Problem 286 - Precalculus

Question

Use the product rule for logarithms to find all \(x\) values such that \(\log _{12}(2 x+6)+\log _{12}(x+2)=2 .\) Show the steps for solving.

Step-by-Step Solution

Answer

The valid solution is \(x = 6\).

Step 1: Apply the Product Rule for Logarithms

The product rule for logarithms states that \(\log_b(A) + \log_b(B) = \log_b(A \cdot B)\). Apply this rule to the given equation: \(\log_{12}(2x+6) + \log_{12}(x+2) = \log_{12}((2x+6)(x+2))\).

Step 2: Simplify the Expression Inside the Logarithm

Expand the expression \((2x+6)(x+2)\): \((2x+6)(x+2) = 2x^2 + 4x + 6x + 12 = 2x^2 + 10x + 12\). Thus, the equation becomes \(\log_{12}(2x^2 + 10x + 12) = 2\).

Step 3: Convert the Logarithmic Equation to Exponential Form

Rewrite the logarithmic equation in its exponential form: \(\log_{12}(2x^2 + 10x + 12) = 2\) becomes \(2x^2 + 10x + 12 = 12^2\).

Step 4: Solve the Resulting Quadratic Equation

Calculate \(12^2 = 144\). Thus, the equation is \(2x^2 + 10x + 12 = 144\). Subtract 144 from both sides to form: \(2x^2 + 10x + 12 - 144 = 0\), which simplifies to \(2x^2 + 10x - 132 = 0\). Divide the entire equation by 2: \(x^2 + 5x - 66 = 0\). Factor the quadratic: \((x - 6)(x + 11) = 0\). Solve for \(x\) to get \(x = 6\) or \(x = -11\).

Step 5: Check for Valid Solutions in Context

The solution must satisfy the original logarithmic conditions. For \(x = 6\), \(2x+6 = 18\) and \(x+2 = 8\), both positive, so \(x = 6\) is valid. For \(x = -11\), \(2x+6 = -16\) and \(x+2 = -9\), both negative, invalid for logarithms. Thus, the only valid solution is \(x = 6\).

Key Concepts

Logarithmic Equations

Logarithmic equations are fundamental in mathematics and appear in various contexts, such as solving problems related to growth or decay processes. They often include terms that consist of log expressions and can be simplified using the rules of logarithms.
In our exercise, we use the product rule for logarithms, which states that the sum of two log expressions with the same base can be combined into a single log expression:

• \(\log_b(A) + \log_b(B) = \log_b(A \cdot B)\)

When applying this to our equation \(\log_{12}(2x+6) + \log_{12}(x+2)\), we combine the logs as follows:

• \(\log_{12}((2x+6)(x+2))\)

This simplification is pivotal, reducing a complicated expression into one easier to handle in subsequent steps.

Quadratic Equations

Once we have reduced our logarithmic equation to a single expression, we transform it into a quadratic form. Quadratic equations take the form \(ax^2 + bx + c = 0\). In our example, after expanding \((2x+6)(x+2)\) and applying exponential transformations, we end up with:

• \(2x^2 + 10x + 12 = 144\)
• Simplified to: \(x^2 + 5x - 66 = 0\)

To solve the quadratic equation, we factor it into:

• \((x - 6)(x + 11) = 0\)

This process, called factoring, allows us to find potential solutions for \(x\) by setting each factor equal to zero.

Exponential Form

Transforming logarithmic expressions into exponential form is a crucial step in solving logarithmic equations. The conversion from logarithmic to exponential involves expressing the log equation \(\log_b(y) = x\) as \(y = b^x\).
In our exercise, we convert the equation \(\log_{12}(2x^2 + 10x + 12) = 2\) into its exponential equivalent:

• \(2x^2 + 10x + 12 = 12^2\)

This reveals the underlying expression to be equal to \(144\), outlining a clear path toward solving for \(x\). Recognizing when to switch forms is crucial for progress, bridging the gap between forms and simplifying the problem-solving process.

Valid Solutions

When solving logarithmic equations, it's essential to verify the validity of solutions by considering the domain of the logarithms. Log expressions require positive arguments due to their definitions.
In our exercise, we find potential solutions \(x = 6\) or \(x = -11\) after factoring the quadratic. However, only \(x = 6\) ensures that both \(2x+6\) and \(x+2\) remain positive:

• For \(x = 6\), \(2x+6 = 18\) and \(x+2 = 8\), which are positive and valid
• For \(x = -11\), \(2x+6 = -16\) and \(x+2 = -9\), which are negative and invalid

Therefore, careful checks ensure that we exclude values that do not satisfy the initial logarithmic conditions, providing only logically valid solutions.