Problem 241 - Precalculus

Question

For the following exercises, use a graphing calculator to find approximate solutions to each equation. $$\ln (x-2)=-\ln (x+1)$$

Step-by-Step Solution

Answer

The solution is approximately \( x \approx 2.3028 \).

Step 1: Understand the Equation

We have the equation \( \ln(x-2) = -\ln(x+1) \). This is a logarithmic equation, and we will manipulate it to find the value of \( x \).

Step 2: Simplify Using Logarithmic Properties

Using the property that \( \ln(a) = -\ln(b) \) implies \( \ln(a) + \ln(b) = 0 \) or \( \ln(ab) = 0 \), we rewrite the equation as:\[ \ln((x-2)(x+1)) = 0 \] because \( -\ln(x+1) \) can be rewritten as \( \ln((x+1)^{-1}) = \ln\left(\frac{1}{x+1}\right) \). Adding both logarithms gives \( \ln(x-2) + \ln(x+1) = \ln((x-2)(x+1)) \).

Step 3: Exponentiate Both Sides

Since \( \ln((x-2)(x+1)) = 0 \), we exponentiate both sides to remove the logarithm:\[ (x-2)(x+1) = e^0 = 1 \]which simplifies the equation to \( (x-2)(x+1) = 1 \).

Step 4: Expand and Rearrange the Equation

Expand and rearrange the equation:\[ x^2 - 2x + x - 2 = 1 \] which simplifies to:\[ x^2 - x - 2 = 1 \] Now, bring all terms to one side:\[ x^2 - x - 3 = 0 \]

Step 5: Use the Quadratic Formula

Use the quadratic formula to find the approximate solutions to the quadratic equation:\[ x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \] where \( a = 1 \), \( b = -1 \), and \( c = -3 \). Calculate:\[ x = \frac{-(-1) \pm \sqrt{(-1)^2 - 4 \cdot 1 \cdot (-3)}}{2 \cdot 1} \]\[ x = \frac{1 \pm \sqrt{1 + 12}}{2} \]\[ x = \frac{1 \pm \sqrt{13}}{2} \]Calculating these, the approximate numerical solutions can be found.

Step 6: Evaluate Solutions Using a Graphing Calculator

Use a graphing calculator to find the numeric values of the equation. The solutions are approximately \( x \approx 2.3028 \) and \( x \approx -1.3028 \). However, note that \( x = -1.3028 \) doesn't satisfy the original equation's domain constraints (\( x-2 > 0 \)) so is not a valid solution.

Key Concepts

Quadratic Formula

The quadratic formula is a powerful tool for solving quadratic equations, which are any equations of the form \( ax^2 + bx + c = 0 \). In this case, we applied it to the equation \( x^2 - x - 3 = 0 \), where \( a = 1 \), \( b = -1 \), and \( c = -3 \). By substituting these values into the formula \( x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \), you can find the solutions for \( x \).
Substituting the values gives us:

• \( x = \frac{-(-1) \pm \sqrt{(-1)^2 - 4 \cdot 1 \cdot (-3)}}{2 \cdot 1} \)
• \( x = \frac{1 \pm \sqrt{13}}{2} \)

This produces two potential solutions, corresponding to the \( + \) and \( - \) in the formula. These solutions give us the approximate values of \( x \) that satisfy the quadratic equation.
In solving logarithmic equations that convert to quadratics, it is crucial to verify these solutions against any domain constraints of the original equation.

Logarithmic Properties

Logarithmic properties are essential when working with logarithmic equations. These properties help simplify and manipulate equations to enable further solutions.
In this context, the equation \( \ln(x-2) = -\ln(x+1) \) utilized the property that \(-\ln(b)\) is equivalent to \(\ln(1/b)\). This allows us to rewrite the original equation:

• Convert \( -\ln(x+1) \) to \( \ln((x+1)^{-1}) = \ln\left(\frac{1}{x+1}\right) \)
• Add the logarithms: \( \ln(x-2) + \ln(x+1) = \ln((x-2)(x+1)) \)

Using another property where \( \ln(a) + \ln(b) = \ln(ab) \), we simplify the expression to \( \ln((x-2)(x+1)) \).
Understanding and applying these properties correctly are vital steps in transitioning from logarithmic to other forms of equations.

Graphing Calculator

A graphing calculator is an indispensable tool when solving complex equations. It not only provides numeric approximations of solutions but also visualizes mathematical functions.
In the step-by-step solution, the graphing calculator was used for:

• Confirming the numeric solutions of \( x = \frac{1 \pm \sqrt{13}}{2} \), resulting in approximate \( x \) values.
• Plotting the functions to understand the behavior of \( \ln(x-2) \) and \( -\ln(x+1) \) visually.

This aids in checking where the functions cross, which corresponds to the solutions of the logarithmic equation.
The graphing calculator is especially helpful because it also brings any potential domain issues to light that might affect the solutions.

Domain Constraints

Domain constraints are essential considerations when solving equations, particularly those involving logarithms. For logarithmic functions, the input value (argument) of the log must be greater than zero.
In our original problem \( \ln(x-2) = -\ln(x+1) \), the constraints are:

• \( x-2 > 0 \Rightarrow x > 2 \)
• \( x+1 > 0 \Rightarrow x > -1 \)

These constraints mean that when solving the quadratic equation, only solutions where \( x > 2 \) satisfy the original equation due to the domain restriction \( x-2 > 0 \).
Without respecting these constraints, one could potentially accept an invalid solution, such as \( x \approx -1.3028 \), which doesn't satisfy \( x > 2 \). Proper understanding ensures only valid solutions are considered. This highlights why domain considerations are crucial in mathematical problem-solving.