Problem 36 - College Algebra with Corequisite Support

Question

For the following exercises, use the one-to-one property of logarithms to solve. \(\ln (10-3 x)=\ln (-4 x)\)

Step-by-Step Solution

Answer

The solution is \( x = -10 \).

Step 1: Understanding the Problem

We need to solve the equation \( \ln(10 - 3x) = \ln(-4x) \) by using the one-to-one property of logarithms. The one-to-one property states that if \( \ln(a) = \ln(b) \), then \( a = b \).

Step 2: Apply the One-to-One Property

According to the one-to-one property of logarithms, since \( \ln(10 - 3x) = \ln(-4x) \), it follows that the arguments must be equal. Thus, we can set up the equation: \( 10 - 3x = -4x \).

Step 3: Solve for x

To solve \( 10 - 3x = -4x \), first add \( 3x \) to both sides to get \( 10 = -4x + 3x \). Simplifying, we have \( 10 = -x \). Solving for \( x \), we obtain \( x = -10 \).

Step 4: Verify the Solution

Substitute \( x = -10 \) back into the original expressions: \( \ln(10 - 3(-10)) = \ln(-4(-10)) \). Simplifying, \( \ln(40) = \ln(40) \), which is true. Therefore, \( x = -10 \) is a valid solution.

Key Concepts

One-to-One Property

The one-to-one property is a fundamental concept in logarithmic equations, making them easier to solve. This property is based on the premise that if two logarithms with the same base are equal, then their arguments must be equal as well. This means that if you have an equation like \( \ln(a) = \ln(b) \), you can infer that \( a = b \).
This property is especially useful because it allows you to eliminate the logarithms entirely, reducing the complexity of the equation. Once the logarithms are removed, you're left with a simple algebraic equation to solve.

• Ensure the logarithms are of the same base. Otherwise, the one-to-one property doesn't apply directly.
• Remember: this property holds for any logarithmic base, not just the natural logarithm \( \ln \).

Understanding and leveraging this property can significantly streamline the problem-solving process when working with logarithmic equations.

Solving Equations

Solving equations in the context of logarithmic functions often involves manipulating the equation to reveal its algebraic form. After applying the one-to-one property, as seen in the exercise, we simplify by equating the arguments of the logarithms: \( 10 - 3x = -4x \).
From this point, solving for \( x \) is straightforward. You need to perform basic algebraic operations:

• Combine like terms by moving variables to one side and constants to the other, in this case add \( 3x \) to both sides for simplification.
• Once combined, you should isolate \( x \) to find its value. Here, \( 10 = -x \), resulting in \( x = -10 \).

Each step simplifies the equation further, making it easier to solve. Practice these steps, as they are common in algebraic manipulation.

Verification of Solutions

Verification is a crucial step in solving equations to ensure the solution found is correct and prevents potential errors. Even though it may seem like an extra step, verification confirms that the value of \( x \) satisfies the original equation.
In the problem, substitute \( x = -10 \) back into the original equations: \( \ln(10-3(-10)) \) and \( \ln(-4(-10)) \). Simplifying both arguments results in \( \ln(40) = \ln(40) \), confirming both sides equal each other.

• Substitute the solution back into the original equations, not the simplified ones.
• Ensure all calculations, especially with logarithms, are correct for an accurate verification.

Through verification, you affirm that your solution is mathematically correct, giving you confidence in your answer.