Problem 55 - College Algebra

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: Identify Properties of Logarithms

The given equation is \( \ln(x-2) = -\ln(x+1) \). First, recall the property of logarithms that states \(-\ln(a) = \ln \left(\frac{1}{a}\right)\). Apply this property to simplify the equation to \( \ln(x-2) = \ln \left(\frac{1}{x+1}\right) \).

Step 2: Set the Expressions Equal

Since the logarithms are equal, we can set their arguments equal to one another: \(x-2 = \frac{1}{x+1}\). This gives us a rational equation to solve.

Step 3: Clear the Fraction

Multiply both sides of the equation \( x - 2 = \frac{1}{x+1} \) by \( x+1 \) to eliminate the fraction. This yields: \( (x-2)(x+1) = 1 \).

Step 4: Expand and Rearrange

Expand the left side: \( x^2 + x - 2x - 2 = 1 \). Simplify to obtain \( x^2 - x - 2 = 1 \). Then, bring all terms to one side to form a quadratic equation: \( x^2 - x - 3 = 0 \).

Step 5: Use Quadratic Formula

Solve the quadratic equation \( x^2 - x - 3 = 0 \) using the quadratic formula \( x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \), where \( a = 1 \), \( b = -1 \), and \( c = -3 \). This gives \( x = \frac{1 \pm \sqrt{1 + 12}}{2} \).

Step 6: Calculate Solutions

Calculate the solutions: \( x = \frac{1 \pm \sqrt{13}}{2} \). Numerically approximate these roots using a calculator to find that \( x \approx 2.3028 \) or \( x \approx -1.3028 \).

Step 7: Confirm Valid Solution

Since logarithms are undefined for non-positive arguments, check which of these solutions satisfy the conditions \( x-2 > 0 \) and \( x+1 > 0 \). Only \( x \approx 2.3028 \) is valid, as it results in positive arguments for both logarithms.

Key Concepts

Properties of Logarithms

Logarithms have some unique properties that make them very useful in solving equations like the one we have here. The key property used in this problem is that

• Negative logarithms can be transformed into reciprocals: o If we have \(-\ln(a)\), it is equivalent to \(\ln\left(\frac{1}{a}\right)\).
• We also know that if two logarithms with the same base are equal, their arguments must be equal: \( \ln(b) = \ln(c) \Rightarrow b = c \).

This property allows us to equate the insides (or arguments) of the logs directly, turning a logarithmic equation into a simpler form. All we do is remove the log functions and set the expressions equal.
This step-by-step approach helps convert complicated logarithmic expressions into easier-to-solve forms, paving the way for the next steps in solving the equation.

Quadratic Formula

The quadratic formula is an essential tool used to solve quadratic equations, which are equations of the form \(ax^2 + bx + c = 0\). This formula is:

• \( x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \)

It allows us to find the roots of the quadratic equation where \a\, \b\, and \c\ are constants. The '±' symbol indicates there are typically two solutions, since a square has two roots.
In the problem we solved, after forming the quadratic equation: \(x^2 - x - 3 = 0\),

• we assigned \a = 1\, \b = -1\, and \c = -3\.
• Plugging these values into the formula, we obtain the solutions for \(x\).
• This includes both the calculation of the discriminant \( b^2 - 4ac \), providing information on the nature and number of roots.

By calculating these values, we discovered that the valid solution is \(x \approx 2.3028\). This shows how the quadratic formula aids in simplifying and solving quadratic equations efficiently.

Rational Equations

Rational equations are equations where the variables appear in the denominators of fractions. Solving rational equations involves finding a common base or clearing fractions to simplify them into a more workable form.
The given equation, once transformed through the properties of logarithms, becomes a rational equation: \(x-2 = \frac{1}{x+1}\).
To solve rational equations effectively:

• Multiply both sides by the common denominator to get rid of the fraction.
• Expand and rearrange terms to form an algebraic equation without fraction.

This leads to a cleaner and more accessible quadratic equation: \( (x-2)(x+1) = 1\) easily expanded to \(x^2 - x - 3 = 0\). This is important to clear fractions before proceeding to solve an equation, as it makes the process of reaching the solution straightforward and less cumbersome.