Problem 37 - College Algebra with Corequisite Support

Question

For the following exercises, use the one-to-one property of logarithms to solve. \(\log _{13}(5 n-2)=\log _{13}(8-5 n)\)

Step-by-Step Solution

Answer

The solution is \( n = 1 \).

Step 1: Apply the One-to-One Property

The one-to-one property of logarithms states that if \( \log_b(x) = \log_b(y) \), then \( x = y \). Since the given equation is \( \log_{13}(5n - 2) = \log_{13}(8 - 5n) \), we can set the arguments equal to each other: \( 5n - 2 = 8 - 5n \).

Step 2: Set Up the Equation

We now have a linear equation from Step 1: \( 5n - 2 = 8 - 5n \). The goal is to solve for \( n \).

Step 3: Combine Like Terms

Add \( 5n \) to both sides to get \( 5n + 5n - 2 = 8 \). This simplifies to \( 10n - 2 = 8 \).

Step 4: Solve for the Variable

Add 2 to both sides: \( 10n = 10 \). Now, divide both sides by 10 to isolate \( n \): \( n = 1 \).

Key Concepts

One-to-One Property

The one-to-one property is a fundamental concept in logarithms that makes solving certain types of equations much simpler. It suggests that if two logarithms with the same base are equal, their arguments must also be equal. This property is expressed as follows: if \( \log_b(x) = \log_b(y) \), then \( x = y \).

This principle is extremely useful when both sides of an equation have logarithms. When you see an equation like \( \log_{13}(5n - 2) = \log_{13}(8 - 5n) \), you can confidently say that the expressions inside the logarithms must be equal: \( 5n - 2 = 8 - 5n \).

• It allows you to "drop" the logs and solve the simpler algebraic equation left behind.
• You don't need to worry about the base if it is the same for both logarithms on either side of the equation.
• This streamlines the process of finding the solution.

The one-to-one property not only simplifies the equation but also makes the often intimidating logarithmic equations more approachable for students.

Solving Linear Equations

Once the one-to-one property has done its job, what remains is a straightforward linear equation. For example, the equation \( 5n - 2 = 8 - 5n \) needs to be solved for \( n \). Solving linear equations involves finding the value of the unknown that makes the equation true.

Here's a simple way to approach it:

• First, combine like terms. Add \( 5n \) to both sides to avoid negative coefficients as much as possible. This will change our equation to \( 10n - 2 = 8 \).
• Next, isolate the variable. Add 2 to both sides to get \( 10n = 10 \).
• Finally, solve for the variable by dividing both sides by 10, yielding \( n = 1 \).

Each step in this process systematically simplifies the equation, making it solvable with basic arithmetic operations. This methodical approach helps prevent errors and provides a clear pathway to finding the solution.

Logarithmic Properties

Understanding logarithmic properties is key to mastering logarithmic equations. These properties give us the tools to manipulate and solve equations involving logs.

Some essential properties include:

• Product Property: \( \log_b(xy) = \log_b(x) + \log_b(y) \). It shows how to split the log of a product into the sum of two logs.
• Quotient Property: \( \log_b\left(\frac{x}{y}\right) = \log_b(x) - \log_b(y) \). It describes how to convert the log of a quotient into a difference of logs.
• Power Property: \( \log_b(x^n) = n \cdot \log_b(x) \). This allows you to pull down an exponent as a coefficient.

These properties allow us to simplify complex logarithmic expressions and can be essential in proving the one-to-one property when you isolate variables behind log symbols.

Moreover, these properties extend beyond numerical usage and are foundational in understanding how logs scale and interrelate within various scientific and engineering contexts. Familiarity with them makes working with logarithms intuitive and manageable.