Problem 329 - Precalculus

Question

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

No valid real solution for the equation.

Step 1: Understanding the One-to-One Property

The one-to-one property of logarithms states that if \( ext{ln}(A) = \text{ln}(B)\), then \(A = B\). This is because the natural logarithm function is injective, meaning each input is mapped to a unique output. Here, we set \(-3x = x^2 - 6x\).

Step 2: Rearranging the Equation

We need to rearrange the equation to form a standard quadratic equation: \(-3x = x^2 - 6x\). Move all terms to one side to get: \(x^2 - 6x + 3x = 0\), simplifying to \(x^2 - 3x = 0\).

Step 3: Factoring the Quadratic Equation

Take out the common factor of \(x\) from the equation. This gives us \(x(x - 3) = 0\). We have factored the quadratic equation ready to solve for \(x\).

Step 4: Solving for x

Using the zero-product property, which states that if \(ab = 0\), then \(a = 0\) or \(b = 0\), we set each factor in \(x(x - 3) = 0\) to zero. This results in solutions \(x = 0\) and \(x = 3\).

Step 5: Checking for Validity in the Original Log Equation

Substitute the possible solutions back into the original logarithmic expressions to check for validity. Substituting \(x = 0\) gives \(\ln(0)\), which is undefined. Substituting \(x = 3\) gives \(\ln(-9)\) and \(\ln(3^2 - 18) = \ln(-9)\), both of which are undefined for real numbers.

Step 6: Conclusion

Since neither potential solution \(x = 0\) nor \(x = 3\) results in a valid expression for \(ln\), there is no valid solution to the equation \(\ln(-3x) = \ln(x^2 - 6x)\) within the real number system.

Key Concepts

Understanding the One-to-One Property of Logarithms

When solving logarithmic equations, the one-to-one property of logarithms is a key concept to understand. This property can be summed up as: If \( \ln(A) = \ln(B) \), then it must follow that \( A = B \). The logic behind this is simple—the natural logarithm (\( \ln \)) is an injective or one-to-one function. This means each input into the logarithm function corresponds to a unique output.
So, in the statement above, if two expressions share the same natural logarithm, those expressions must be equal.
You can use this principle to move from an equation involving logarithms to a simpler, often more solvable, algebraic equation. In our exercise example where \( \ln(-3x) = \ln(x^2 - 6x) \), applying the one-to-one property allows us to equate \( -3x \) directly to \( x^2 - 6x \).
This simplification is the key first step in solving logarithmic equations, allowing us to shift the focus from handling tricky logarithmic expressions to tackling standard algebraic forms.

The Role of Quadratic Equations

Quadratic equations show up everywhere in algebra, and solving them is fundamental. These equations typically take the standard form of \( ax^2 + bx + c = 0 \). In our exercise, by applying the one-to-one property, we transformed the equation \( \ln(-3x) = \ln(x^2 - 6x) \) into a quadratic equation: \( x^2 - 3x = 0 \).
Once in this form, the equation can be solved using methods such as factoring, completing the square, or using the quadratic formula.

• Factoring is often the first method attempted, as it is straightforward and intuitive for simpler quadratic equations.
• Completing the square involves rearranging the equation into a perfect square trinomial.
• The quadratic formula, \( x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \), works for all quadratics but may be more complex.

For this specific problem, factoring \( x^2 - 3x = 0 \) was readily doable. By factoring out an \( x \), we get \( x(x - 3) = 0 \), making it easy to find the potential solutions.
Recognizing the characteristics of quadratic equations and knowing how to effectively solve them is crucial to overcoming more complex algebraic problems.

Applying the Zero-Product Property

The zero-product property is a fundamental rule that assists in solving equations like quadratics once they are factored. The property states simply: if the product of two numbers is zero, then at least one of the numbers must be zero.
This is given mathematically as: If \( ab = 0 \), then \( a = 0 \) or \( b = 0 \).
In the context of our exercise, after transforming the original logarithmic expression into a factored quadratic form \( x(x - 3) = 0 \), the zero-product property comes into play.
We split the equation into \( x = 0 \) or \( x - 3 = 0 \), resulting in possible solutions \( x = 0 \) and \( x = 3 \).
While these solutions work algebraically in the context of the quadratic equation, we must verify them with the original context of the logarithm equation. Unfortunately, in our problem set, substituting these potential solutions back does not yield valid logarithmic expressions, leading us to determine that no real solution exists.
The zero-product property thus is invaluable in solving equations but must be applied with context in mind, especially for equations initially rooted in other mathematical functions like logarithms.