Problem 327 - Precalculus

Question

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 of Logarithms

The one-to-one property of logarithms states that if \( \log_b(x) = \log_b(y) \), then \( x = y \). Since both sides of the equation have the same logarithm base (13), set the arguments equal to each other: \( 5n - 2 = 8 - 5n \).

Step 2: Simplify the Equation

Add \(5n\) to both sides to move all terms involving \(n\) to one side of the equation. This results in \( 10n - 2 = 8 \).

Step 3: Isolate the Variable Term

Add 2 to both sides of the equation to isolate the term with \(n\). This gives \(10n = 10\).

Step 4: Solve for the Variable

Divide both sides of the equation by 10 to solve for \(n\). This results in \(n = 1\).

Key Concepts

Solving Logarithmic Equations

When solving logarithmic equations, one of the key concepts to understand is how to deal with the properties of logarithms to find the unknown variable. Logarithmic equations are equations in which the unknown variable appears within a logarithm. Here's a step-by-step approach:

• Identify the logarithmic expression within the equation. This might be a log on both sides or a log and a number.
• Determine the appropriate logarithmic property to apply. In many cases, you'll use the one-to-one property, which allows you to set the arguments of the logs equal to each other when the bases are the same.
• After simplifying the equation, use algebraic techniques to solve for the unknown variable.

Breaking down the equation through these steps ensures you fully leverage the powerful properties of logarithms to find your solution. Understanding this step-by-step procedure helps in mastering the art of solving logarithmic equations effectively.

Logarithmic Properties

Logarithmic properties are fundamental tools that make solving logarithmic equations a lot easier. These properties allow us to manipulate logarithms in a way that simplifies expressions and equations:

• One-to-One Property: This property is crucial when solving equations. If you have \( \log_b(x) = \log_b(y) \), the one-to-one property allows you to equate the arguments directly, leading to the equation \(x = y\). This is because logarithmic functions are "one-to-one functions," meaning each input has exactly one output.
• Product, Quotient, and Power Properties: These allow you to break down more complex expressions into simpler parts or combine multiple logarithms into one. They're especially helpful in algebraic manipulation when simplifying or solving equations.
• Change of Base Formula: While not directly used often in solving simple equations, it's important as it allows conversions between different logarithmic bases.

These properties are like a toolkit, each providing a different function that can be applied when working with logs. The one-to-one property is particularly useful for easily converting a logarithmic equation to a linear one, as seen in our exercise.

Algebraic Manipulation

Algebraic manipulation involves using various algebraic techniques and properties to rearrange equations and isolate the variable of interest. It's a critical step in solving logarithmic equations once the logarithmic properties have been applied. Here's how it generally works:

• Start by identifying the terms that contain the variable. Your goal is to rearrange the equation so that all terms involving the variable are on one side, while constant terms are on the other.
• Use addition or subtraction to eliminate terms from one side of the equation. This can involve getting rid of constants or unwanted coefficients.
• Divide or multiply to isolate the term containing the variable. This often involves clearing coefficients by performing the inverse operation.
• Once the variable is isolated, perform any necessary arithmetic to solve for its value.

In our example, we used these techniques after applying the one-to-one property to set up the simplest linear equation. Remember, precision in these steps is pivotal to ensuring that you arrive at the correct solution efficiently.