Problem 54 - Precalculus Mathematics for Calculus

Question

Solve the logarithmic equation for \(x .\) $$\ln (x-1)+\ln (x+2)=1$$

Step-by-Step Solution

Answer

\(x = \frac{-1 + \sqrt{9+4e}}{2}\) if it's greater than 1.

Step 1: Use the Property of Logarithms

The given equation is \( \ln (x-1) + \ln (x+2) = 1 \). Utilize the logarithmic property \( \ln a + \ln b = \ln(ab) \) to combine the logarithms on the left side of the equation. This gives us \( \ln((x-1)(x+2)) = 1 \).

Step 2: Exponentiate Both Sides

Exponentiate both sides to eliminate the natural logarithm. The equation becomes \( (x-1)(x+2) = e^1 \). Since \( e^1 = e \), we rewrite the equation as \( (x-1)(x+2) = e \).

Step 3: Expand the Quadratic Expression

Expand the quadratic expression \((x-1)(x+2)\) to simplify. This results in \(x^2 + x - 2\). The equation is now \(x^2 + x - 2 = e\).

Step 4: Rearrange to Form a Quadratic Equation

Rearrange the equation to form a standard quadratic equation: \(x^2 + x - (2 + e) = 0\).

Step 5: Solve the Quadratic Equation Using the Quadratic Formula

Apply the quadratic formula \(x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\), where \(a = 1\), \(b = 1\), and \(c = -(2+e)\). Calculate the discriminant: \(b^2 - 4ac = 1^2 - 4(1)(-2-e) = 1 + 8 + 4e = 9 + 4e\). Substitute into the formula: \(x = \frac{-1 \pm \sqrt{9 + 4e}}{2}\).

Step 6: Consider Validity of Solutions

Check the domain of the original logarithmic equation. \((x-1) > 0\) and \((x+2) > 0\) imply that \(x > 1\). Only solutions greater than 1 satisfy the original equation.

Key Concepts

Properties of Logarithms

Logarithms are an essential part of mathematical language. They transform multiplicative relationships into additive ones. This ability is especially helpful in solving equations. One a key property of logarithms is:

• \( \ln a + \ln b = \ln(ab) \)

This property allows us to combine the logarithms into a single log term of the product.
In the problem, \( \ln (x-1) + \ln (x+2) = \ln((x-1)(x+2)) \) becomes useful. This step is crucial because it simplifies the problem into a single equation, making it easier to solve. Understanding how to use these properties effectively is essential for tackling various logarithmic equations. Whenever you see multiple logarithm terms added, you might need this property!

Quadratic Formula

The quadratic formula provides a dependable method to solve any quadratic equation of the form

• \( ax^2 + bx + c = 0 \)

This formula helps to find the values of \( x \) where the equation equals zero. It is given by: \[ x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \]
In the context of the problem, after using properties of logarithms and exponentiation, the equation is transformed into \( x^2 + x - (2+e) = 0 \). Identifying \( a = 1 \), \( b = 1 \), and \( c = -(2+e) \), we substitute these values into the formula to find the possible values of \( x \).
Learning the quadratic formula is straightforward, but be sure to understand the meaning of each term to apply it correctly. It is a fundamental tool for solving quadratic equations.

Exponentiation

Exponentiation is the process of raising a number to a power. This concept is used to "undo" logarithmic functions.
In our exercise, after using the property of logarithms, we have an equation: \( \ln((x-1)(x+2)) = 1 \). The next step is to "exponentiate" both sides to eliminate the logarithm. We use the base \( e \) because we are dealing with the natural logarithm \( \ln \).

• This step gives us: \( (x-1)(x+2) = e^1 \) which is simply \( e \).

Exponentiation transforms our logarithmic equation into a quadratic one, making further calculations possible. Understanding when and how to use exponentiation will open new doors to solve logarithmic equations correctly.

Discriminant Calculation

The discriminant is a specific part of the quadratic formula that helps determine the nature of the roots of a quadratic equation. Found under the square root in the quadratic formula, it is calculated as:

• \( b^2 - 4ac \)

The value of this discriminant tells us:

• If \( b^2 - 4ac > 0 \), there are two distinct real roots.
• If \( b^2 - 4ac = 0 \), there is one real root (also called a double root).
• If \( b^2 - 4ac < 0 \), there are no real roots (only complex roots).

In our exercise, we calculate the discriminant for \( x^2 + x - (2+e) = 0 \), which is \( 9 + 4e \). Since this expression is positive, we know there will be two real solutions. Calculating and analyzing the discriminant is crucial for understanding the roots fully. This knowledge helps predict how many solutions we will actually have before going through the entire quadratic formula.