Using the property that \( e^{-x} = \frac{1}{e^x} \), we know the reciprocal property of exponentials holds. Now, consider the division of two exponential expressions: \( \frac{e^{x_1}}{e^{x_2}} \). This can be rewritten using negative exponents: \( e^{x_1} \times e^{-x_2} \), which is equivalent to \( e^{x_1 + (-x_2)} \), simplifying to \( e^{x_1-x_2} \).Thus, \( \frac{e^{x_1}}{e^{x_2}} = e^{x_1-x_2} \).

Key Concepts

Properties of Exponents

Exponents have specific rules, or properties, that simplify their manipulation and computation. One such fundamental property is the multiplication of exponential terms. When we multiply powers with the same base, we add their exponents together. For example, if you have \(e^{x_1} \times e^{x_2}\), it simplifies to \(e^{x_1 + x_2}\). This rule makes handling exponential functions more straightforward, especially in complex calculations.
Another essential property involves raising an exponential term to another power. In this case, you multiply the exponents. Remember these properties as you work with exponential functions:

• When bases are multiplied: Add the exponents (e.g., \(a^m \times a^n = a^{m+n}\)).
• When an exponent is raised by another exponent: Multiply the exponents (e.g., \((a^m)^n = a^{mn}\)).

Recognizing and applying these properties simplifies many algebraic expressions.

Negative Exponents

Negative exponents might seem confusing at first, but they represent a concept of reciprocals. When you see an expression like \(e^{-x}\), it actually means \(\frac{1}{e^x}\). This tells us that raising a number to a negative exponent is the same as taking the reciprocal of that number raised to the corresponding positive exponent.
Why does this matter? Understanding negative exponents allows you to simplify and solve equations more easily, especially those involving fractions. For instance, in the expression \(e^{x_1} \times e^{-x_1} = 1\), the negative exponent \(e^{-x_1}\) can be rearranged to show \(\frac{1}{e^{x_1}}\), confirming the reciprocal relationship.
Always remember:

• \(a^{-n} = \frac{1}{a^n}\).
• \(e^{-x} = \frac{1}{e^x}\).

This concept is particularly useful when deriving division rules for exponentials.

Division of Exponents

Dividing exponents is a common operation in algebra, especially with exponential functions. The key rule is to subtract the exponents of matching bases. Say you have the expression \(\frac{e^{x_1}}{e^{x_2}}\), you can rewrite it as \(e^{x_1-x_2}\). This rule stems from the reciprocal nature of negative exponents. When handled appropriately, it will help simplify complex expressions into a more manageable form.
Let's break it down:

• \(\frac{a^m}{a^n} = a^{m-n}\): When you divide like bases, you subtract the exponents.
• This property allows the expression \(e^{x_1} \div e^{x_2}\) to be simplified as \(e^{x_1 - x_2}\).

Understanding and applying this division property of exponents is crucial for solving many algebraic problems efficiently.