Problem 39 - College Algebra

Question

Use the quotient rule for logarithms to fi \(\mathrm{d}\) all \(\mathrm{x}\) values such that \(\log _{6}(x+2)-\log _{6}(x-3)=1\). Show the steps for solving.

Step-by-Step Solution

Answer

The solution is \( x = 4 \).

Step 1: Apply the Quotient Rule for Logarithms

The expression \( \log_{6}(x+2) - \log_{6}(x-3) = 1 \) can be simplified using the quotient rule for logarithms, which states \( \log_{b}(A) - \log_{b}(B) = \log_{b}\left(\frac{A}{B}\right) \). Thus, \( \log_{6}(x+2) - \log_{6}(x-3) \) becomes \( \log_{6}\left(\frac{x+2}{x-3}\right) \).

Step 2: Set the Logarithmic Equation Equal to the Exponential Form

We now have the equation \( \log_{6}\left(\frac{x+2}{x-3}\right) = 1 \). According to the property of logarithms, \( \log_{b}(A) = C \) means \( A = b^{C} \). Therefore, \( \frac{x+2}{x-3} = 6^{1} \), which simplifies to \( \frac{x+2}{x-3} = 6 \).

Step 3: Solve the Rational Equation

To solve the equation \( \frac{x+2}{x-3} = 6 \), multiply both sides by \( x-3 \) to eliminate the fraction: \( x+2 = 6(x-3) \). Expand the right-hand side: \( x + 2 = 6x - 18 \).

Step 4: Rearrange and Solve for \(x\)

Rearrange the equation by moving \( x \) to one side and the constant terms to the other: \( 2 + 18 = 6x - x \) to get \( 20 = 5x \). Solve for \( x \) by dividing both sides by 5: \( x = \frac{20}{5} \).

Step 5: Simplify the Solution

Simplify \( x = \frac{20}{5} \) to find \( x = 4 \).

Key Concepts

Quotient Rule for Logarithms

Logarithms have several handy properties that can simplify complex expressions or aid in solving equations. One important property is the quotient rule for logarithms. This rule states that, if you have two logarithms with the same base being subtracted, such as \( \log_{b}(A) - \log_{b}(B) \), you can rewrite this expression as a single logarithm: \( \log_{b}\left(\frac{A}{B}\right) \).
This simplification is crucial when dealing with equations containing logarithms. It helps combine and reduce multiple log terms into one, making it easier to transform them into other forms, like exponential equations. For the given exercise, using the quotient rule for \( \log_{6}(x+2) - \log_{6}(x-3) = 1 \), we simplify to \( \log_{6}\left(\frac{x+2}{x-3}\right) = 1 \). This step is the foundation for solving the problem effectively.

• Always remember to check the base of the logarithms; they must be the same to apply this rule.
• Use the rule to combine terms only when it's clear it will lead to a simpler form.

Exponential Form

Once you've simplified a logarithmic equation, the next step often involves rewriting it in exponential form. This transformation is possible thanks to the definition of a logarithm: if \( \log_{b}(A) = C \), then \( A = b^{C} \). This link between logarithms and exponents is a powerful tool.
In the exercise, after applying the quotient rule, we reach \( \log_{6}\left(\frac{x+2}{x-3}\right) = 1 \). To solve for \( x \), we convert this into exponential form: \( \frac{x+2}{x-3} = 6^1 \). Converting to exponential form simplifies logarithmic problems into algebraic ones, where you can apply familiar algebraic techniques.

• Check the exponent carefully. In this example, it's simply \( 1 \), resulting in a base raised to power one, retaining the base's value.
• Be consistent with logical steps; transitioning from log to exponential form needs meticulous attention to detail.

Rational Equation Solving

Once the equation is expressed in exponential form, it often simplifies to a rational equation. This is commonly seen in exercises involving equations from logarithmic transformations. A rational equation is an equation involving fractions where the numerator and/or the denominator contain algebraic expressions.
For our equation \( \frac{x+2}{x-3} = 6 \), solving it requires clearing the fraction by multiplying both sides by \( x-3 \), resulting in \( x+2 = 6(x-3) \). The fraction disappears, and you end up with a linear equation: \( x + 2 = 6x - 18 \).
This equation can be solved using basic algebraic methods. Rearrange terms to isolate the variable \( x \) on one side. Moving terms carefully, \( 20 = 5x \), and then solve for \( x \) by dividing both sides by \( 5 \): \( x = 4 \).

• Always check your solution back in the original equation to ensure no extraneous solutions are introduced during clearing fractions.
• Isolating the variable correctly is key to obtaining the correct solution.