Problem 39 - College Algebra with Corequisite Support

Question

For the following exercises, use the one-to-one property of logarithms to solve. \(\ln (-3 x)=\ln \left(x^{2}-6 x\right)\)

Step-by-Step Solution

Answer

The solution is \( x = 3 \).

Step 1: Apply the One-to-One Property

The one-to-one property of logarithms states that if \( \ln(a) = \ln(b) \), then \( a = b \). Using this property, we set the inside of the logarithms equal to each other: \( -3x = x^2 - 6x \).

Step 2: Rearrange the Equation

To solve the equation \( -3x = x^2 - 6x \), we first move all terms to one side of the equation to set it to zero: \( 0 = x^2 - 6x + 3x \). Simplifying this gives \( x^2 - 3x = 0 \).

Step 3: Factor the Equation

The quadratic equation \( x^2 - 3x = 0 \) can be factored by taking \( x \) common: \( x(x - 3) = 0 \).

Step 4: Solve for x

Set each factor equal to zero to solve for \( x \). First, \( x = 0 \), and second, \( x - 3 = 0 \) which gives \( x = 3 \).

Step 5: Check for Extraneous Solutions

Substitute \( x = 0 \) and \( x = 3 \) back into the original equation \( \ln(-3x) = \ln(x^2 - 6x) \). Since \( \ln(-3(3)) = \ln(3^2 - 6(3)) \) holds, but \( \ln(-3(0)) = \ln(0^2 - 6(0)) \) does not (because ln(0) is undefined), \( x = 3 \) is the only valid solution.

Key Concepts

Understanding the One-to-One Property

In the world of logarithms, the one-to-one property is a fundamental principle that helps simplify equations. Simply put, this property says that if two logarithms with the same base are equal, then their arguments must be equal as well. For example, if \( \ln(a) = \ln(b) \), we can assert that \( a = b \). This is because a logarithm is essentially a way to define an exponent, and identical exponents on the same base must be derived from equal values.

In this exercise, the one-to-one property allows us to directly equate \( -3x = x^2 - 6x \). This step significantly simplifies the problem by reducing the complexity of dealing with logarithmic expressions, allowing us to focus on solving an algebraic equation instead.

Recognizing Extraneous Solutions

Extraneous solutions are solutions that emerge during the process of solving an equation but do not fit the original problem. They often occur when operations like squaring both sides are used, or when considering domains of functions, such as logarithms.

In logarithmic equations, we must ensure the argument inside the logarithm remains valid. For instance, a logarithm argument must be strictly positive.

In our specific problem, we found potential solutions, \( x = 0 \) and \( x = 3 \). By substituting back into the original equation \( \ln(-3x) = \ln(x^2 - 6x) \), we discovered that \( x = 0 \) was extraneous due to the undefined nature of \( \ln(0) \). This validation step is crucial to ensure that the solutions not only solve the algebraic equation but also align with the constraints of the logarithmic form.

Factoring Quadratic Equations

Factoring is a key technique used to solve quadratic equations. A quadratic equation generally looks like \( ax^2 + bx + c = 0 \). In this context, factoring breaks down the equation into products that set each factor to zero, reflecting the roots of the equation.

For instance, the quadratic equation \( x^2 - 3x = 0 \) can be factored by taking \( x \) as a common factor, leading to \( x(x - 3) = 0 \). Factoring this way simplifies the process because it directly identifies the potential solutions. Each factor - in this case, \( x \) and \( x - 3 \) - is then set to zero separately:

• \( x = 0 \)
• \( x - 3 = 0 \)

Both results are subjected to additional checks to confirm their validity within the context of the original equation.

Implementing a Step-by-Step Solution

Breaking down problems into smaller steps makes complex equations more manageable. The step-by-step approach ensures a comprehensive and structured path while solving problems, minimizing errors and enhancing understanding. Let's revisit the key steps implemented in solving our logarithmic equation:

• First, use the one-to-one property of logarithms to simplify the equation \( \ln(-3x) = \ln(x^2 - 6x) \) to \( -3x = x^2 - 6x \).
• Next, rearrange and set all terms on one side to form a quadratic equation, resulting in \( x^2 - 3x = 0 \).
• Factor this equation by taking common factors, breaking it down to \( x(x - 3) = 0 \).
• Solve for \( x \) by considering each factor, giving potential solutions \( x = 0 \) and \( x = 3 \).
• Finally, check each solution against the original equation to weed out any extraneous solutions. Through substitution, confirm that \( x = 3 \) is valid while \( x = 0 \) is not.

Each step builds on the last and consolidates the solution process, ensuring a fashion-forward resolution aligned with mathematical principles.