Problem 38 - College Algebra with Corequisite Support

Question

For the following exercises, use the one-to-one property of logarithms to solve. \(\log (x+3)-\log (x)=\log (74)\)

Step-by-Step Solution

Answer

The solution is \(x = \frac{3}{73}\).

Step 1: Apply the Logarithmic Difference Property

The logarithmic difference property states that \( \log_a(b) - \log_a(c) = \log_a\left(\frac{b}{c}\right) \). Apply this property to \( \log(x+3) - \log(x) \) to get: \[ \log\left(\frac{x+3}{x}\right) = \log(74) \]

Step 2: Use the One-to-One Property of Logarithms

The one-to-one property of logarithms states that if \( \log_a(b) = \log_a(c) \), then \( b = c \). Apply this property to the equation \( \log\left(\frac{x+3}{x}\right) = \log(74) \): \[ \frac{x+3}{x} = 74 \]

Step 3: Solve the Resulting Equation

To solve the equation \( \frac{x+3}{x} = 74 \), multiply both sides by \( x \) to eliminate the fraction: \[ x+3 = 74x \]Next, subtract \( x \) from both sides:\[ 3 = 73x \]Finally, divide both sides by 73 to solve for \( x \):\[ x = \frac{3}{73} \]

Key Concepts

One-to-One Property of Logarithms

The one-to-one property of logarithms is a powerful tool in solving logarithmic equations. This property states that if two logarithmic expressions with the same base are equal, then their arguments must be equal as well. In simpler terms, if

• \( \log_a(b) = \log_a(c) \)

then

• \( b = c \)

This principle is essential because it allows us to remove the logarithmic part of the equation and focus on solving the simpler equation that results from comparing the arguments directly. For instance, in the given exercise, once we use the one-to-one property, it becomes easier to solve the equation since it reduces to \( \frac{x+3}{x} = 74 \). Thus, understanding and applying the one-to-one property can make seemingly complex logarithmic equations much more manageable.

Logarithmic Difference Property

The logarithmic difference property simplifies expressions involving the subtraction of two logarithms. This property can be expressed as:

• \( \log_a(b) - \log_a(c) = \log_a\left(\frac{b}{c}\right) \)

Essentially, it allows you to combine two logarithms into a single logarithm, which is especially useful when dealing with solved variables. In the provided equation, \( \log(x+3) - \log(x) \) becomes \( \log\left(\frac{x+3}{x}\right) \), thanks to the logarithmic difference property.
This transformation is valuable because it not only simplifies the equation but also sets it up perfectly for applying the one-to-one property. By reducing the number of logarithmic terms, it makes the entire process of solving the equation more straightforward, ensuring clarity and ease.

Solving Equations with Logarithms

Solving equations with logarithms involves combining several properties and concepts to find the unknown variable. Begin by using properties like the logarithmic difference property to simplify the equation. This reduces the equation to a form where direct applications like the one-to-one property can be used.

For example, once the original equation \( \log(x+3) - \log(x) = \log(74) \) is simplified using the difference property, it reduces to \( \log\left(\frac{x+3}{x}\right) = \log(74) \). We then apply the one-to-one property, which allows us to link the arguments, resulting in the equation:

• \( \frac{x+3}{x} = 74 \)

This equation can now be solved using basic algebraic manipulations. Multiply by \( x \) to remove the fraction, then rearrange and solve for \( x \). Thus, by systematically applying these key properties, solving the equation becomes a sequence of simple steps, stripping away complexity and focusing on algebraic solutions.