Problem 35 - Calculus with Concepts in Calculus

Question

$$ \text { Solve } \ln x+\ln (3 x-1)=0 \text { for } x \text { . } $$

Step-by-Step Solution

Answer

x = \(\frac{1 + \sqrt{13}}{6}\) is the valid solution.

Step 1: Apply Logarithm Properties

Use the property of logarithms that states \( \ln a + \ln b = \ln(ab) \). This allows us to combine the logarithms: \( \ln x + \ln(3x - 1) = \ln[x(3x - 1)] \).

Step 2: Set the Equation Equal to Zero

The equation now becomes \( \ln[x(3x - 1)] = 0 \). The logarithm is zero when the argument is 1. Thus, we set \( x(3x - 1) = 1 \).

Step 3: Solve the Quadratic Equation

Rewrite the equation from Step 2 as \( 3x^2 - x = 1 \) and then rearrange to \( 3x^2 - x - 1 = 0 \). This is a quadratic equation, which we solve using the quadratic formula: \( x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \). Here, \( a = 3 \), \( b = -1 \), and \( c = -1 \).

Step 4: Compute the Discriminant

Calculate the discriminant \( b^2 - 4ac \): \( (-1)^2 - 4(3)(-1) = 1 + 12 = 13 \).

Step 5: Solve for x

Substitute the discriminant and coefficients back into the quadratic formula: \( x = \frac{-(-1) \pm \sqrt{13}}{2(3)} = \frac{1 \pm \sqrt{13}}{6} \). This gives two potential solutions: \( x = \frac{1 + \sqrt{13}}{6} \) and \( x = \frac{1 - \sqrt{13}}{6} \).

Step 6: Validate the Solutions

Since logarithms are undefined for non-positive numbers, the solution must ensure \( x > 0 \) and \( 3x - 1 > 0 \). Calculate: \( x = \frac{1 - \sqrt{13}}{6} \) is negative and invalid. \( x = \frac{1 + \sqrt{13}}{6} > 0 \); check \( 3x - 1 > 0 \) can be done by verifying the inequality holds for the positive root.

Key Concepts

Quadratic Equations

Quadratic equations are fundamental in algebra and come in the standard form: \[ ax^2 + bx + c = 0 \] where \(a\), \(b\), and \(c\) are constants and \(x\) represents the variable. These equations are solved to find the values of \(x\) that make the equation true.

• **Standard Form:** It's important to first ensure the quadratic equation is in the standard form before attempting to solve it.
• **Methods of Solving:** There are various methods to solve quadratic equations, such as factoring, completing the square, and using the quadratic formula.

When using the quadratic formula: \[ x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \] it helps in finding the roots when the equation does not factor easily.This formula provides two solutions, known as the roots of the equation, which can be real or complex numbers. Always check the solutions against the original problem context to ensure they are valid.

Logarithm Properties

Logarithms are the inverse operations of exponents, providing solutions to equations involving exponential terms. Understanding logarithm properties allows us to manipulate and solve logarithmic equations effectively. Here's a key property used in solving logarithmic equations:

• **Product Property:** This states that \( \ln a + \ln b = \ln(ab) \). It helps in merging multiple logarithmic terms into a single entity, simplifying the equation momentarily.

Utilizing this property allows the equation \( \ln x + \ln (3x-1) = 0 \) to be combined into \( \ln[x(3x-1)] = 0 \). The advantage is apparent as a multiplication inside the log can be set to an equation such as \( x(3x-1) = 1 \), making it easier to transition to a solvable quadratic form.

Discriminant

The discriminant is a component of the quadratic formula that indicates the nature of the roots of a quadratic equation. It is expressed as:\[ b^2 - 4ac \] Evaluating the discriminant provides insight into the type of solutions you can expect:

• **Positive Discriminant:** Indicates two distinct real roots.
• **Zero Discriminant:** Results in one real root, signifying a perfect square.
• **Negative Discriminant:** Leads to complex or imaginary roots.

In the original problem, the discriminant calculation was performed as: \[ (-1)^2 - 4\times3\times(-1) = 13 \] This positive result confirms that the quadratic equation has two distinct real roots. However, always remember to verify applicability to the original logarithmic context, where only positive real numbers are valid.