Problem 287 - Precalculus

Question

Use the quotient rule for logarithms to find all \(x\) 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 of Logarithms

The quotient rule for logarithms states that \( \log_b(a/c) = \log_b(a) - \log_b(c) \). Given the expression \( \log_{6}(x+2) - \log_{6}(x-3) = 1 \), apply this rule: \( \log_{6}\left(\frac{x+2}{x-3}\right) = 1 \).

Step 2: Eliminate the Logarithm

Convert the logarithmic equation into an exponential form. Since the base is 6, we rewrite the equation as: \( \frac{x+2}{x-3} = 6^1 \). This simplifies to: \( \frac{x+2}{x-3} = 6 \).

Step 3: Solve the Resulting Equation

Clear the fraction by multiplying both sides by \( x - 3 \). This gives: \( x + 2 = 6(x - 3) \). Expanding the right side, we have: \( x + 2 = 6x - 18 \).

Step 4: Isolate the Variable \(x\)

Rearrange the equation to get all \( x \) terms on one side: \( x - 6x = -18 - 2 \). Simplify to: \( -5x = -20 \).

Step 5: Solve for \(x\)

Divide both sides by -5 to solve for \( x \): \( x = 4 \).

Step 6: Verify the Solution

Substitute \( x = 4 \) back into the original logarithmic expression to ensure validity: \( \log_{6}(4+2) - \log_{6}(4-3) = \log_{6}(6) - \log_{6}(1) = 1 \). This confirms the solution is correct.

Key Concepts

Quotient Rule of Logarithms

Understanding the quotient rule of logarithms is essential for simplifying complex logarithmic expressions. This rule states that the logarithm of a quotient is equal to the difference of the logarithms. In mathematical terms, it can be expressed as \( \log_b(a/c) = \log_b(a) - \log_b(c) \). This rule helps us transform subtraction into division, making our calculations more manageable. For example, in the expression \( \log_{6}(x+2) - \log_{6}(x-3) \), we can apply the quotient rule to condense it into a single logarithmic statement: \( \log_{6}\left(\frac{x+2}{x-3}\right) \). This is particularly useful when trying to solve for unknown values, as it simplifies the mathematical landscape into more deterministic forms. By understanding and applying the quotient rule, students can break down subtraction within logs to reveal straightforward relationships between numbers. This forms a solid foundation for solving more complex equations involving logarithms.

Exponential Form

Once we have simplified a logarithmic expression using rules like the quotient rule, the next step often involves rewriting the equation in exponential form. This is a key transformation because it shifts the relationship from a logarithmic to an algebraic format, which can be easier to manipulate. To convert the equation \( \log_{6}\left(\frac{x+2}{x-3}\right) = 1 \) into exponential form, you use the base of the logarithm (which is 6, in this case) and set it as the base of an exponential equation. The expression becomes \( \frac{x+2}{x-3} = 6^1 \). Here, 1 is the exponent taken directly from the logarithmic equality. This transition into exponential form is crucial as it removes the complexity of logs, allowing us to work with straightforward algebraic expressions. It equates a fraction to a numeric value and prepares the ground for further algebraic manipulation. This skill is useful in solving equations efficiently.

Solving Equations

Once the logarithmic equation is transformed into an exponential form, it simplifies the process of solving for the variable, which in this problem is \( x \). Consider the example \( \frac{x+2}{x-3} = 6 \). Solving such equations often involves clearing the fraction by multiplying both sides by the denominator. Here, we multiply both sides by \( x-3 \), leading to the equation \( x+2 = 6(x-3) \). Expanding this, you get \( x+2 = 6x - 18 \). To isolate \( x \), rearrange the terms by moving all \( x \) terms to one side and constant terms to the other. This gives \( x - 6x = -18 - 2 \), simplifying to \( -5x = -20 \). Finally, divide each side by \(-5\) to solve for \( x \), yielding \( x = 4 \). Understanding the algebraic steps involved in isolating variables prepares students to tackle increasingly challenging problems. Mastery of these techniques builds confidence in solving both simple and complex equations.

Verification of Solutions

Verifying your solutions is a critical step in mathematical problem-solving to ensure that the solution works within the original parameters of the equation. After calculating that \( x = 4 \), it's crucial to verify this result by substituting it back into the original equation. For the original equation \( \log_{6} (x+2) - \log_{6} (x-3) = 1 \), substitute \( x = 4 \):

• First, calculate \( \log_{6} (4+2) = \log_{6} (6) \).
• Then, calculate \( \log_{6} (4-3) = \log_{6} (1) \).

Now, recall that any logarithm of 1 is zero, so \( \log_{6}(1) = 0 \). Therefore, the equation reevaluates to \( \log_{6}(6) - 0 = 1 \), which confirms our solution is correct. Verification not only ensures correctness but also strengthens the problem-solving process, helping to prevent errors and confirm understanding of the relationships between mathematical expressions.