Problem 42 - Precalculus : Building Concepts and Connections

Question

In Exercises \(31-46,\) write each expression as a logarithm of a single quantity and then simplify if possible. Assume that each variable expression is defined for appropriate values of the variable(s). Do not use a calculator. $$\ln \left(x^{2}-9\right)-\ln (x+3)$$

Step-by-Step Solution

Answer

The simplified single logarithmic expression is \(ln |x-3|\)

Step 1: Application of Logarithm Difference Rule

Apply the logarithm difference rule to the given expression, which says the difference of two logarithms is equal to the log of the quotient of their arguments. \[ln \left(x^{2}-9\right)-ln (x+3) = \ln \frac{x^{2}-9}{x+3}\]

Step 2: Simplifying the expression

To simplify the expression, recognize that the numerator is a difference of squares, which can be factored into a product of \((x-3)(x+3)\). Thus, we have \[ln \frac{(x-3)(x+3)}{x+3}\]. Now, cancel the term \(x+3\) in the numerator with that in the denominator.

Step 3: Final Result

After cancelling, the result will be \[ln|x-3|\]. Note that the absolute sign has been added. Typically, the logarithm of negative numbers is undefined, but given that the domain for these difference expressions are variable, adding the absolute value will ensure the solution stays within the domain.

Key Concepts

Logarithm Difference Rule

The logarithm difference rule is an essential concept in simplifying expressions involving logarithms. This rule states that the logarithm of a difference, specifically the difference of two logarithms, can be transformed into the logarithm of a quotient. Basically, whenever you have an expression like \( \ln(a) - \ln(b) \), you can simplify it to \( \ln\left(\frac{a}{b}\right) \).
This rule is fundamental because it allows you to combine multiple logarithms into one, making calculations more straightforward.
In our original exercise, we applied the logarithm difference rule to \( \ln(x^2-9) - \ln(x+3) \) and simplified it to \( \ln\left(\frac{x^2 - 9}{x + 3}\right) \).

• The difference transforms into a single logarithmic expression through division.
• This action sets the stage for further simplification of complex expressions.

Factoring Polynomials

Factoring polynomials is a crucial step when working with algebraic expressions, especially when simplifying logarithmic holds. In our exercise, we encountered the expression \( x^2 - 9 \). This particular expression is identifiable as a "difference of squares".
Difference of squares can be simplified using the formula \( a^2 - b^2 = (a-b)(a+b) \). Let's illustrate this concept with the expression \( x^2 - 9 \):

• Recognize it as \( x^2 - 3^2 \).
• Apply the difference of squares rule to rewrite it as \((x - 3)(x + 3)\).

This method is immensely useful as it breaks down complex polynomials into simpler, solvable components. In the exercise, this allowed further reduction within the context of a logarithmic expression.

Simplifying Expressions

Simplifying expressions is a line of reasoning we often utilize to make mathematical expressions easier to interpret and solve. In the context of our exercise, simplifying was necessary after applying the logarithm difference rule and factoring polynomials.
With the expression \( \ln\left(\frac{(x-3)(x+3)}{x+3}\right) \) in hand, one critical simplification step involved canceling terms that appear in both the numerator and the denominator. Here's the detailed breakdown of the simplification process:

• Given \((x + 3)\) appears in both the numerator and denominator, you can cancel them out to simplify.
• The simplified form becomes \( \ln|x - 3| \).

Note the introduction of absolute value signs around \((x - 3)\). This addition ensures the expression remains valid over the domain when the original difference-based expression could lead to an undefined logarithm for negative inputs.