Problem 45 - College Algebra

Question

For the following exercises, solve each equation for \(x\). $$ \ln (x)+\ln (x-3)=\ln (7 x) $$

Step-by-Step Solution

Answer

The solution is \( x = 10 \).

Step 1: Use the properties of logarithms

We start by using the property of logarithms that states \( \ln a + \ln b = \ln (a \cdot b) \). Apply this property to combine the left-hand side: \( \ln (x) + \ln (x-3) = \ln (x(x-3)) = \ln (x^2 - 3x) \). Therefore, the equation becomes \( \ln (x^2 - 3x) = \ln (7x) \).

Step 2: Set the arguments equal

Since the logarithms of both sides of the equation are equal, their arguments must be equal. Thus, we set \( x^2 - 3x = 7x \).

Step 3: Solve the quadratic equation

Rearrange the equation to get a standard quadratic form: \( x^2 - 3x - 7x = 0 \) simplifies to \( x^2 - 10x = 0 \). Factor the quadratic equation: \( x(x - 10) = 0 \).

Step 4: Find the solutions

Set each factor in the factored equation equal to zero to find the solutions. For \( x = 0 \) and \( x - 10 = 0 \), solve to get \( x = 0 \) and \( x = 10 \) respectively. However, since \( \ln (x) \) is undefined for \( x = 0 \), \( x = 0 \) is not a valid solution. Therefore, \( x = 10 \) is the only solution.

Key Concepts

Properties of Logarithms

Logarithms have several properties that can simplify and solve equations. One such property is the product rule, which states that adding logarithms of two numbers is equivalent to the logarithm of their product:

• \( \ln a + \ln b = \ln (a \cdot b) \)

This property is particularly useful in reducing complex logarithmic expressions. For example, if we encounter an equation \( \ln x + \ln (x-3) \), we can use the product rule to simplify it to \( \ln (x(x-3)) = \ln(x^2 - 3x) \).
Notice that this transformation makes the equation more straightforward, allowing us to handle a single logarithmic term instead of two. Understanding and applying properties of logarithms like this one is crucial for efficiently solving logarithmic equations.
Effectively mastering these properties not only aids in algebraic manipulation but also provides pathways to solving more complex equations later on.

Quadratic Equations

In mathematics, quadratic equations are polynomial equations of the second degree. These equations are typically presented in the standard form:

• \( ax^2 + bx + c = 0 \)

To solve a quadratic equation means to find the values of \( x \) that satisfy the equation.
This usually involves manipulation to set all terms on one side of the equation, presenting it in its standard form. In solving our logarithmic equation, once we equalized the arguments, we had the equation \( x^2 - 10x = 0 \).
This equation already looks like a quadratic equation because it is set to equal zero and is a second-degree polynomial. Recognizing quadratic structures in such scenarios is essential.
It enables you to employ algebraic techniques such as the quadratic formula or factoring to find solutions. But remember, not every potential solution is valid in the context of the original logarithmic equation.

Factoring

Factoring is a mathematical process used to break down expressions into products of simpler factors. When dealing with quadratic equations, factoring is a common technique used to find solutions if the quadratic can be easily decomposed into factors.
A simple trinomial like \( x^2 - 10x \) is factored by identifying common elements and expressing the equation as a product of two or more expressions. In this example, each term shares a common factor of \( x \). Thus, it is factored as:

• \( x(x - 10) \)

This means setting each factor equal to zero gives potential solutions: \( x = 0 \) and \( x - 10 = 0 \).
However, considering the original logarithm constraints, \( \ln(x) \) is undefined at \( x = 0 \), so we discard \( x = 0 \) as a valid answer.
Ultimately, factoring simplifies the process of finding and verifying potential solutions, especially in contexts where direct computation or more complex methods are unnecessary.